Class 11: Transformation Formulae – Exercise 8.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Express each of the following as the product of sines and cosines:

$\displaystyle \text{i) } \sin 12 x + \sin 4x \hspace{1.0cm} \text{ii) } \sin 5x - \sin x \hspace{1.0cm} \text{iii) } \cos 12 x + \cos 8x$

$\displaystyle \text{iv) } \cos 12 x - \cos 4x \hspace{1.0cm} \text{v) } \sin 2x + \cos 4x$

Answer:

$\displaystyle \text{i) } \sin 12 x + \sin 4x = 2 \sin \Big( \frac{12x + 4x}{2} \Big) \cos \Big( \frac{12x - 4x}{2} \Big) = 2 \sin 8x \cos 4x$

$\displaystyle \text{ii) } \sin 5x - \sin x = 2 \cos \Big( \frac{5x + x}{2} \Big) \sin \Big( \frac{5x - x}{2} \Big) = 2 \cos 3x \sin 2x$

$\displaystyle \text{iii) } \cos 12 x + \cos 8x = 2 \cos \Big( \frac{12x + 8x}{2} \Big) \cos \Big( \frac{12x - 8x}{2} \Big) = 2 \cos 10x \cos 2x$

$\displaystyle \text{iv) } \cos 12 x - \cos 4x = -2 \sin \Big( \frac{12x + 4x}{2} \Big) \sin \Big( \frac{12x - 4x}{2} \Big) = -2 \sin 8x \sin 4x$

$\displaystyle \text{v) } \sin 2x + \cos 4x = \sin 2x + \sin (90^{\circ} - 4x)$

$\displaystyle = 2 \sin \Big( \frac{2x + 90^{\circ} - 4x}{2} \Big) \cos \Big( \frac{2x - 90^{\circ} + 4x}{2} \Big)$

$\displaystyle = 2 \sin (45^{\circ} - x) \cos ( 3x - 45^{\circ})$

$\displaystyle = 2 \sin (45^{\circ} - x) \cos ( 45^{\circ}- 3x)$

$\displaystyle \\$

Question 2: Prove that:

$\displaystyle \text{i) } \sin 38^{\circ} + \sin 22^{\circ} = \sin 82^{\circ} \hspace{1.0cm} \text{ii) } \cos 100^{\circ} + \cos 20^{\circ} = \cos 40^{\circ}$

$\displaystyle \text{iii) } \sin 50^{\circ} + \sin 10^{\circ} = \cos 20^{\circ} \hspace{1.0cm} \text{iv) } \sin 23^{\circ} + \sin 37^{\circ} = \cos 7^{\circ}$

$\displaystyle \text{v) } \sin 105^{\circ} + \cos 105^{\circ} = \cos 45^{\circ} \hspace{1.0cm} \text{vi) } \sin 40^{\circ} + \sin 20^{\circ} = \cos 10^{\circ}$

Answer:

$\displaystyle \text{i) } \text{LHS } =\sin 38^{\circ} + \sin 22^{\circ}$

$\displaystyle =2 \sin \Big( \frac{38^{\circ}+22^{\circ}}{2} \Big) \cos \Big( \frac{38^{\circ}-22^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 30^{\circ} \cos 8^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 8^{\circ}$

$\displaystyle = \cos (90^{\circ}-82^{\circ}) = \sin 82^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } =\cos 100^{\circ} + \cos 20^{\circ}$

$\displaystyle =2 \cos \Big( \frac{100^{\circ}+20^{\circ}}{2} \Big) \cos \Big( \frac{100^{\circ}-20^{\circ}}{2} \Big)$

$\displaystyle = 2 \cos 60^{\circ} \cos 40^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 40^{\circ}$

$\displaystyle = \cos 40^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{iii) } \text{LHS } =\sin 50^{\circ} + \sin 10^{\circ}$

$\displaystyle =2 \sin \Big( \frac{50^{\circ}+10^{\circ}}{2} \Big) \cos \Big( \frac{50^{\circ}-10^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 30^{\circ} \cos 20^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 20^{\circ}$

$\displaystyle = \cos 20^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{iv) } \text{LHS } =\sin 23^{\circ} + \sin 37^{\circ}$

$\displaystyle =2 \sin \Big( \frac{23^{\circ}+37^{\circ}}{2} \Big) \cos \Big( \frac{23^{\circ}-37^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 30^{\circ} \cos (-7^{\circ})$

$\displaystyle = 2 \times \frac{1}{2} \cos 7^{\circ}$

$\displaystyle = \cos 7^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{v) } \text{LHS } =\sin 105^{\circ} + \cos 105^{\circ}$

$\displaystyle =\sin 105^{\circ} + \cos (90^{\circ}+ 15^{\circ})$

$\displaystyle =\sin 105^{\circ} - \sin 15^{\circ}$

$\displaystyle =2 \sin \Big( \frac{105^{\circ}-15^{\circ}}{2} \Big) \cos \Big( \frac{105^{\circ}+15^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 45^{\circ} \cos 60^{\circ}$

$\displaystyle = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2}$

$\displaystyle = \frac{1}{\sqrt{2}} = \cos 45^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{vi) } \text{LHS } =\sin 40^{\circ} + \sin 20^{\circ}$

$\displaystyle =2 \sin \Big( \frac{40^{\circ}+20^{\circ}}{2} \Big) \cos \Big( \frac{20^{\circ}-40^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 30^{\circ} \cos (-10^{\circ})$

$\displaystyle = 2 \times \frac{1}{2} \cos 10^{\circ}$

$\displaystyle = \cos 10^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 3: Prove that:

$\displaystyle \text{i) } \cos 55^{\circ} + \cos 65^{\circ} + \cos 175^{\circ} = 0 \hspace{1.0cm} \text{ii) } \sin 50^{\circ} - \sin 70^{\circ} + \sin 10^{\circ} = 0$

$\displaystyle \text{iii) } \cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ} = 0 \hspace{1.0cm} \text{v) } \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ} = 0$

$\displaystyle \text{v) } \sin \frac{5\pi}{18} - \cos \frac{4\pi}{9} = \sqrt{3} \sin \frac{\pi}{9} \hspace{1.0cm} \text{vi) } \cos \frac{\pi}{12} - \sin \frac{\pi}{12} = \frac{1}{\sqrt{2}}$

$\displaystyle \text{vii) } \sin 80^{\circ} - \cos 70^{\circ} = \cos 50^{\circ} \hspace{1.0cm} \text{viii) } \sin 51^{\circ} + \cos 81^{\circ} = \cos 21^{\circ}$

Answer:

$\displaystyle \text{i) } \text{LHS } = \cos 55^{\circ} + \cos 65^{\circ} + \cos 175^{\circ}$

$\displaystyle = \cos 55^{\circ} + \cos 65^{\circ} + \cos ( 180^{\circ} - 5^{\circ})$

$\displaystyle = \cos 55^{\circ} + \cos 65^{\circ} - \cos 5^{\circ}$

$\displaystyle = 2 \cos \frac{55^{\circ}+65^{\circ}}{2} \cos \frac{55^{\circ}-65^{\circ}}{2} - \cos 5^{\circ}$

$\displaystyle =2 \cos 60^{\circ} \cos 5^{\circ} - \cos 5^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 5^{\circ} - \cos 5^{\circ}$

$\displaystyle = \cos 5^{\circ} - \cos 5^{\circ} = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \sin 50^{\circ} - \sin 70^{\circ} + \sin 10^{\circ}$

$\displaystyle = 2 \sin \frac{50^{\circ}-70^{\circ}}{2} \cos \frac{50^{\circ}+70^{\circ}}{2} + \sin 10^{\circ}$

$\displaystyle =2 \sin (-10^{\circ}) \cos 60^{\circ} + \sin 10^{\circ}$

$\displaystyle = -2 \times \frac{1}{2} \sin 10^{\circ} + \sin 10^{\circ}$

$\displaystyle = -\sin 10^{\circ} + \sin 10^{\circ} = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{iii) } \text{LHS } = \cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ}$

$\displaystyle = 2 \cos \frac{80^{\circ}+40^{\circ}}{2} \cos \frac{80^{\circ}-40^{\circ}}{2} - \cos 20^{\circ}$

$\displaystyle =2 \cos 60^{\circ} \cos 20^{\circ} - \cos 20^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 20^{\circ} - \cos 20^{\circ}$

$\displaystyle = \cos 20^{\circ} - \cos 20^{\circ} = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{iv) } \text{LHS } = \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}$

$\displaystyle = 2 \cos \frac{20^{\circ}+100^{\circ}}{2} \cos \frac{20^{\circ}-100^{\circ}}{2} + \cos (180^{\circ} - 40^{\circ})$

$\displaystyle =2 \cos 60^{\circ} \cos 40^{\circ} - \cos 40^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 40^{\circ} - \cos 40^{\circ}$

$\displaystyle = \cos 40^{\circ} - \cos 40^{\circ} = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{v) } \text{LHS } = \sin \frac{5\pi}{18} - \cos \frac{4\pi}{9}$

$\displaystyle = \sin 50^{\circ} - \cos 80^{\circ} = \sin 50^{\circ} - \cos ( 90^{\circ}-10^{\circ}) = \sin 50^{\circ} - \sin 10^{\circ}$

$\displaystyle = 2 \sin \frac{50^{\circ}-10^{\circ}}{2} \cos \frac{50^{\circ}+10^{\circ}}{2}$

$\displaystyle =2 \sin 20^{\circ} \cos 30^{\circ}$

$\displaystyle = 2 \times \frac{\sqrt{3}}{2} \sin 20^{\circ}$

$\displaystyle = \sqrt{3} \sin 20^{\circ} = \sqrt{3} \sin \frac{\pi}{9} = \text{ RHS. Hence proved. }$

$\displaystyle \text{vi) } \text{LHS } = \cos \frac{\pi}{12} - \sin \frac{\pi}{12}$

$\displaystyle = \cos 15^{\circ} - \sin 15^{\circ} = \cos 15^{\circ} - \sin ( 90^{\circ}-75^{\circ}) = \cos 15^{\circ} - \cos 75^{\circ}$

$\displaystyle = 2 \sin \frac{15^{\circ}+75^{\circ}}{2} \sin \frac{75^{\circ}-15^{\circ}}{2}$

$\displaystyle = 2 \sin 45^{\circ} \sin 30^{\circ}$

$\displaystyle = 2 \times \frac{1}{\sqrt{2}} \sin 30^{\circ}$

$\displaystyle = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{1}{\sqrt{2}} = \text{ RHS. Hence proved. }$

$\displaystyle \text{vii) } \text{LHS } = \sin 80^{\circ} - \cos 70^{\circ} = \sin 80^{\circ} - \cos ( 90^{\circ}-20^{\circ}) = \sin 80^{\circ} - \sin 20^{\circ}$

$\displaystyle = 2 \sin \frac{80^{\circ}-20^{\circ}}{2} \cos \frac{80^{\circ}+20^{\circ}}{2}$

$\displaystyle =2 \sin 30^{\circ} \cos 50^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 50^{\circ}$

$\displaystyle = \cos 50^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{viii) } \text{LHS } = \sin 51^{\circ} + \cos 81^{\circ} = \sin 51^{\circ} + \cos ( 90^{\circ}-9^{\circ}) = \sin 51^{\circ} + \sin 9^{\circ}$

$\displaystyle = 2 \sin \frac{51^{\circ}+9^{\circ}}{2} \cos \frac{51^{\circ}-9^{\circ}}{2}$

$\displaystyle =2 \sin 30^{\circ} \cos 21^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 21^{\circ}$

$\displaystyle = \cos 21^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 4: Prove that:

$\displaystyle \text{i) } \cos \Big( \frac{3\pi}{4} +x \Big) - \cos \Big( \frac{3\pi}{4} -x \Big) = -\sqrt{2} \sin x$

$\displaystyle \text{ii) } \cos \Big( \frac{\pi}{4} +x \Big) + \cos \Big( \frac{\pi}{4} -x \Big) = -\sqrt{2} \cos x$

Answer:

$\displaystyle \text{i) } \text{LHS } = \cos \Big( \frac{3\pi}{4} +x \Big) - \cos \Big( \frac{3\pi}{4} -x \Big)$

$\displaystyle = - \Big[ \cos \Big( \frac{3\pi}{4} -x \Big) - \cos \Big( \frac{3\pi}{4} +x \Big) \Big]$

$\displaystyle = - \Big[ -2 \sin \Big( \frac{\frac{3\pi}{4} - x + \frac{3\pi}{4} + x}{2} \Big) \sin \Big( \frac{\frac{3\pi}{4} - x - \frac{3\pi}{4} - x}{2} \Big) \Big]$

$\displaystyle = - \Big[ - 2 \sin \frac{3\pi}{4} \sin (-x) \Big]$

$\displaystyle = - 2 \sin \Big( \frac{\pi}{2} + \frac{\pi}{4} \Big) \sin x$

$\displaystyle = -2 \cos \frac{\pi}{4} \sin x = - \sqrt{2} \sin x = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \cos \Big( \frac{\pi}{4} +x \Big) + \cos \Big( \frac{\pi}{4} -x \Big)$

$\displaystyle = 2 \cos \Big( \frac{\frac{\pi}{4}+ x + \frac{\pi}{4} - x}{2} \Big) \cos \Big( \frac{\frac{\pi}{4}+ x - \frac{\pi}{4} + x}{2} \Big)$

$\displaystyle = 2 \cos \frac{\pi}{4} \cos x$

$\displaystyle = \sqrt{2} \cos x = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 5: Prove that:

$\displaystyle \text{i) } \sin 65^{\circ} + \cos 65^{\circ} = \sqrt{2} \cos 20^{\circ} \hspace{1.0cm} \text{ii) } \sin 47^{\circ} +\cos 77^{\circ} = \cos 17^{\circ}$

Answer:

$\displaystyle \text{i) } \text{LHS } = \sin 65^{\circ} + \cos 65^{\circ}$

$\displaystyle = \sin 65^{\circ} + \cos ( 90^{\circ}-25^{\circ})$

$\displaystyle = \sin 65^{\circ} + \sin 25^{\circ}$

$\displaystyle = 2 \sin \Big( \frac{65^{\circ}+ 25^{\circ}}{2} \Big) \cos \Big( \frac{65^{\circ}- 25^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 45^{\circ} \cos 20^{\circ}$

$\displaystyle = 2 \times \frac{1}{\sqrt{2}} \cos 20^{\circ}$

$\displaystyle = \sqrt{2} \cos 20^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \sin 47^{\circ} +\cos 77^{\circ}$

$\displaystyle \text{LHS } = \sin 47^{\circ} + \cos 77^{\circ}$

$\displaystyle = \sin 47^{\circ} + \cos ( 90^{\circ}-13^{\circ})$

$\displaystyle = \sin 47^{\circ} + \sin 13^{\circ}$

$\displaystyle = 2 \sin \Big( \frac{47^{\circ}+ 13^{\circ}}{2} \Big) \cos \Big( \frac{47^{\circ}- 13^{\circ}}{2} \Big)$

$\displaystyle = 2 \sin 30^{\circ} \cos 17^{\circ}$

$\displaystyle = 2 \times \frac{1}{2} \cos 17^{\circ}$

$\displaystyle = \cos 17^{\circ} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 6: Prove that:

$\displaystyle \text{i) } \cos 3A + \cos 5A + \cos 7A + \cos 15A = 4 \cos 4A \cos 5A \cos 6A$

$\displaystyle \text{ii) } \cos A + \cos 3A + \cos 5A + \cos 7A = 4 \cos A \cos 2A \cos 4A$

$\displaystyle \text{iii) } \sin A + \sin 2A + \sin 4A + \sin 5A = 4 \cos \frac{A}{2} \cos \frac{3A}{2} \sin 3A$

$\displaystyle \text{iv) } \sin 3A + \sin 2A - \sin A = 4 \sin A \cos \frac{A}{2} \cos \frac{3A}{2}$

$\displaystyle \text{v) } \cos 20^{\circ} \cos 100^{\circ} + \cos 100^{\circ} \cos 140^{\circ} - \cos 140^{\circ} \cos 200^{\circ} = - \frac{3}{4}$

$\displaystyle \text{vi) } \sin \frac{x}{2} \sin \frac{7x}{2} + \sin \frac{3x}{2} \sin \frac{11x}{2} = \sin 2x \sin 5x$

$\displaystyle \text{vii) } \cos x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 4x \sin \frac{7x}{2}$

Answer:

$\displaystyle \text{i) } \text{LHS } = \cos 3A + \cos 5A + \cos 7A + \cos 15A$

$\displaystyle = [ \cos 5A + \cos 3A] + [ \cos 15A + \cos 7A ]$

$\displaystyle = 2 \cos \Big( \frac{5A + 3A}{2} \Big) \cos \Big( \frac{5A - 3A}{2} \Big) + 2 \cos \Big( \frac{15A + 7A}{2} \Big) \cos \Big( \frac{15A - 7A}{2} \Big)$

$\displaystyle = 2 \cos 4A \cos A + 2 \cos 11 A \cos 4A$

$\displaystyle = 2 \cos 4A [ \cos 11A + \cos A ]$

$\displaystyle = 2 \cos 4A \Big[ 2 \cos \Big( \frac{11A + A}{2} \Big) \cos \Big( \frac{11A - A}{2} \Big) \Big]$

$\displaystyle = 2 \cos 4A [ 2 \cos 6A \cos 5A ]$

$\displaystyle = 4 \cos 4A \cos 5A \cos 6A = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \cos A + \cos 3A + \cos 5A + \cos 7A$

$\displaystyle = [ \cos 3A + \cos A] + [ \cos 7A + \cos 5A ]$

$\displaystyle = 2 \cos \Big( \frac{3A + A}{2} \Big) \cos \Big( \frac{3A - A}{2} \Big) + 2 \cos \Big( \frac{7A + 5A}{2} \Big) \cos \Big( \frac{7A - 5A}{2} \Big)$

$\displaystyle = 2 \cos 2A \cos A + 2 \cos 6 A \cos A$

$\displaystyle = 2 \cos A [ \cos 6A + \cos 2A ]$

$\displaystyle = 2 \cos A \Big[ 2 \cos \Big( \frac{6A + 2A}{2} \Big) \cos \Big( \frac{6A - 2A}{2} \Big) \Big]$

$\displaystyle = 2 \cos A [ 2 \cos 4A \cos 2A ]$

$\displaystyle = 4 \cos A \cos 2A \cos 4A = \text{ RHS. Hence proved. }$

$\displaystyle \text{iii) } \text{LHS } = \sin A + \sin 2A + \sin 4A + \sin 5A$

$\displaystyle = [ \sin 2A + \sin A] + [ \sin 5A + \sin 4A ]$

$\displaystyle = 2 \sin \Big( \frac{2A + A}{2} \Big) \cos \Big( \frac{2A - A}{2} \Big) + 2 \sin \Big( \frac{5A + 4A}{2} \Big) \cos \Big( \frac{5A - 4A}{2} \Big)$

$\displaystyle = 2 \sin \frac{3A}{2} \cos \frac{A}{2} + 2 \sin \frac{9A}{2} \cos \frac{A}{2}$

$\displaystyle = 2 \cos \frac{A}{2} \Big[ \sin \frac{3A}{2} + \sin \frac{9A}{2} \Big]$

$\displaystyle = 2 \cos \frac{A}{2} \Big[ 2 \sin \Big( \frac{\frac{9A}{2} + \frac{3A}{2}}{2} \Big) \cos \Big( \frac{\frac{9A}{2} - \frac{3A}{2}}{2} \Big) \Big]$

$\displaystyle = 2 \cos \frac{A}{2} [ 2 \sin 3A \sin \frac{3A}{2} ]$

$\displaystyle = 4 \cos \frac{A}{2} \sin \frac{3A}{2} \sin 3A = \text{ RHS. Hence proved. }$

$\displaystyle \text{iv) } \text{LHS } = \sin 3A + \sin 2A - \sin A$

$\displaystyle = [ \sin 3A - \sin A ] + \sin 2A$

$\displaystyle = 2 \sin \Big( \frac{3A- A}{2} \Big) \cos \Big( \frac{3A+ A}{2} \Big) + \sin 2A$

$\displaystyle = 2 \sin A \cos 2A + \sin 2A$

$\displaystyle = 2 \sin A \cos 2A + 2\sin A \cos A$

$\displaystyle = 2 \sin A [ \cos 2A + \cos A ]$

$\displaystyle = 2 \sin A \Big[ 2 \cos \Big( \frac{2A+ A}{2} \Big) \cos \Big( \frac{2A- A}{2} \Big) \Big]$

$\displaystyle = 4 \sin A \cos \frac{3A}{2} \cos \frac{A}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \text{v) } \text{LHS } = \cos 20^{\circ} \cos 100^{\circ} + \cos 100^{\circ} \cos 140^{\circ} - \cos 140^{\circ} \cos 200^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ 2\cos 20^{\circ} \cos 100^{\circ} + 2\cos 100^{\circ} \cos 140^{\circ} - 2\cos 140^{\circ} \cos 200^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos (100^{\circ}+20^{\circ})+ \cos (100^{\circ}-20^{\circ}) + \cos (140^{\circ}+100^{\circ})+ \\ \cos (140^{\circ}-100^{\circ}) - \cos (200^{\circ}+140^{\circ}) -\cos (200^{\circ}-140^{\circ}) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 120^{\circ}+ \cos 80^{\circ} + \cos 240^{\circ}+ \cos 40^{\circ} - \cos 340^{\circ} -\cos 60^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos (90^{\circ}+30^{\circ})+ \cos 80^{\circ} + \cos (180^{\circ}+60^{\circ})+ \cos 40^{\circ} - \cos (360^{\circ}-20^{\circ}) -\cos 60^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ -\sin 30^{\circ} + [ \cos 80^{\circ} + \cos 40^{\circ} ] - \cos 60^{\circ} - \cos 20^{\circ} -\cos 60^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ -\sin 30^{\circ} + 2 \cos 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} - \cos 20^{\circ} -\cos 60^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ - \frac{1}{2} + 2 \times \frac{1}{2} \cos 20^{\circ} - \frac{1}{2} - \cos 20^{\circ} - \frac{1}{2} \Big]$

$\displaystyle = \frac{1}{2} \Big[ - \frac{3}{2} \Big] = - \frac{3}{4} = \text{ RHS. Hence proved. }$

$\displaystyle \text{vi) } \text{LHS } = \sin \frac{x}{2} \sin \frac{7x}{2} + \sin \frac{3x}{2} \sin \frac{11x}{2}$

$\displaystyle = \frac{1}{2} \Big[ 2\sin \frac{x}{2} \sin \frac{7x}{2} + 2\sin \frac{3x}{2} \sin \frac{11x}{2} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos \Big( \frac{7x}{2} - \frac{x}{2} \Big) - \cos \Big( \frac{7x}{2} + \frac{x}{2} \Big) +\cos \Big( \frac{11x}{2} - \frac{3x}{2} \Big) - \cos \Big( \frac{11x}{2} + \frac{3x}{2} \Big) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 3x - \cos 4x + \cos 4x - \cos 7x \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 3x - \cos 7x \Big]$

$\displaystyle = - \frac{1}{2} \Big[ \cos 7x - \cos 3x \Big]$

$\displaystyle = - \frac{1}{2} \Big[ -2 \sin \Big( \frac{7x+3x}{2} \Big) \sin \Big( \frac{7x-3x}{2} \Big) \Big]$

$\displaystyle = \sin 5x \sin 2x = \text{ RHS. Hence proved. }$

$\displaystyle \text{vii) } \text{LHS } = \cos x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2}$

$\displaystyle = \frac{1}{2} \Big[ 2\cos x \cos \frac{x}{2} - 2\cos 3x \cos \frac{9x}{2} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos \Big( x + \frac{x}{2} \Big) + \cos \Big( x - \frac{x}{2} \Big) - \cos \Big( \frac{9x}{2} + 3x \Big) - \cos \Big( \frac{9x}{2} - 3x \Big) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos \frac{3x}{2} + \cos \frac{x}{2} - \cos \frac{15x}{2} - \cos \frac{3x}{2} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos \frac{x}{2} - \cos \frac{15x}{2} \Big]$

$\displaystyle = \frac{1}{2} \Big[ - 2 \sin \Big( \frac{\frac{x}{2}+\frac{15x}{2} }{2} \Big) \sin \Big( \frac{\frac{x}{2}-\frac{15x}{2} }{2} \Big) \Big]$

$\displaystyle = \sin 4x \sin \frac{7x}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 7: Prove that:

$\displaystyle \text{i) } \frac{\sin A + \sin 3A}{\cos A - \cos 3A} = \cot A \hspace{1.0cm} \text{ii) } \frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A} = \cot 8A$

$\displaystyle \text{iii) } \frac{\sin A - \sin B}{\cos A + \cos B} = \tan \Big( \frac{A-B}{2} \Big) \hspace{1.0cm} \text{iv) } \frac{\sin A + \sin B}{\sin A - \sin B} = \tan \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big)$

$\displaystyle \text{v) } \frac{\cos A + \cos B}{\cos B - \cos A} = \cot \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big)$

Answer:

$\displaystyle \text{i) } \text{LHS } = \frac{\sin A + \sin 3A}{\cos A - \cos 3A}$

$\displaystyle = \frac{2 \sin \Big( \frac{A+3A}{2} \Big) \cos \Big( \frac{A-3A}{2} \Big) }{-2 \sin \Big( \frac{A+3A}{2} \Big) \sin \Big( \frac{A-3A}{2} \Big)}$

$\displaystyle = \frac{-2 \sin 2A \cos (-A) }{2 \sin 2A \sin (-A) } = \frac{- \cos ( -A)}{\sin ( -A)} = \frac{\cos A}{\sin A} = \cot A = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}$

$\displaystyle = \frac{2 \sin \Big( \frac{9A-7A}{2} \Big) \cos \Big( \frac{9A+7A}{2} \Big) }{-2 \sin \Big( \frac{7A+9A}{2} \Big) \sin \Big( \frac{7A-9A}{2} \Big)}$

$\displaystyle = \frac{-2 \sin A \cos 8A }{2 \sin 8A \sin (-A) } = \frac{- \sin A \cos 8A}{- \sin A \sin 8A } = \frac{\cos 8A}{\sin 8A} = \cot 8A = \text{ RHS. Hence proved. }$

$\displaystyle \text{iii) } \text{LHS } = \frac{\sin A - \sin B}{\cos A + \cos B}$

$\displaystyle = \frac{2 \cos \Big( \frac{A+B}{2} \Big) \sin \Big( \frac{A-B}{2} \Big) }{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big)}$

$\displaystyle = \frac{ \sin \Big( \frac{A-B}{2} \Big) }{ \cos \Big( \frac{A-B}{2} \Big)} = \tan \Big( \frac{A-B}{2} \Big) = \text{ RHS. Hence proved. }$

$\displaystyle \text{iv) } \text{LHS } = \frac{\sin A + \sin B}{\sin A - \sin B}$

$\displaystyle = \frac{2 \sin \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \sin \Big( \frac{A-B}{2} \Big) \cos \Big( \frac{A+B}{2} \Big)}$

$\displaystyle = \tan \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big) = \text{ RHS. Hence proved. }$

$\displaystyle \text{v) } \text{LHS } = \frac{\cos A + \cos B}{\cos B - \cos A}$

$\displaystyle = \frac{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{-2 \sin \Big( \frac{B+A}{2} \Big) \sin \Big( \frac{B-A}{2} \Big)}$

$\displaystyle = \frac{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \sin \Big( \frac{A+B}{2} \Big) \sin \Big( \frac{A-B}{2} \Big)}$

$\displaystyle = \cot \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big) = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 8: Prove that:

$\displaystyle \text{i) } \frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A} = \tan 3A$

$\displaystyle \text{ii) } \frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A} = \frac{\cos 5A}{\cos 3A}$

$\displaystyle \text{iii) } \frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A$

$\displaystyle \text{iv) } \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A$

$\displaystyle \text{v) } \frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A} = \cot 6A$

$\displaystyle \text{vi) } \frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A$

$\displaystyle \text{vii) } \frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A} = \tan 8A$

$\displaystyle \text{viii) } \frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A} = \tan 2A$

$\displaystyle \text{ix) } \frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A} = \tan 5A$

$\displaystyle \text{x) } \frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}$

$\displaystyle \text{xi) } \frac{\sin (\theta + \phi) -2 \sin \theta + \sin (\theta - \phi)}{\cos (\theta + \phi) -2 \cos \theta + \cos (\theta - \phi)} = \tan \theta$

Answer:

$\displaystyle \text{i) } \text{LHS } = \frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}$

$\displaystyle = \frac{(\sin 5A + \sin A) + \sin 3A}{(\cos 5A + \cos A) + \cos 3A}$

$\displaystyle = \frac{2 \sin \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + \sin 3A } {2 \cos \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + \cos 3A }$

$\displaystyle = \frac{ 2 \sin 3A \cos 2A + \sin 3A}{ 2 \cos 3A \cos 2A+ \cos 3A}$

$\displaystyle = \frac{ \sin 3A ( 2 \cos 2A + 1)}{\cos 3A ( 2 \cos 2A + 1)}$

$\displaystyle = \tan 3A = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}$

$\displaystyle = \frac{(\cos 3A + \cos 7A) + 2\cos 5A}{(\cos A + \cos 5A) + 2\cos 3A}$

$\displaystyle = \frac{2 \cos \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big) + 2\cos 5A } {2 \cos \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + 2\cos 3A }$

$\displaystyle = \frac{ 2 \cos 5A \cos 2A + 2\cos 5A}{ 2 \cos 3A \cos 2A+ 2\cos 3A}$

$\displaystyle = \frac{ 2\cos 5A ( \cos 2A + 1)}{2\cos 3A ( \cos 2A + 1)}$

$\displaystyle = \frac{\cos 5A}{\cos 3A} = \text{ RHS. Hence proved. }$

$\displaystyle \text{iii) } \text{LHS } = \frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 5A}$

$\displaystyle = \frac{(\cos 4A + \cos 2A) + \cos 3A}{(\sin 4A + \sin 2A) + \sin 3A}$

$\displaystyle = \frac{2 \cos \Big( \frac{4A+2A}{2} \Big) \cos \Big( \frac{4A-2A}{2} \Big) + \cos 3A } {2 \sin \Big( \frac{4A+2A}{2} \Big) \cos \Big( \frac{4A-2A}{2} \Big) + \sin 3A }$

$\displaystyle = \frac{ 2 \cos 3A \cos A + \cos 3A}{ 2 \sin 3A \cos A+ \sin 3A}$

$\displaystyle = \frac{ \cos 3A ( 2\cos A + 1)}{\sin 3A ( 2\cos A + 1)}$

$\displaystyle = \frac{\cos 3A}{\cos 3A} = \cot 3A \text{ RHS. Hence proved. }$

$\displaystyle \text{iv) } \text{LHS } = \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}$

$\displaystyle = \frac{ (\sin 9A + \sin 3A) + (\sin 7A + \sin 5A) }{ (\cos 9A + \cos 3A) + (\cos 7A + \cos 5A) }$

$\displaystyle = \frac{ 2 \sin \Big( \frac{9A+3A}{2} \Big) \cos \Big( \frac{9A-3A}{2} \Big) + 2 \sin \Big( \frac{7A+5A}{2} \Big) \cos \Big( \frac{7A-5A}{2} \Big) }{ 2 \cos \Big( \frac{9A+3A}{2} \Big) \cos \Big( \frac{9A-3A}{2} \Big) + 2 \cos \Big( \frac{7A+5A}{2} \Big) \cos \Big( \frac{7A-5A}{2} \Big) }$

$\displaystyle = \frac{2 \sin 6A \cos 3A + 2 \sin 6A \cos 2A}{2 \cos 6A \cos 3A + 2 \cos 6A \cos 2A}$

$\displaystyle = \frac{2 \sin 6A (\cos 3A + \cos 2A) }{2 \cos 6A (\cos 3A + \cos 2A)}$

$\displaystyle = \tan 6A = \text{ RHS. Hence proved. }$

$\displaystyle \text{v) } \text{LHS } = \frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{ \cos 4A + \cos 7A - \cos 5A - \cos 8A}$

$\displaystyle = \frac{ - (\sin 7A - \sin 5A) + (\sin 8A - \sin 4A) }{ - (\cos 7A - \cos 5A) - (\cos 8A - \cos 4A) }$

$\displaystyle = \frac{ -2 \sin \Big( \frac{7A-5A}{2} \Big) \cos \Big( \frac{7A+5A}{2} \Big) + 2 \sin \Big( \frac{8A-4A}{2} \Big) \cos \Big( \frac{8A+4A}{2} \Big) }{- 2 \sin \Big( \frac{7A+5A}{2} \Big) \sin \Big( \frac{7A-5A}{2} \Big) +2 \sin \Big( \frac{8A+4A}{2} \Big) \sin \Big( \frac{8A-4A}{2} \Big) }$

$\displaystyle = \frac{-2 \sin A \cos 6A + 2 \sin 2A \cos 6A}{-2 \sin 6A \sin A + 2 \sin 6A \sin 2A}$

$\displaystyle = \frac{2 \cos 6A (-\sin A + \sin 2A) }{2 \sin 6A (-\sin A + \sin 2A) }$

$\displaystyle = \cot 6A = \text{ RHS. Hence proved. }$

$\displaystyle \text{vi) } \text{LHS } = \frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}$

$\displaystyle = \frac{2\sin 5A \cos 2A - 2\sin 6A \cos A}{2\sin A \sin 2A - 2\cos 2A \cos 3A}$

$\displaystyle = \frac{ \sin ( 5A +2A) + \sin (5A - 2A) - [ \sin ( 6A +A) + \sin (6A - A) ] }{ \cos ( 2A - A) - \cos ( 2A + A) - [ \cos ( 3A + 2A) + \cos ( 3A - 2 A) ] }$

$\displaystyle = \frac{ \sin 7A + \sin 3A - [ \sin 7A + \sin 5A ] }{ \cos A - \cos 3A - [ \cos 5A + \cos A ] }$

$\displaystyle = \frac{ \sin 3A - \sin 5A }{ - \cos 3A - \cos 5A }$

$\displaystyle = \frac{ -( \sin 5A - \sin 3A) }{ - (\cos 5A + \cos 3A) }$

$\displaystyle = \frac{2 \sin \Big( \frac{5A-3A}{2} \Big) \cos \Big( \frac{5A+3A}{2} \Big)}{2 \cos \Big( \frac{5A+3A}{2} \Big) \cos \Big( \frac{5A-3A}{2} \Big)}$

$\displaystyle = \frac{\sin A \cos 4A}{\cos 4A \cos A}$

$\displaystyle = \frac{\sin A }{\cos A} = \tan A =$ RHS

$\displaystyle \text{vii) } \text{LHS } = \frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A}$

$\displaystyle = \frac{2\sin 11A \sin A + 2\sin 7A \sin 3A}{2\cos 11A \sin A + 2\cos 7A \sin 3A}$

$\displaystyle = \frac{ \cos ( 11A -A) - \cos (11A + A) + \cos ( 7A -3A) - \cos (7A + 3A) }{ \sin ( 11A + A) - \sin ( 11A - A) + \sin ( 7A + 3A) - \sin ( 7A - 3 A) }$

$\displaystyle = \frac{ \cos 10A - \cos 12A + \cos 4A - \cos 10A }{ \sin 12A - \sin 10A + \sin 10A - \sin 4A }$

$\displaystyle = \frac{ - ( \cos 12A - \cos 4A) }{ \sin 12A - \sin 4A }$

$\displaystyle = \frac{ - \Big[ -2 \sin \Big( \frac{12A+4A}{2} \Big) \cos \Big( \frac{12A-4A}{2} \Big) \Big] }{2 \sin \Big( \frac{12A-4A}{2} \Big) \cos \Big( \frac{12A+4A}{2} \Big)}$

$\displaystyle = \frac{2\sin 8A \sin 4A}{2\sin 4A \cos 8A}$

$\displaystyle = \frac{\sin 8A }{\cos 8A} = \tan 8A =$ RHS

$\displaystyle \text{viii) } \text{LHS } = \frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A}$

$\displaystyle = \frac{2\sin 3A \cos 4A - 2\sin A \cos 2A}{2\sin 4A \sin A + 2\cos 6A \cos A}$

$\displaystyle = \frac{ \sin ( 4A +3A) - \sin (4A -3A) - [ \sin ( 2A+A) - \sin (2A -A) ] }{ \cos ( 4A - A) - \cos ( 4A + A) + \cos ( 6A + A) + \cos ( 6A - A) }$

$\displaystyle = \frac{ \sin 7A - \sin A - \sin 3A + \sin A }{ \cos 3A - \cos 5A + \cos 7A + \cos 5A }$

$\displaystyle = \frac{ \sin 7A - \sin 3A }{ \cos 3A + \cos 7A }$

$\displaystyle = \frac{2 \sin \Big( \frac{7A-3A}{2} \Big) \cos \Big( \frac{7A+3A}{2} \Big)}{2 \cos \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big)}$

$\displaystyle = \frac{2\sin 2A \cos 5A}{2\cos 5A \cos 2A}$

$\displaystyle = \frac{\sin 2A }{\cos 2A} = \tan 2A =$ RHS

$\displaystyle \text{ix) } \text{LHS } = \frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$

$\displaystyle = \frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}$

$\displaystyle = \frac{ \cos ( 2A -A) - \cos (2A +A) + \cos ( 6A-3A) - \cos (6A+3A) }{ \sin ( 2A + A) - \sin ( 2A - A) + \sin ( 6A + 3A) - \sin ( 6A - 3A) }$

$\displaystyle = \frac{ \cos A - \cos 3A + \cos 3A - \cos 9A }{ \sin 3A - \sin A + \sin 9A - \sin 3A }$

$\displaystyle = \frac{ \cos A - \cos 9A }{ - \sin A + \sin 9A }$

$\displaystyle = \frac{ - (\cos 9A - \cos A) }{ ( \sin 9A - \sin A) }$

$\displaystyle = \frac{- \Big[ -2 \sin \Big( \frac{9A+A}{2} \Big) \sin \Big( \frac{9A-A}{2} \Big) \Big] }{2 \sin \Big( \frac{9A-A}{2} \Big) \cos \Big( \frac{9A+A}{2} \Big)}$

$\displaystyle = \frac{2\sin 5A \sin 4A}{2\sin 4A \cos 5A}$

$\displaystyle = \frac{\sin 5A }{\cos 5A} = \tan 5A =$ RHS

$\displaystyle \text{x) } \text{LHS } = \frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A}$

$\displaystyle = \frac{ (\sin A + \sin 5A )+ 2\sin 3A }{ (\sin 3A + \sin 7A) + 2\sin 5A }$

$\displaystyle = \frac{2 \sin \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + 2\sin 3A } {2 \sin \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big) + 2\sin 5A }$

$\displaystyle = \frac{ 2 \sin 3A \cos 2A + 2 \sin 3A}{ 2 \sin 5A \cos 2A+ 2\sin 5A}$

$\displaystyle = \frac{ \sin 3A ( 1+ 2\cos A )}{\sin 5A ( 1+ 2\cos A )}$

$\displaystyle = \frac{\sin 3A}{\sin 5A} = \cot 3A \text{ RHS. Hence proved. }$

$\displaystyle \text{xi) } \text{LHS } = \frac{\sin (\theta + \phi) -2 \sin \theta + \sin (\theta - \phi)}{\cos (\theta + \phi) -2 \cos \theta + \cos (\theta - \phi)} = \tan \theta$

$\displaystyle = \frac{ [ \sin (\theta + \phi) + \sin (\theta - \phi) ] -2 \sin \theta}{ [ \cos (\theta + \phi) + \cos (\theta - \phi) ] -2 \cos \theta }$

$\displaystyle = \frac{ 2 \sin \Big( \frac{\theta + \phi+\theta - \phi}{2} \Big) \cos \Big( \frac{\theta + \phi-\theta + \phi}{2} \Big) -2 \sin \theta}{ 2 \cos \Big( \frac{\theta + \phi+\theta - \phi}{2} \Big) \cos \Big( \frac{\theta + \phi-\theta + \phi}{2} \Big) -2 \cos \theta }$

$\displaystyle = \frac{2\sin \theta \cos \phi - 2 \sin \theta}{2\cos \theta \cos \phi - 2 \cos \theta}$

$\displaystyle = \frac{2\sin \theta ( \cos \phi - 1) }{2\cos \theta ( \cos \phi - 1)}$

$\displaystyle = \frac{\sin \theta }{\cos \theta}$

$\displaystyle = \tan \theta = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 9: Prove that:

$\displaystyle \text{i) } \sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma) = 4 \sin \Big( \frac{\alpha + \beta}{2} \Big) \sin \Big( \frac{\beta + \gamma}{2} \Big) \sin \Big( \frac{\gamma+ \alpha}{2} \Big)$

$\displaystyle \text{ii) } \cos ( A+B+C) + \cos (A-B+C) + \cos ( A + B - C) + \cos (-A+B+C) = 4 \cos A \cos B \cos C$

Answer:

$\displaystyle \text{i) } \text{LHS } = \sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma)$

$\displaystyle = [ \sin \alpha + \sin \beta ] + [ \sin \gamma - \sin (\alpha + \beta + \gamma) ]$

$\displaystyle = 2 \sin \Big( \frac{\alpha + \beta}{2} \Big) \cos \Big( \frac{\alpha - \beta}{2} \Big) + 2 \sin \Big( \frac{\gamma - (\alpha + \beta + \gamma)}{2} \Big) \cos \Big( \frac{\gamma + (\alpha + \beta + \gamma)}{2} \Big)$

$\displaystyle = 2 \sin \Big( \frac{\alpha + \beta}{2} \Big) \cos \Big( \frac{\alpha - \beta}{2} \Big) + 2 \sin \Big( \frac{ - (\alpha + \beta )}{2} \Big) \cos \Big( \frac{\alpha + \beta + 2\gamma}{2} \Big)$

$\displaystyle = 2 \sin \Big( \frac{\alpha + \beta}{2} \Big) \Big[ \cos \Big( \frac{\alpha - \beta}{2} \Big) - \cos \Big( \frac{\alpha + \beta + 2\gamma)}{2} \Big) \Big]$

$\displaystyle = 2 \sin \Big( \frac{\alpha + \beta}{2} \Big) \Big[ - 2 \sin \Big( \frac{\frac{\alpha - \beta}{2} +\frac{\alpha + \beta + 2\gamma}{2}}{2} \Big) \sin \Big( \frac{\frac{\alpha - \beta}{2} - \frac{\alpha + \beta + 2\gamma}{2}}{2} \Big) \Big]$

$\displaystyle = 2 \sin \Big( \frac{\alpha + \beta}{2} \Big)\Big[ -2 \sin \Big( \frac{\alpha + \gamma}{2} \Big) \sin \Big( \frac{-(\beta + \gamma)}{2} \Big) \Big]$

$\displaystyle = 4 \sin \Big( \frac{\alpha + \beta}{2} \Big)\Big[ \sin \Big( \frac{\alpha + \gamma}{2} \Big) \sin \Big( \frac{\beta + \gamma}{2} \Big) \Big]$

$\displaystyle = 4 \sin \Big( \frac{\alpha + \beta}{2} \Big)\Big[ \sin \Big( \frac{\beta + \gamma}{2} \Big) \sin \Big( \frac{\alpha + \gamma}{2} \Big) \Big] = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \cos ( A+B+C) + \cos (A-B+C) + \cos ( A + B - C) + \cos (-A+B+C)$

$\displaystyle = [ \cos ( A+B+C) + \cos (A-B+C) ] + [ \cos ( A + B - C) + \cos (-A+B+C) ]$

$\displaystyle = 2 \cos \Big( \frac{A+B+C + A - B + C}{2} \cos \frac{A + B + C - A + B - C}{2} \Big) + 2 \cos \Big( \frac{A+B-C - A + B + C}{2} \cos \frac{A + B - C + A - B - C}{2} \Big)$

$\displaystyle = 2 \cos ( A + C) \cos B + 2 \cos B \cos ( A -C)$

$\displaystyle = 2 \cos B [ \cos ( A + C) + \cos ( A - C) ]$

$\displaystyle = 2 \cos B \Big[ 2 \cos \Big( \frac{A + C + A - C}{2} \Big) \cos \Big( \frac{A + C - A +C}{2} \Big) \Big]$

$\displaystyle = 4 \cos B \cos A \cos C$

$\displaystyle = 4 \cos A \cos B \cos C = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 10: If $\displaystyle \cos A + \cos B = \frac{1}{2}$ and $\displaystyle \sin A + \sin B = \frac{1}{4}$ , prove that:

$\displaystyle \tan \Big( \frac{A+B}{2} \Big) = \frac{1}{2}$

Answer:

$\displaystyle \text{Given: } \cos A + \cos B = \frac{1}{2}$ and $\displaystyle \sin A + \sin B = \frac{1}{4}$

Dividing one by another

$\displaystyle \Rightarrow \frac{\sin A + \sin B}{\cos A + \cos B} = \frac{\frac{1}{4}}{\frac{1}{2}}$

$\displaystyle \Rightarrow \frac{2 \sin \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big)} = \frac{1}{2}$

$\displaystyle \Rightarrow \frac{ \sin \Big( \frac{A+B}{2} \Big) }{ \cos \Big( \frac{A+B}{2} \Big) } = \frac{1}{2}$

$\displaystyle \Rightarrow \tan \Big( \frac{A+B}{2} \Big) = \frac{1}{2}$

$\displaystyle \\$

Question 11: If $\displaystyle \mathrm{cosec} A + \sec A = \mathrm{cosec} B + \sec B$ , then prove that $\displaystyle \tan A \tan B = \cot \Big( \frac{A+B}{2} \Big)$

Answer:

$\displaystyle \text{Given: } \mathrm{cosec} A + \sec A = \mathrm{cosec} B + \sec B$

$\displaystyle \Rightarrow \sec B - \sec A = \mathrm{cosec} B - \mathrm{cosec} A$

$\displaystyle \Rightarrow \frac{1}{\cos A} - \frac{1}{\cos B} = \frac{1}{\sin B} - \frac{1}{\sin A}$

$\displaystyle \Rightarrow \frac{\cos B - \cos A}{\cos A \cos B} = \frac{\sin A - \sin B}{\sin A \sin B}$

$\displaystyle \Rightarrow \frac{\sin A \sin B}{\cos A \cos B} = \frac{\sin A - \sin B}{\cos B - \cos A}$

$\displaystyle \Rightarrow \tan A \tan B = \frac{2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}}{2 \sin \frac{B-A}{2} \sin \frac{B+A}{2}}$

$\displaystyle \Rightarrow \tan A \tan B = \cot \frac{A+B}{2}$

$\displaystyle \\$

$\displaystyle \text{Question 12: If } \sin 2A = \lambda \sin 2B , \text{ prove that } \frac{\tan (A+B)}{\tan (A-B)} = \frac{\lambda+1}{\lambda - 1}$

Answer:

$\displaystyle \text{Given } \sin 2A = \lambda \sin 2B$

$\displaystyle \Rightarrow \lambda = \frac{\sin 2A}{\sin 2B}$

Applying componendo and dividendo

$\displaystyle \frac{\lambda+1}{\lambda-1} = \frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B}$

$\displaystyle \Rightarrow \frac{\lambda+1}{\lambda-1} = \frac{2 \sin ( A+B) \cos ( A - B) }{2 \sin ( A-B) \cos ( A + B) }$

$\displaystyle \Rightarrow \frac{\lambda+1}{\lambda-1} = \frac{\tan (A+B) }{\tan (A-B) }$

Hence proved.

$\displaystyle \\$

Question 13: Prove that:

$\displaystyle \text{i) } \frac{\cos (A+B+C) + \cos (-A + B + C) +\cos (A-B+C) + \cos (A+B-C) }{\sin (A+B+C) + \sin (-A + B + C) +\sin (A-B+C) + \sin (A+B-C)} = \cot C$

$\displaystyle \text{ii) } \sin (B-C) \cos (A-D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D) = 0$

Answer:

$\displaystyle \text{i) } \text{LHS } = \frac{\cos (A+B+C) + \cos (-A + B + C) +\cos (A-B+C) + \cos (A+B-C) }{\sin (A+B+C) + \sin (-A + B + C) +\sin (A-B+C) + \sin (A+B-C)}$

$\displaystyle = \frac{2 \cos \Big( \frac{A+B+C-A+B+C}{2} \Big) \cos \Big( \frac{A+B+C+A-B-C}{2} \Big) + 2 \cos \Big( \frac{A-B+C+A+B-C}{2} \Big) \cos \Big( \frac{A-B+C-A-B+C}{2} \Big) }{2 \sin \Big( \frac{A+B+C-A+B+C}{2} \Big) \cos \Big( \frac{A+B+C+A-B-C}{2} \Big) + 2 \sin \Big( \frac{A-B+C+A+B-C}{2} \Big) \cos \Big( \frac{A-B+C-A-B+C}{2} \Big) }$

$\displaystyle = \frac{2 \cos (B+C) \cos A + 2 \cos A \cos (C-B) }{2 \sin (B+C) \cos A + 2 \sin (C-B) \cos A}$

$\displaystyle = \frac{2 \cos A [ \cos ( B+C) + \cos (C-B) ]}{2 \cos A [ \sin ( B + C) + \sin (C-B) ]}$

$\displaystyle = \frac{ \cos ( B+C) + \cos (C-B) }{ \sin ( B + C) + \sin (C-B) }$

$\displaystyle = \frac{ 2 \cos \Big( \frac{B+C+C-B}{2} \Big) \cos \Big( \frac{B+C-C+B}{2} \Big) }{ 2 \sin \Big( \frac{B+C+C-B}{2} \Big) \cos \Big( \frac{B+C-C+B}{2} \Big) }$

$\displaystyle = \frac{2 \cos C \cos B}{ 2 \sin C \cos B}$

$\displaystyle = \frac{\cos C}{\sin C} = \cot C$

$\displaystyle \text{ii) } \text{LHS } = \sin (B-C) \cos (A-D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D)$

$\displaystyle = \frac{1}{2} [ 2\sin (B-C) \cos (A-D) + 2\sin (C-A) \cos (B-D) + 2\sin (A-B) \cos (C-D) ]$

$\displaystyle = \frac{1}{2} [ \sin ( B-C+A-D ) + \sin ( B - C - A + D) + \sin ( C - A + B - D) + \\ \sin ( C - A - B + D) + \sin (A - B + C - D) + \sin ( A - B - C + D) ]$

$\displaystyle = \frac{1}{2} [ \sin ( A + B - C - D ) + \sin ( B+D - A - C) + \sin ( B+C - A - D) \\ - \sin ( A + B - C - D ) - \sin (B + D - A - C) - \sin ( B+C-A-D) ]$

$\displaystyle = \frac{1}{2} [ 0 ] = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 14: If $\displaystyle \frac{\cos (A-B) }{\cos (A+B)} + \frac{\cos (C+D)}{\cos (C-D)} = 0 , \text{ prove that } \tan A \tan B \tan C \tan D = -1$

Answer:

$\displaystyle \text{Given } \frac{\cos (A-B) }{\cos (A+B)} + \frac{\cos (C+D)}{\cos (C-D)} = 0$

$\displaystyle \Rightarrow \frac{\cos ( A - B) \cos ( C - D) + \cos ( C+D) \cos (A+B) }{\cos ( A+ B) \cos ( C+D)} = 0$

$\displaystyle \Rightarrow \cos ( A - B) \cos ( C - D) + \cos ( C+D) \cos (A+B) = 0$

$\displaystyle \Rightarrow \cos ( A - B) \cos ( C - D) = - \cos ( C+D) \cos (A+B)$

$\displaystyle \Rightarrow (\cos A \cos B + \sin A \sin B) (\cos C \cos D + \sin C \sin D) = - ( \cos C \cos D - \sin C \sin D)(\cos A \cos B - \sin A \sin B)$

Dividing both sides by $\displaystyle \cos A \cos B \cos C \cos D$

$\displaystyle \Rightarrow \frac{ (\cos A \cos B + \sin A \sin B) (\cos C \cos D + \sin C \sin D)}{\cos A \cos B \cos C \cos D} = - \frac{ ( \cos C \cos D - \sin C \sin D)(\cos A \cos B - \sin A \sin B)}{\cos A \cos B \cos C \cos D}$

$\displaystyle \Rightarrow \frac{ (\cos A \cos B + \sin A \sin B)}{\cos A \cos B } \frac{ (\cos C \cos D + \sin C \sin D)}{ \cos C \cos D} = - \frac{ ( \cos C \cos D - \sin C \sin D)}{\cos A \cos B } \frac{ (\cos A \cos B - \sin A \sin B)}{ \cos C \cos D}$

$\displaystyle \Rightarrow ( 1 + \tan A \tan B) ( 1 + \tan C \tan D) = - ( 1- \tan C \tan D) ( 1 - \tan A \tan B)$

$\displaystyle \Rightarrow ( 1 + \tan A \tan B) ( 1 + \tan C \tan D) = ( \tan C \tan D - 1) ( 1 - \tan A \tan B)$

$\displaystyle \Rightarrow 1 + \tan A \tan B + \tan C \tan D + \tan A \tan B \tan C \tan D = \tan C \tan D - \tan A \tan B \tan C \tan D - 1 + \tan A \tan B$

$\displaystyle \Rightarrow 2 \tan A \tan B \tan C \tan D = -2$

$\displaystyle \Rightarrow \tan A \tan B \tan C \tan D = -1$

Hence proved.

$\displaystyle \\$

Question 15: If $\displaystyle \cos (\alpha + \beta) \sin (\gamma + \delta) = \cos (\alpha - \beta) \sin (\gamma - \delta)$ , then prove that $\displaystyle \cot \alpha \cot \beta \cot \gamma = \cot \delta$

Answer:

$\displaystyle \text{Given } \cos (\alpha + \beta) \sin (\gamma + \delta) = \cos (\alpha - \beta) \sin (\gamma - \delta)$

$\displaystyle \Rightarrow [ \cos \alpha \cos \beta - \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta + \cos \gamma \sin \delta ] = [ \cos \alpha \cos \beta + \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta - \cos \gamma \sin \delta ]$

Dividing both sides by $\displaystyle \sin \alpha \sin \beta \sin \gamma \sin \delta$

$\displaystyle \Rightarrow \frac{ [ \cos \alpha \cos \beta - \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta + \cos \gamma \sin \delta ]}{\sin \alpha \sin \beta \sin \gamma \sin \delta} = \frac{ [ \cos \alpha \cos \beta + \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta - \cos \gamma \sin \delta ]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}$

$\displaystyle \Rightarrow (\cot \alpha \cot \beta - 1 ) ( \cot \delta + \cot \gamma) = (\cot \alpha \cot \beta + 1 ) ( \cot \delta - \cot \gamma)$

$\displaystyle \Rightarrow \cot \alpha \cot \beta \cot \delta - \cot \delta + \cot \alpha \cot \beta \cot \gamma - \cot \gamma = \cot \alpha \cot \beta \cot \delta + \cot \delta - \cot \alpha \cot \beta \cot \gamma - \cot \gamma$

$\displaystyle \Rightarrow 2 \cot \alpha \cot \beta \cot \gamma = 2 \cot \delta$

$\displaystyle \Rightarrow \cot \alpha \cot \beta \cot \gamma = \cot \delta$

Hence proved.

$\displaystyle \\$

Question 16: If $\displaystyle y \sin \phi = x \sin (2 \theta + \phi)$ , prove that $\displaystyle (x+y) \cot (\theta + \phi)= (y-x) \cot \theta$

Answer:

$\displaystyle \text{Given } y \sin \phi = x \sin (2 \theta + \phi)$

$\displaystyle \Rightarrow \frac{y}{x} = \frac{\sin (2 \theta + \phi)}{\sin \phi}$

Applying componendo and dividendo

$\displaystyle \frac{y-x}{y+x} = \frac{\sin (2 \theta + \phi) - \sin \phi}{ \sin (2 \theta + \phi) + \sin \phi}$

$\displaystyle \Rightarrow \frac{y-x}{y+x} = \frac{2 \sin \Big( \frac{2 \theta + \phi - \phi}{2} \Big) \cos \Big( \frac{2 \theta + \phi + \phi}{2} \Big)}{ 2 \sin \Big( \frac{2 \theta + \phi + \phi}{2} \Big) \cos \Big( \frac{2 \theta + \phi - \phi}{2} \Big)}$

$\displaystyle \Rightarrow \frac{y-x}{y+x} = \frac{\sin \theta \cos ( \theta + \phi) }{\sin ( \theta + \phi) \cos \theta}$

$\displaystyle \Rightarrow \frac{y-x}{y+x} = \frac{\cot ( \theta + \phi) }{\cos \theta}$

$\displaystyle \Rightarrow (y-x) \cos \theta = (y+x) \cot ( \theta + \phi)$

Hence proved.

$\displaystyle \\$

Question 17: If $\displaystyle \cos (A+B) \sin (C-D) = \cos (A-B) \sin (C+D)$ , prove that $\displaystyle \tan A \tan B \tan C + \tan D = 0$

Answer:

$\displaystyle \text{Given } \cos (A+B) \sin (C-D) = \cos (A-B) \sin (C+D)$

$\displaystyle \Rightarrow (\cos A \cos B - \sin A \sin B) ( \sin C \cos D - \cos C \sin D) = (\cos A \cos B + \sin A \sin B) ( \sin C \cos D + \cos C \sin D)$

Divide both sides by $\displaystyle \cos A \cos B \cos C\cos D$ we get

$\displaystyle \frac{(\cos A \cos B - \sin A \sin B) ( \sin C \cos D - \cos C \sin D)}{\cos A \cos B \cos C\cos D} = \frac{(\cos A \cos B + \sin A \sin B) ( \sin C \cos D + \cos C \sin D)}{\cos A \cos B \cos C\cos D}$

$\displaystyle \Rightarrow ( 1 - \tan A \tan B)(\tan C - \tan D) = (1 + \tan A \tan B) (\tan C + \tan D)$

$\displaystyle \Rightarrow \tan C - \tan D - \tan A \tan B \tan C + \tan A \tan B \tan D = \tan C + \tan D + \tan A \tan B \tan C + \tan A \tan B \tan D$

$\displaystyle \Rightarrow -2 \tan D = 2 \tan A \tan B \tan C$

$\displaystyle \Rightarrow \tan A \tan B \tan C + \tan D = 0$

Hence proved.

$\displaystyle \\$

Question 18: If $\displaystyle x \cos \theta = y \cos \Big( \theta+ \frac{2\pi}{3} \Big) = z \cos \Big( \theta + \frac{4\pi}{3} \Big) , \text{ prove that } xy + yz + zx = 0$

Answer:

$\displaystyle \text{Given } x \cos \theta = y \cos \Big( \theta+ \frac{2\pi}{3} \Big) = z \cos \Big( \theta + \frac{4\pi}{3} \Big) = k$

$\displaystyle \Rightarrow x = \frac{k}{\cos \theta}$

$\displaystyle \Rightarrow y = \frac{k}{\cos \Big( \theta + \frac{2\pi}{3} \Big) }$

$\displaystyle \Rightarrow z = \frac{k}{\cos \Big( \theta + \frac{4\pi}{3} \Big) }$

$\displaystyle \therefore xy + yz + zx$

$\displaystyle = \Big[ \frac{k}{\cos \theta} \frac{k}{\cos \big( \theta + \frac{2\pi}{3} \big) } + \frac{k}{\cos \big( \theta + \frac{2\pi}{3} \big) } \frac{k}{\cos \big( \theta + \frac{4\pi}{3} \big) } + \frac{k}{\cos \big( \theta + \frac{4\pi}{3} \big) } \frac{k}{\cos \theta} \Big]$

$\displaystyle = k^2 \Big[ \frac{1}{\cos \theta} \frac{1}{\cos \big( \theta + \frac{2\pi}{3} \big) } + \frac{1}{\cos \big( \theta + \frac{2\pi}{3} \big) } \frac{1}{\cos \big( \theta + \frac{4\pi}{3} \big) } + \frac{1}{\cos \big( \theta + \frac{4\pi}{3} \big) } \frac{1}{\cos \theta} \Big]$

$\displaystyle = k^2 \Big[ \frac{\cos \big( \theta + \frac{4\pi}{3} \big) + \cos \theta + \cos \big( \theta + \frac{2\pi}{3} \big)}{\cos \big( \theta+ \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)} \Big]$

$\displaystyle = k^2 \Big[ \frac{ \cos \theta \cos \frac{4\pi}{3} - \sin \theta \sin \frac{4\pi}{3} + \cos \theta + \cos \theta \cos \frac{2\pi}{3} - \sin \theta \sin \frac{2\pi}{3}}{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)} \Big]$

$\displaystyle = k^2 \Big[ \frac{ \cos \theta (-\frac{1}{2}) - \sin \theta (-\frac{\sqrt{3}}{2}) + \cos \theta + \cos \theta (-\frac{1}{2}) - \sin \theta (\frac{\sqrt{3}}{2}) }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)} \Big]$

$\displaystyle = k^2 \Big[ \frac{ -\cos \theta + (\frac{\sqrt{3}}{2}) \sin \theta + \cos \theta - (\frac{\sqrt{3}}{2}) \sin \theta }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)} \Big]$

$\displaystyle = k^2 \Big[ \frac{ 0 }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)} \Big]$

$\displaystyle = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 19: If $\displaystyle m \sin \theta = n \sin (\theta + 2 \alpha) , \text{ prove that } \tan (\theta + \alpha) \cot \alpha = \frac{m+n}{m-n}$

Answer:

$\displaystyle \text{Given: } m \sin \theta = n \sin (\theta + 2 \alpha)$

$\displaystyle \frac{m}{n} = \frac{\sin (\theta + 2 \alpha)}{\sin \theta}$

Applying componendo and dividendo

$\displaystyle \frac{m+n}{m-n} = \frac{\sin (\theta + 2 \alpha) +\sin \theta }{\sin (\theta + 2 \alpha) -\sin \theta}$

$\displaystyle \frac{m+n}{m-n} = \frac{2 \sin \big( \frac{\theta + 2 \alpha + \theta}{2} \big) \cos \big( \frac{\theta + 2 \alpha - \theta}{2} \big) }{2 \cos \big( \frac{\theta + 2 \alpha + \theta}{2} \big) \sin \big( \frac{\theta + 2 \alpha - \theta}{2} \big)}$

$\displaystyle \frac{m+n}{m-n} = \frac{2 \sin ( \theta + \alpha) \cos \alpha}{2 \cos ( \theta + \alpha) \sin \alpha}$

$\displaystyle \frac{m+n}{m-n} = \frac{\tan ( \theta + \alpha) }{\tan \alpha}$

$\displaystyle \tan ( \theta + \alpha) = \frac{m+n}{m-n} \tan \alpha$

Hence proved.

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: Transformation Formulae – Exercise 8.2

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