Question 4 Exercise 8.3

Solutions of Question 4 of Exercise 8.3 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

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

Solution. \begin{align*} LHS &= \cos 80^\circ \cos 60^\circ \cos 40^\circ \cos 20^\circ \\ &= \cos 80^\circ \left(\frac{1}{2}\right) \cos 40^\circ \cos 20^\circ \\ &= \frac{1}{2} \left( \cos 80^\circ \cos 40^\circ \right) \cos 20^\circ \\ &= \frac{1}{4} \left(2 \cos 80^\circ \cos 40^\circ \right) \cos 20^\circ \\ &= \frac{1}{4} \left( \cos (80^\circ + 40^\circ) + \cos (80^\circ - 40^\circ) \right) \cos 20^\circ \\ &=\frac{1}{4}\left( \cos {{120}^{\circ }}+\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }} \\ & =\frac{1}{4}\left( -\frac{1}{2}+\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }} \\ & =-\frac{1}{8}\cos {{20}^{\circ }}+\frac{1}{4}\cos {{40}^{\circ }}\cos {{20}^{\circ }} \\ & =-\frac{1}{8} \cos {{20}^{\circ }} +\frac{1}{8}\left( 2\cos {{40}^{\circ }}\cos {{20}^{\circ }} \right) \\ & =-\frac{1}{8}\cos {{20}^{\circ }}+\frac{1}{8}\left( \cos (40+20)+\cos (40-20) \right)\\ & =-\frac{1}{8}\cos {{20}^{\circ }}+\frac{1}{8}\left( \cos 60+\cos 20 \right)\\ & =-\frac{1}{8}\cos {{20}^{\circ }}+\frac{1}{8}\left( \frac{1}{2}+\cos 20 \right)\\ & =-\frac{1}{8}\cos {{20}^{\circ }}+\frac{1}{16}+\frac{1}{8}\cos {{20}^{\circ }} \\ & =\frac{1}{16} = RHS \end{align*}

Prove that: $\sin 70^{\circ} \sin 50^{\circ} \sin 30^{\circ} \sin 10^{\circ}=\dfrac{1}{16}$

Solution.

Do youself as above.

Prove that: $\sin \dfrac{\pi}{9} \sin \dfrac{2\pi}{9} \sin \dfrac{3\pi}{9} \sin \dfrac{4\pi}{9}=\dfrac{3}{16}$

Solution.

\begin{align*} LHS &= =\sin \frac{\pi }{9}\sin \frac{2\pi }{9}\sin 3\frac{\pi }{9}\sin \frac{4\pi }{9} \\ &= \sin \frac{{{180}^{\circ }}}{9}\sin \frac{2({{180}^{\circ }})}{9}\sin 3\frac{({{180}^{\circ }})}{9}\sin \frac{4({{180}^{\circ }})}{9} \quad \because \quad \pi ={{180}^{\circ }} \\ & =\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }} \\ &=\sin {{20}^{\circ }}\sin {{40}^{\circ }}\frac{\sqrt{3}}{2}\sin {{80}^{\circ }} \\ &=\frac{\sqrt{3}}{2}\sin {{80}^{\circ }}\sin {{40}^{\circ }}\sin {{20}^{\circ }} \\ &=-\frac{\sqrt{3}}{4}\left( -2\sin {{80}^{\circ }}\sin {{40}^{\circ }} \right)\sin {{20}^{\circ }} \\ &=-\frac{\sqrt{3}}{4}\left( \cos (80+40)-\cos (80-40) \right)\sin {{20}^{\circ }} \\ &=-\frac{\sqrt{3}}{4}\left( \cos {{120}^{\circ }}-\cos {{40}^{\circ }} \right)\sin {{20}^{\circ }} \\ &=-\frac{\sqrt{3}}{4}\left( -\frac{1}{2}-\cos {{40}^{\circ }} \right)\sin {{20}^{\circ }} \\ &=\frac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\frac{\sqrt{3}}{4}\cos {{40}^{\circ }}\sin {{20}^{\circ }}\\ &=\frac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\frac{\sqrt{3}}{8}\left( 2\cos {{40}^{\circ }}\sin {{20}^{\circ }} \right) \\ &=\frac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\frac{\sqrt{3}}{8}\left( \sin (40+20)-\sin (40-20) \right) \\ &=\frac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\frac{\sqrt{3}}{8}\left( \sin {{60}^{\circ }}-\sin {{20}^{\circ }} \right) \\ &=\frac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\frac{\sqrt{3}}{8}\left( \frac{\sqrt{3}}{2}-\sin {{20}^{\circ }} \right) \\ &=\frac{\sqrt{3}}{8}\sin {{20}^{\circ }}+\frac{3}{16}-\frac{\sqrt{3}}{8}\sin {{20}^{\circ }} \\ &=\frac{3}{16} = RHS \end{align*}

< 3(xi, xii & xiii)