Question

cos pi/9*cos2pi/9*cos3pi/9*cos4pi/9=1/16

Answer 1

$\textbf{Given:}$

$\mathsf{cos\dfrac{\pi}{9}{\times}cos\dfrac{2\pi}{9}{\times}cos\dfrac{3\pi}{9}{\times}cos\dfrac{4\pi}{9}}$

$\textbf{To prove:}$

$\mathsf{cos\dfrac{\pi}{9}{\times}cos\dfrac{2\pi}{9}{\times}cos\dfrac{3\pi}{9}{\times}cos\dfrac{4\pi}{9}=\dfrac{1}{16}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{cos\dfrac{\pi}{9}{\times}cos\dfrac{2\pi}{9}{\times}cos\dfrac{3\pi}{9}{\times}cos\dfrac{4\pi}{9}}$

$\mathsf{=cos\,20^\circ{\times}cos\,40^\circ{\times}cos\,60^\circ{\times}cos\,80^\circ}$

$\mathsf{=cos\,20^\circ{\times}cos\,40^\circ{\times}\dfrac{1}{2}{\times}cos\,80^\circ}$

$\mathsf{=\dfrac{1}{2}[cos\,20^\circ{\times}cos\,40^\circ{\times}cos\,80^\circ]}$

$\mathsf{=\dfrac{1}{2}[cos(60^\circ-20^\circ){\times}cos\,20^\circ{\times}cos(60^\circ+20^\circ)]}$

$\mathsf{Using\;the\;identity,}$

$\boxed{\mathsf{cos(60-A)\;cosA\;cos(60+A)=\frac{1}{4}cos\,3A}}$

$\mathsf{=\dfrac{1}{2}{\times}\dfrac{1}{4}cos\,3(20^\circ)}$

$\mathsf{=\dfrac{1}{2}{\times}\dfrac{1}{4}cos\,60^\circ}$

$\mathsf{=\dfrac{1}{2}{\times}\dfrac{1}{4}{\times}\dfrac{1}{2}}$

$\mathsf{=\dfrac{1}{16}}$

$\implies\boxed{\mathsf{cos\dfrac{\pi}{9}{\times}cos\dfrac{2\pi}{9}{\times}cos\dfrac{3\pi}{9}{\times}cos\dfrac{4\pi}{9}=\dfrac{1}{16}}}$

$\textbf{Find more:}$

PROVE THAT:

cos 12 cos 24 cos 36 cos 48 Cos 60 cos 72 cos 84 equal = 1 / 128​

https://brainly.in/question/9849513

Prove that : 8 sin 20 sin 40 sin 80 = (3)^1/2

https://brainly.in/question/2394832