Class 11: Values of Trigonometric Functions at Multiples and Submultiples of an Angle – Exercise 9.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: } \sin^2 \frac{2\pi}{5} - \sin^2 \frac{\pi}{3} = \frac{\sqrt{5}-1}{8}$

Answer:

$\displaystyle \text{LHS } = \sin^2 72^{\circ} - \sin^2 60^{\circ}$

$\displaystyle = \sin^2 ( 90^{\circ}-18^{\circ}) - \Big( \frac{\sqrt{3}}{2} \Big)^2$

$\displaystyle = \cos^2 18^{\circ} - \frac{3}{4}$

$\displaystyle \because \cos 18^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$

$\displaystyle = \Big( \frac{\sqrt{10+2\sqrt{5}}}{4} \Big)^2 - \frac{3}{4}$

$\displaystyle = \frac{10+2\sqrt{5}}{16} - \frac{3}{4}$

$\displaystyle = \frac{10 + 2 \sqrt{5}-12}{16} = \frac{2\sqrt{5} - 2 }{16} = \frac{\sqrt{5} - 1}{8} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 2: } \sin^2 24^{\circ} - \sin^2 6^{\circ} = \frac{\sqrt{5}-1}{8}$

Answer:

$\displaystyle \text{LHS } = \sin^2 24^{\circ} - \sin^2 6^{\circ}$

$\displaystyle \because \sin ( A+B) \sin ( A - B) = \sin^2 A - \sin^2 B$

$\displaystyle = \sin ( 24^{\circ} +6^{\circ}) \sin (24^{\circ}-6^{\circ})$

$\displaystyle = \sin 30^{\circ} \sin 18^{\circ}$

$\displaystyle \because \sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$

$\displaystyle = \frac{1}{2} \Big( \frac{\sqrt{5}-1}{4} \Big)$

$\displaystyle = \frac{\sqrt{5}-1}{8} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 3: } \sin^2 42^{\circ} - \cos^2 78^{\circ} = \frac{\sqrt{5}+1}{8}$

Answer:

$\displaystyle \text{LHS } = \sin^2 42^{\circ} - \cos^2 78^{\circ}$

$\displaystyle = \sin^2 (90^{\circ}-48^{\circ}) - \cos^2 (90^{\circ}-12^{\circ})$

$\displaystyle = \cos^2 48^{\circ} - \sin^2 12^{\circ}$

$\displaystyle = \cos ( 48^{\circ} + 12^{\circ}) \cos ( 48^{\circ} - 12^{\circ})$

$\displaystyle = \cos 60^{\circ} \cos 36^{\circ}$

$\displaystyle \because \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$

$\displaystyle = \frac{1}{2} \times \frac{\sqrt{5}+1}{4}$

$\displaystyle = \frac{\sqrt{5}+1}{8} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 4: } \cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ} = \frac{1}{8}$

Answer:

$\displaystyle \text{LHS } = \cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}$

$\displaystyle = \frac{1}{2} [ 2 \cos 78^{\circ} \cos 42^{\circ} ] \cos 36^{\circ}$

$\displaystyle = \frac{1}{2} [ \cos 120^{\circ} + \cos 36^{\circ} ] \cos 36^{\circ}$

$\displaystyle = \frac{1}{2} \Big[ - \frac{1}{2} + \frac{\sqrt{5}+ 1}{4} \Big] \Big( \frac{\sqrt{5}+ 1}{4} \Big)$

$\displaystyle = \frac{1}{2} \Big[ \frac{\sqrt{5}- 1}{4} \Big] \Big( \frac{\sqrt{5}+ 1}{4} \Big)$

$\displaystyle = \frac{1}{2} \times \frac{5-1}{16} = \frac{1}{8} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 5: } \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{16}$

Answer:

$\displaystyle \text{LHS } = \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{2\sin \frac{\pi}{15}} \Big[ 2 \sin \frac{\pi}{15} \cos \frac{\pi}{15} \Big] \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{4\sin \frac{\pi}{15}} \Big[ 2 \sin \frac{2\pi}{15} \cos \frac{2\pi}{15} \Big] \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{8\sin \frac{\pi}{15}} \Big[ 2 \sin \frac{4\pi}{15} \cos \frac{4\pi}{15} \Big] \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{16\sin \frac{\pi}{15}} \Big[ 2 \sin \frac{8\pi}{15} \cos \frac{7\pi}{15} \Big]$

$\displaystyle = \frac{1}{16\sin \frac{\pi}{15}} \Big[ \sin \big( \frac{8\pi}{15} + \frac{7\pi}{15} \big) + \sin \big( \frac{8\pi}{15} - \frac{7\pi}{15} \big) \Big]$

$\displaystyle = \frac{1}{16\sin \frac{\pi}{15}} \Big[ \sin \pi + \sin \frac{\pi}{15} \Big]$

$\displaystyle = \frac{1}{16\sin \frac{\pi}{15}} \Big[ \sin \frac{\pi}{15} \Big]$

$\displaystyle = \frac{1}{16} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 6: } \cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ} = \frac{1}{16}$

Answer:

$\displaystyle \text{LHS } = \cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}$

$\displaystyle = \frac{1}{4} ( 2 \cos 6^{\circ} \cos 66^{\circ})( 2 \cos 42^{\circ} \cos 78^{\circ})$

$\displaystyle = \frac{1}{4} ( \cos 72^{\circ} + \cos 60^{\circ}) ( \cos 120^{\circ} + \cos 36^{\circ})$

$\displaystyle = \frac{1}{4} \big(\sin 18^{\circ} + \frac{1}{2} \big) \big( - \frac{1}{2} + \frac{\sqrt{5}+ 1}{4} \big )$

$\displaystyle = \frac{1}{4} \big ( \frac{\sqrt{5}-1}{4} + \frac{1}{2} \big) \big( + \frac{\sqrt{5}+ 1}{4} - \frac{1}{2} \big )$

$\displaystyle = \frac{1}{4} \big ( \frac{\sqrt{5}+1}{4} \big) \big( \frac{\sqrt{5}-1}{4} \big )$

$\displaystyle = \frac{1}{4} \times \frac{4}{16} = \frac{1}{16} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ} = \frac{1}{16}$

Answer:

$\displaystyle \text{LHS } = \sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ}$

$\displaystyle = \frac{1}{4} ( 2 \sin 6^{\circ} \sin 66^{\circ})( 2 \sin 42^{\circ} \sin 78^{\circ})$

$\displaystyle = \frac{1}{4} ( \cos 60^{\circ} - \cos 72^{\circ}) ( \cos 36^{\circ} - \cos 120^{\circ})$

$\displaystyle = \frac{1}{4} \big( \frac{1}{2} - \sin 18^{\circ} \big) \big( \frac{\sqrt{5}+ 1}{4} + \frac{1}{2} \big )$

$\displaystyle = \frac{1}{4} \big( \frac{1}{2} - \frac{\sqrt{5}- 1}{4} \big) \big( \frac{\sqrt{5}+ 1}{4} + \frac{1}{2} \big )$

$\displaystyle = \frac{1}{4} \big ( \frac{3 - \sqrt{5}}{4} \big) \big( \frac{3 + \sqrt{5}}{4} \big )$

$\displaystyle = \frac{1}{4} \times \frac{4}{16} = \frac{1}{16} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 8: } \cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ} = \frac{1}{16}$

Answer:

$\displaystyle \text{LHS } = \cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}$

$\displaystyle = \frac{1}{2} \cos 36^{\circ} \cos 60^{\circ} ( 2 \cos 42^{\circ} \cos 78^{\circ})$

$\displaystyle = \frac{1}{2} \Big( \frac{\sqrt{5}+1}{4} \Big) \times \frac{1}{2} ( \cos 120^{\circ} + \cos 36^{\circ})$

$\displaystyle = \Big( \frac{\sqrt{5}+1}{16} \Big) \Big( - \frac{1}{2} + \frac{\sqrt{5}+1}{4} \Big)$

$\displaystyle = \Big( \frac{\sqrt{5}+1}{16} \Big) \Big( \frac{\sqrt{5}-1}{4} \Big)$

$\displaystyle = \frac{4}{64} = \frac{1}{16} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 9: } \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \frac{3\pi}{5} \sin \frac{4\pi}{5} = \frac{5}{16}$

Answer:

$\displaystyle \text{LHS } = \sin 36^{\circ} \sin 72^{\circ} \sin 108^{\circ} \sin 144^{\circ}$

$\displaystyle = \sin 36^{\circ} \sin 72^{\circ} \sin (180^{\circ} - 72^{\circ}) \sin ( 180^{\circ} - 36^{\circ})$

$\displaystyle = \sin 36^{\circ} \sin 72^{\circ} \sin 72^{\circ} \sin 36^{\circ}$

$\displaystyle = ( 2 \sin 36^{\circ} \sin 72^{\circ} )^2$

$\displaystyle = ( 2 \sin 36^{\circ} \cos 18^{\circ} )^2$

$\displaystyle = \Big( \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \times \frac{\sqrt{10 + 2 \sqrt{5}}}{4} \Big)$

$\displaystyle = \frac{100 - 20}{16 \times 16} = \frac{80}{16 \times 16} = \frac{5}{16} =\text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$ $\displaystyle = \frac{1}{128}$

Answer:

$\displaystyle \text{LHS } = \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{2 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{\pi}{15} \cos \frac{\pi}{15} \Big] \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{4 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{2\pi}{15} \cos \frac{2\pi}{15} \Big] \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{8 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{4\pi}{15} \cos \frac{4\pi}{15} \Big] \cos \frac{3\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{16 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ 2 \sin \frac{3\pi}{15} \cos \frac{3\pi}{15} \Big] \sin \frac{8\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ 2 \sin \frac{6\pi}{15} \cos \frac{6\pi}{15} \Big] \sin \frac{8\pi}{15} \cos \frac{7\pi}{15}$

$\displaystyle = \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ \sin \frac{12\pi}{15} \sin \frac{8\pi}{15} \cos \frac{7\pi}{15} \Big]$

$\displaystyle = \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ \sin ( \pi - \frac{3\pi}{15}) \sin (\pi - \frac{7\pi}{15} ) \cos \frac{7\pi}{15} \Big]$