Question 5, Exercise 10.3

Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove that $\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$.

We know that
$2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$
$$\begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos \left( 80+40 \right)+\cos \left( 80-40 \right) \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos {{120}^{\circ }}+\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( -\dfrac{1}{2}+\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=-\dfrac{1}{8}\cos {{20}^{\circ }}+\dfrac{1}{4}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=-\dfrac{1}{8}+\dfrac{1}{8}\left( 2\cos {{40}^{\circ }}\cos {{20}^{\circ }} \right)\\ &=-\dfrac{1}{8}\cos {{20}^{\circ }}+\dfrac{1}{8}\left( \cos (40+20)+\cos (40-20) \right)\\ &=-\dfrac{1}{8}\cos {{20}^{\circ }}+\dfrac{1}{8}\left( \cos 60+\cos 20 \right)\\ &=-\dfrac{1}{8}\cos {{20}^{\circ }}+\dfrac{1}{8}\left( \dfrac{1}{2}+\cos 20 \right)\\ &=-\dfrac{1}{8}\cos {{20}^{\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\cos {{20}^{\circ }}\\ &=\dfrac{1}{16}=R.H.S.\end{align}$$

Prove the identity $\sin \dfrac{\pi }{9}\sin \dfrac{2\pi }{9}\sin \dfrac{\pi }{3}\sin \dfrac{4\pi }{9}=\dfrac{3}{16}$.

We know that
$2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ $$\begin{align}L.H.S.&=\sin \dfrac{\pi }{9}\sin \dfrac{2\pi }{9}\sin \dfrac{\pi }{3}\sin \dfrac{4\pi }{9}\\ &=\sin \dfrac{{{180}^{\circ }}}{9}\sin \dfrac{2({{180}^{\circ }})}{9}\sin \dfrac{({{180}^{\circ }})}{3}\sin \dfrac{4({{180}^{\circ }})}{9}\quad \because \,\,\,\,\pi ={{180}^{\circ }}\\ &=\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}\\ &=\sin {{20}^{\circ }}\sin {{40}^{\circ }}\dfrac{\sqrt{3}}{2}\sin {{80}^{\circ }}\\ &=\dfrac{\sqrt{3}}{2}\sin {{80}^{\circ }}\sin {{40}^{\circ }}\sin {{20}^{\circ }}\\ &=-\dfrac{\sqrt{3}}{4}\left( -2\sin {{80}^{\circ }}\sin {{40}^{\circ }} \right)\sin {{20}^{\circ }}\\ &=-\dfrac{\sqrt{3}}{4}\left( \cos (80+40)-\cos (80-40) \right)\sin {{20}^{\circ }}\\ &=-\dfrac{\sqrt{3}}{4}\left( \cos {{120}^{\circ }}-\cos {{40}^{\circ }} \right)\sin {{20}^{\circ }}\\ &=-\dfrac{\sqrt{3}}{4}\left( -\dfrac{1}{2}-\cos {{40}^{\circ }} \right)\sin {{20}^{\circ }}\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\dfrac{\sqrt{3}}{4}\cos {{40}^{\circ }}\sin {{20}^{\circ }}\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\dfrac{\sqrt{3}}{8}\left( 2\cos {{40}^{\circ }}\sin {{20}^{\circ }} \right)\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\dfrac{\sqrt{3}}{8}\left( \sin (40+20)-\sin (40-20) \right)\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\dfrac{\sqrt{3}}{8}\left( \sin {{60}^{\circ }}-\sin {{20}^{\circ }} \right)\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\dfrac{\sqrt{3}}{8}\left( \dfrac{\sqrt{3}}{2}-\sin {{20}^{\circ }} \right)\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\dfrac{3}{16}-\dfrac{\sqrt{3}}{8}\sin {{20}^{\circ }}\\ &=\dfrac{3}{16}=R.H.S.\end{align}$$

Prove the identity $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$.

We know that
$2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ $$\begin{align}L.H.S.&=\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}\\ &=\sin {{10}^{\circ }}\left( \dfrac{1}{2} \right)\sin {{50}^{\circ }}\sin {{70}^{\circ }}\\ &=\dfrac{1}{2}\sin {{50}^{\circ }}\sin {{70}^{\circ }}\sin {{10}^{\circ }}\\ &=-\dfrac{1}{4}\sin {{50}^{\circ }}\left( -2\sin {{70}^{\circ }}\sin {{10}^{\circ }} \right)\\ &=-\dfrac{1}{4}\sin {{50}^{\circ }}\left( \cos \left( {{70}^{\circ }}+{{10}^{\circ }} \right)-\cos \left( {{70}^{\circ }}-{{10}^{\circ }} \right) \right)\\ &=-\dfrac{1}{4}\sin {{50}^{\circ }}\left( \cos {{80}^{\circ }}-\cos {{60}^{\circ }} \right)\\ &=-\dfrac{1}{4}\sin {{50}^{\circ }}\left( \cos {{80}^{\circ }}-\dfrac{1}{2} \right)\\ &=-\dfrac{1}{4}\cos {{80}^{\circ }}\sin {{50}^{\circ }}+\dfrac{1}{8}\sin {{50}^{\circ }}\\ &=-\dfrac{1}{8}\left( 2\cos {{80}^{\circ }}\sin {{50}^{\circ }} \right)+\dfrac{1}{8}\sin {{50}^{\circ }}\\ &=-\dfrac{1}{8}\left( \sin \left( {{80}^{\circ }}+{{50}^{\circ }} \right)-\sin \left( {{80}^{\circ }}-{{50}^{\circ }} \right) \right)+\dfrac{1}{8}\sin {{50}^{\circ }}\\ &=-\dfrac{1}{8}\left( \sin {{130}^{\circ }}-\sin {{30}^{\circ }} \right)+\dfrac{1}{8}\sin {{50}^{\circ }}\\ &=-\dfrac{1}{8}\left( \sin \left( 2({{90}^{\circ }})-{{50}^{\circ }} \right)-\dfrac{1}{2} \right)+\dfrac{1}{8}\sin {{50}^{\circ }}\\ &=-\dfrac{1}{8}\left( \sin {{50}^{\circ }}-\dfrac{1}{2} \right)+\dfrac{1}{8}\sin {{50}^{\circ }},\quad \therefore \,\,\,\sin \left( 2(90)-\theta \right)=\sin \theta \\ &=-\dfrac{1}{8}\sin {{50}^{\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\sin {{50}^{\circ }}\\ &=\dfrac{1}{16}=R.H.S.\end{align}$$

< Question 4