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.
Question 5(i)
Prove that $$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$
Solution
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}
Question 5(ii)
Prove the identity $$\sin \dfrac{\pi }{9}\sin \dfrac{2\pi }{9}\sin \dfrac{\pi }{3}\sin \dfrac{4\pi }{9}=\dfrac{3}{16}.$$
Solution
We have an identities: $$2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right) \ldots (1)$$ and $$2\cos \alpha \sin \beta =\sin \left( \alpha +\beta \right)-\cos \left( \alpha -\beta \right) \ldots (2).$$ Now \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^\circ+40^\circ)-\cos (80^\circ-40^\circ) \right)\sin {{20}^{\circ }} \quad \text{by using (1)}\\ &=-\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^\circ+20^\circ)-\sin(40^\circ-20^\circ) \right) \quad \text{by using (2)}\\ &=\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}
Question 5(iii)
Prove the identity $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$.
Solution
We have an identities: $$2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right) \ldots (1)$$ and $$2\cos \alpha \sin \beta =\sin \left( \alpha +\beta \right)-\cos \left( \alpha -\beta \right) \ldots (2).$$ Now \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) \quad \text{by using (1)}\\ &=-\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 }} \quad \text{by using (2)}\\ &=-\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}
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