Question 1: Express each of the following as a sum or difference of sine and cosines:

$\displaystyle \text{ i) } 2 \sin 3x \cos x \hspace{1.0cm} \text{ ii) } 2 \cos 3x \sin 2x \hspace{1.0cm} \text{ iii) } 2 \sin 4x \sin 3x \hspace{1.0cm} \text{ iv) } 2 \cos 7x \cos 3x$

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

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

$\displaystyle \text{ iii) } 2 \sin 4x \sin 3x = \cos ( 4x - 3x) - \cos (4x + 3x) = \cos x - \cos 7x$

$\displaystyle \text{ iv) } 2 \cos 7x \cos 3x = \cos ( 7x + 3x) + \cos (7x - 3x) = \cos 10 x + \cos 4x$

$\displaystyle \\$

Question 2: Prove that:

$\displaystyle \text{ i) } 2 \sin \frac{5\pi}{12} \sin \frac{\pi}{12} = \frac{1}{2} \hspace{1.0cm} \text{ ii) } 2 \cos \frac{5\pi}{12} \cos \frac{\pi}{12} = \frac{1}{2} \hspace{1.0cm} \text{ iii) } 2 \sin \frac{5\pi}{12} \cos \frac{\pi}{12} = \frac{\sqrt{3}+2}{2}$

$\displaystyle \text{ i) } 2 \sin \frac{5\pi}{12} \sin \frac{\pi}{12}$

$\displaystyle = \cos \Big( \frac{5\pi}{12} - \frac{\pi}{12} \Big) - \cos \Big( \frac{5\pi}{12} + \frac{\pi}{12} \Big)$

$\displaystyle = \cos \frac{4\pi}{12} - \cos \frac{6\pi}{12}$

$\displaystyle = \cos \frac{\pi}{3} - \cos \frac{\pi}{2}$

$\displaystyle = \frac{1}{2} - 0$

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

$\displaystyle \text{ ii) } 2 \cos \frac{5\pi}{12} \cos \frac{\pi}{12}$

$\displaystyle = \cos \Big( \frac{5\pi}{12} + \frac{\pi}{12} \Big) + \cos \Big( \frac{5\pi}{12} - \frac{\pi}{12} \Big)$

$\displaystyle = \cos \frac{\pi}{2} + \cos \frac{\pi}{3}$

$\displaystyle = 0 + \frac{1}{2}$

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

$\displaystyle \text{ iii) } 2 \sin \frac{5\pi}{12} \cos \frac{\pi}{12}$

$\displaystyle = \sin \Big( \frac{5\pi}{12} + \frac{\pi}{12} \Big) + \sin \Big( \frac{5\pi}{12} - \frac{\pi}{12} \Big)$

$\displaystyle = \sin \frac{\pi}{2} + \sin \frac{\pi}{3}$

$\displaystyle = 1 + \frac{\sqrt{3}}{2}$

$\displaystyle = \frac{2 + \sqrt{3}}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 3: Show that:

$\displaystyle \text{ i) } \sin 50^{\circ} \cos 85^{\circ} = \frac{1 - \sqrt{2} \sin 35^{\circ}}{2\sqrt{2}} \hspace{1.0cm} \text{ ii) } \sin 25^{\circ} \cos 115^{\circ} = \frac{1}{2} ( \sin 40^{\circ} - 1)$

$\displaystyle \text{i) } \text{LHS } = \sin 50^{\circ} \cos 85^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ 2 \sin 50^{\circ} \cos 85^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin (50^{\circ} + 85^{\circ}) + \sin (50^{\circ} - 85^{\circ}) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin 135^{\circ} + \sin ( - 35^{\circ}) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin (90^{\circ} + 45^{\circ}) - \sin 35^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 45^{\circ} - \sin 35^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \frac{1}{\sqrt{2}} - \sin 35^{\circ} \Big]$

$\displaystyle = \frac{1 - \sqrt{2} \sin 35^{\circ}}{2 \sqrt{2}} = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \sin 25^{\circ} \cos 115^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ 2 \sin 25^{\circ} \cos 115^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin (25^{\circ} + 115^{\circ}) + \sin (25^{\circ} - 115^{\circ}) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin 140^{\circ} + \sin ( - 90^{\circ}) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin (90^{\circ} + 50^{\circ}) - \sin 90^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 50^{\circ} - 1 \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos (90^{\circ} - 40^{\circ}) - 1 \Big]$

$\displaystyle = \frac{1}{2} \Big[ \sin 40^{\circ} - 1 \Big] \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 4: Prove that } 4 \cos x \cos \Big( \frac{\pi}{3} +x \Big) \cos \Big( \frac{\pi}{3} -x \Big) = \cos 3x$

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

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

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

$\displaystyle = 2 \cos x \Big[ \cos \frac{2\pi}{3} + \cos 2x \Big]$

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

$\displaystyle = 2 \cos x \Big[ - \sin \frac{\pi}{6} + \cos 2x \Big]$

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

$\displaystyle = - \cos x + 2 \cos x \cos 2x$

$\displaystyle = - \cos x + [ \cos (2x + x) + \cos ( 2x - x) ]$

$\displaystyle = - \cos x + \cos 3x + \cos x$

$\displaystyle = \cos 3x =$ RHS

$\displaystyle \\$

Question 5: Prove that:

$\displaystyle \text{ i) } \cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ} = \frac{3}{16} \hspace{1.0cm} \text{ ii) } \cos 40^{\circ} \cos 80^{\circ} \cos 160^{\circ} = - \frac{1}{8}$

$\displaystyle \text{ iii) } \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} = \frac{\sqrt{3}}{8} \hspace{1.0cm} \text{ iv) } \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} = \frac{1}{8}$

$\displaystyle \text{ v) } \tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ} = 3 \hspace{1.0cm} \text{ vi) } \tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 80^{\circ} = 1$

$\displaystyle \text{ vii) } \sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ} = \frac{\sqrt{3}}{16} \hspace{1.0cm} \text{ viii) } \sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ} = \frac{3}{16}$

$\displaystyle \text{i) } \text{LHS } = \cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{2} \Big[ (\cos 10^{\circ} \cos 50^{\circ}) \cos 70^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ (2 \cos 50^{\circ} \cos 10^{\circ}) \cos 70^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ \Big( \cos (50^{\circ}+10^{\circ}) + \cos (50^{\circ}-10^{\circ}) \Big) \cos 70^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ ( \cos 60^{\circ} + \cos 40^{\circ} ) \cos 70^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ ( \frac{1}{2} + \cos 40^{\circ} ) \cos 70^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \cos 70^{\circ} + \frac{\sqrt{3}}{4} \cos 40^{\circ} \cos 70^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{8} \cos 70^{\circ} + \frac{\sqrt{3}}{8} (2\cos 40^{\circ} \cos 70^{\circ})$

$\displaystyle = \frac{\sqrt{3}}{8} \cos 70^{\circ} + \frac{\sqrt{3}}{8} ( \cos (70^{\circ}+40^{\circ}) + \cos (70^{\circ}-40^{\circ}) )$

$\displaystyle = \frac{\sqrt{3}}{8} \cos 70^{\circ} + \frac{\sqrt{3}}{8} \cos 110 + \frac{\sqrt{3}}{8} \cos 30^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 70^{\circ} + \cos 110 + \cos 30 \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 70^{\circ} + \cos (180 - 70) + \cos 30 \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 70^{\circ} - \cos 70^{\circ} + \frac{\sqrt{3}}{2} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \frac{\sqrt{3}}{2} \Big]$

$\displaystyle = \frac{3}{16} = \text{ RHS. Hence proved. }$

$\displaystyle \text{ii) } \text{LHS } = \cos 40^{\circ} \cos 80^{\circ} \cos 160^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ 2 \cos 40^{\circ} \cos 80^{\circ} \cos 160^{\circ} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 80^{\circ} ( 2 \cos 160^{\circ} \cos 40^{\circ}) \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 80^{\circ} [ \cos ( 160^{\circ}+40^{\circ}) + \cos ( 160^{\circ}-40^{\circ}) ] \Big]$

$\displaystyle = \frac{1}{2} \Big[ \cos 80^{\circ} ( \cos 200^{\circ} + \cos 120^{\circ} ) \Big]$

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

$\displaystyle = \frac{1}{2} \Big[ \cos 80^{\circ} ( \cos 20^{\circ} + \cos 60^{\circ} ) \Big]$

$\displaystyle = - \frac{1}{2} \cos 80^{\circ} \cos 20^{\circ} + \frac{1}{2} \cos 80^{\circ} \cos 60^{\circ}$

$\displaystyle = - \frac{1}{4} [ 2 \cos 80^{\circ} \cos 20^{\circ} + 2 \cos 80^{\circ} \cos 60^{\circ} ]$

$\displaystyle = - \frac{1}{4} [ \cos (80^{\circ}+20^{\circ}) + \cos (80^{\circ}-20^{\circ})+ 2 \cos 80^{\circ} \times \frac{1}{2} ]$

$\displaystyle = - \frac{1}{4} [ \cos 100^{\circ} + \cos 60^{\circ}+ \cos 80^{\circ} ]$

$\displaystyle = - \frac{1}{4} [ \cos (180^{\circ} - 80^{\circ}) + \cos 60^{\circ}+ \cos 80^{\circ} ]$

$\displaystyle = - \frac{1}{4} [ -\cos 80^{\circ} + \cos 60^{\circ}+ \cos 80^{\circ} ]$

$\displaystyle = - \frac{1}{4} [ \cos 60^{\circ} ]$

$\displaystyle = - \frac{1}{8} = \text{ RHS. Hence proved. }$

$\displaystyle \text{iii) } \text{LHS } = \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ 2 \sin 20^{\circ} \sin 40^{\circ} \Big] \sin 80^{\circ}$

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

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

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

$\displaystyle = \frac{1}{2} \Big[ \cos 20^{\circ} \sin 80^{\circ} \Big] - \frac{1}{4} \sin 80^{\circ}$

$\displaystyle = \frac{1}{4} \Big[ 2 \cos 20^{\circ} \sin 80^{\circ} \Big] - \frac{1}{4} \sin 80^{\circ}$

$\displaystyle = \frac{1}{4} \Big[ \sin (80^{\circ}+20^{\circ}) + \sin (80^{\circ}-20^{\circ}) \Big] - \frac{1}{4} \sin 80^{\circ}$

$\displaystyle = \frac{1}{4} \Big[ \sin 100^{\circ} + \sin 60^{\circ} \Big] - \frac{1}{4} \sin 80^{\circ}$

$\displaystyle = \frac{1}{4} \Big[ \sin (180^{\circ} - 80^{\circ}) + \sin 60^{\circ} \Big] - \frac{1}{4} \sin 80^{\circ}$

$\displaystyle = \frac{1}{4} \sin 80^{\circ} + \frac{\sqrt{3}}{8} - \frac{1}{4} \sin 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{8} = \text{ RHS. Hence proved. }$

$\displaystyle \text{iv) } \text{LHS } = \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ 2 \cos 20^{\circ} \cos 40^{\circ} \Big] \cos 80^{\circ}$

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

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

$\displaystyle = \frac{1}{2} \Big[ \frac{1}{2} + \cos 20^{\circ} \Big] \cos 80^{\circ}$

$\displaystyle = \frac{1}{4} \Big[ \cos 80^{\circ} + 2\cos 20^{\circ} \cos 80^{\circ} \Big]$

$\displaystyle = \frac{1}{4} \Big[ \cos 80^{\circ} + \cos ( 80^{\circ} + 20^{\circ}) + \cos ( 80^{\circ} - 20^{\circ}) \Big]$

$\displaystyle = \frac{1}{4} \Big[ \cos 80^{\circ} + \cos 100^{\circ} + \cos 60^{\circ} \Big]$

$\displaystyle = \frac{1}{4} \Big[ \cos 80^{\circ} + \cos (180^{\circ} - 80^{\circ}) + \cos 60^{\circ} \Big]$

$\displaystyle = \frac{1}{4} \Big[ \cos 80^{\circ} - \cos 80^{\circ} + \frac{1}{2} \Big]$

$\displaystyle = \frac{1}{8} = \text{ RHS. Hence proved. }$

$\displaystyle \text{v) } \text{LHS } = \tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}$

$\displaystyle = \sqrt{3} [ \tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ} ]$

$\displaystyle = \sqrt{3} \Big[ \frac{\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} }{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ [ \cos ( 40^{\circ}-20^{\circ}) - \cos ( 40^{\circ} + 20^{\circ}) ] \sin 80^{\circ} }{ [ \cos ( 20^{\circ}+40^{\circ}) + \cos ( 40^{\circ} - 20^{\circ}) ] \cos 80^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ [ \cos 20^{\circ} - \cos 60^{\circ} ] \sin 80^{\circ} }{ [ \cos 60^{\circ} + \cos 20^{\circ} ] \cos 80^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ [ 2 \cos 20^{\circ} - 1 ] \sin 80^{\circ} }{ [ 1 + 2 \cos 20^{\circ} ] \cos 80^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ 2 \cos 20^{\circ} \sin 80^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} + 2 \cos 20^{\circ} \cos 80^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ \sin (80+20) + \sin (80^{\circ}-20^{\circ}) - \sin 80^{\circ} }{ \cos 80^{\circ} + \cos (80^{\circ}+20^{\circ}) + \cos (80^{\circ}-20^{\circ}) } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ \sin 100^{\circ} + \sin 60^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} + \cos 100^{\circ} + \cos 60^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ \sin (180^{\circ}-80^{\circ}) + \sin 60^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} + \cos (180^{\circ}-80^{\circ}) + \cos 60^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \Big[ \frac{ \sin 80^{\circ} + \sin 60^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} - \cos 80^{\circ} + \cos 60^{\circ} } \Big]$

$\displaystyle = \sqrt{3} \tan 60^{\circ}$

$\displaystyle = \sqrt{3} \times \sqrt{3} = 3 = \text{ RHS. Hence proved. }$

$\displaystyle \text{vi) } \text{LHS } = \tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 80^{\circ}$

$\displaystyle = \frac{1}{\sqrt{3}} [ \tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ} ]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} }{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ (2 \sin 20^{\circ} \sin 40^{\circ}) \sin 80^{\circ} }{ (2 \cos 20^{\circ} \cos 40^{\circ}) \cos 80^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ [ \cos ( 40^{\circ}-20^{\circ}) - \cos ( 40^{\circ} + 20^{\circ}) ] \sin 80^{\circ} }{ [ \cos ( 20^{\circ}+40^{\circ}) + \cos ( 40^{\circ} - 20^{\circ}) ] \cos 80^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ [ \cos 20^{\circ} - \cos 60^{\circ} ] \sin 80^{\circ} }{ [ \cos 60^{\circ} + \cos 20^{\circ} ] \cos 80^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ [ 2 \cos 20^{\circ} - 1 ] \sin 80^{\circ} }{ [ 1 + 2 \cos 20^{\circ} ] \cos 80^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ 2 \cos 20^{\circ} \sin 80^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} + 2 \cos 20^{\circ} \cos 80^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ \sin (80+20) + \sin (80^{\circ}-20^{\circ}) - \sin 80^{\circ} }{ \cos 80^{\circ} + \cos (80^{\circ}+20^{\circ}) + \cos (80^{\circ}-20^{\circ}) } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ \sin 100^{\circ} + \sin 60^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} + \cos 100^{\circ} + \cos 60^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ \sin (180^{\circ}-80^{\circ}) + \sin 60^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} + \cos (180^{\circ}-80^{\circ}) + \cos 60^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \Big[ \frac{ \sin 80^{\circ} + \sin 60^{\circ} - \sin 80^{\circ} }{ \cos 80^{\circ} - \cos 80^{\circ} + \cos 60^{\circ} } \Big]$

$\displaystyle = \frac{1}{\sqrt{3}} \tan 60^{\circ}$

$\displaystyle = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1 = \text{ RHS. Hence proved. }$

$\displaystyle \text{vii) } \text{LHS } = \sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{2} \Big[ \sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{2} \Big[ \sin (90^{\circ} - 80^{\circ}) \sin (90^{\circ} - 40^{\circ}) \sin (90^{\circ} - 20^{\circ}) \Big]$

$\displaystyle = \frac{\sqrt{3}}{2} \Big[ \cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ \cos 80^{\circ} (2 \cos 40^{\circ} \cos 20^{\circ}) \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ \cos (40^{\circ} + 20^{\circ}) + \cos (40^{\circ} - 20^{\circ}) \Big] \cos 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ \cos 60^{\circ}+ \cos 20^{\circ} \Big] \cos 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 80^{\circ}+ 2 \cos 20^{\circ} \cos 80^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 80^{\circ}+ \cos ( 80^{\circ} + 20^{\circ}) + \cos (80^{\circ} - 20^{\circ}) \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 80^{\circ}+ \cos 100^{\circ} + \cos 60^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 80^{\circ}+ \cos (180 - 80^{\circ}) + \cos 60^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ \cos 80^{\circ} - \cos 80^{\circ} + \cos 60^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \cos 60^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{16} = \text{ RHS. Hence proved. }$

$\displaystyle \text{viii) } \text{LHS } = \sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{2} \Big[ \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ (2 \sin 20^{\circ} \sin 40^{\circ}) \sin 80^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ \cos (40^{\circ} - 20^{\circ}) - \cos (40^{\circ} + 20^{\circ}) \Big] \sin 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ - \cos 60^{\circ}+ \cos 20^{\circ} \Big] \sin 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ - \frac{1}{2}+ \cos 20^{\circ} \Big] \sin 80^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ - \frac{1}{2} \sin 80^{\circ} + \cos 20^{\circ} \sin 80^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ - \sin 80^{\circ} + 2 \cos 20^{\circ} \sin 80^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ - \sin 80^{\circ} + \sin (80^{\circ} + 20^{\circ}) + \sin ( 80^{\circ} - 20^{\circ}) \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ - \sin 80^{\circ} + \sin 100^{\circ} + \sin 60^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ - \sin 80^{\circ} + \sin (180^{\circ} - 80^{\circ}) + \sin 60^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \Big[ - \sin 80^{\circ} + \sin 80^{\circ} + \sin 60^{\circ} \Big]$

$\displaystyle = \frac{\sqrt{3}}{8} \sin 60^{\circ}$

$\displaystyle = \frac{\sqrt{3}}{16} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 6: Show that:

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 7: Prove that: $\displaystyle \tan x \tan \Big( \frac{\pi}{3} - x \Big) \tan \Big( \frac{\pi}{3} + x \Big) = \tan 3x$

$\displaystyle \text{LHS } = \tan x \tan (60-x) \tan (60+x)$

$\displaystyle = \frac{2 \sin x \sin ( 60-x) \sin ( 60+x)}{2 \cos x \cos ( 60-x) \cos ( 60+x)}$

$\displaystyle = \frac{\sin x }{\cos x} \Big[ \frac{2 \sin ( 60-x) \sin ( 60+x)}{2 \cos ( 60-x) \cos ( 60+x)} \Big]$

$\displaystyle = \frac{\sin x }{\cos x} \Big[ \frac{\cos ( 60-x- 60 - x) - \cos ( 60 - x + 60 + x) }{\cos ( 60 - x + 60 + x) + \cos ( 60 - x - 60 - x)} \Big]$

$\displaystyle = \frac{\sin x }{\cos x} \Big[ \frac{\cos ( -2x) - \cos 120^{\circ} }{\cos 120^{\circ} + \cos ( -2x)} \Big]$

$\displaystyle = \frac{\sin x }{\cos x} \Big[ \frac{\cos 2x - \cos 120^{\circ} }{\cos 120^{\circ} + \cos 2x } \Big]$

$\displaystyle = \frac{\sin x }{\cos x} \Big[ \frac{\cos 2x - \cos (90^{\circ} + 30^{\circ}) }{\cos (90^{\circ} + 30^{\circ}) + \cos 2x } \Big]$

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

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

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

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

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

$\displaystyle = \frac{\sin 3x}{\cos 3x}$

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

$\displaystyle \\$

Question 8: If $\displaystyle \alpha + \beta = \frac{\pi}{2}$ , show that the maximum value of $\displaystyle \cos \alpha \cos \beta$ is $\displaystyle \frac{1}{2}$

Let $\displaystyle x = \cos \alpha \cos \beta$

$\displaystyle \Rightarrow x = \frac{1}{2} \Big[ 2 \cos \alpha \cos \beta \Big]$

$\displaystyle \Rightarrow x = \frac{1}{2} \Big[ \cos (\alpha + \beta) + \cos (\alpha - \beta) \Big]$

$\displaystyle \Rightarrow x = \frac{1}{2} \Big[ \cos 90^{\circ} + \cos (\alpha - \beta) \Big]$

$\displaystyle \Rightarrow x = \frac{1}{2} \Big[ 0+ \cos (\alpha - \beta) \Big]$

$\displaystyle \Rightarrow x = \frac{1}{2} \cos (\alpha - \beta)$

Now we know$\displaystyle -1 \leq \cos (\alpha - \beta) \leq 1$

$\displaystyle \Rightarrow - \frac{1}{2} \leq \frac{1}{2} \cos (\alpha - \beta) \leq \frac{1}{2}$

$\displaystyle \Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2}$

$\displaystyle \text{Therefore the max value of } x \text{ is } \frac{1}{2}$