Question

Prove that $\cos 20°\times \cos 40°\times \cos 80°=\frac{1}{8}$

Answer 1

$L.H.S.=\cos 20°\times \cos 40°\times \cos 80°$

Multiplying and dividing by $2$

$=\frac{1}{2}(2\cos 20°\times \cos 40°\times \cos 80°)$

We know that, $\left(2\cos a\cos b=\cos \left(a+b\right)+\cos \left(a-b\right)\right)$

$=\frac{1}{2}\left[\cos \left(20°+80°\right)+\cos \left(20°-80°\right)\right]\cos 40°$

$=\frac{1}{2}\left[\cos \left(100°\right)+\cos \left(-60°\right)\right].\cos 40°$

$=\frac{1}{2}\left[\cos \left(100°\right)+\frac{1}{2}\right].\cos 40°$

$=\frac{1}{4}\cos 40°+\frac{1}{2}(\cos 40°.\cos 100°)$

Multiplying and dividing by $2$.

$=\frac{2}{2}\times \frac{1}{4}\cos 40°+\frac{1}{2\times 2}(2\cos 40°.\cos 100°)$

$=\frac{1}{4}\cos 40°+\frac{1}{4}\left[(\cos \left(40°+100°\right)+\cos \left(40°-100°\right)\right]$

$=\frac{1}{4}\cos 40°+\frac{1}{4}(\cos 140°+\cos \left(-60°\right))$…………………..(As, $\left(2\cos a\cos b=\cos \left(a+b\right)+\cos \left(a-b\right)\right)$)

$=\frac{1}{4}\cos 40°+\frac{1}{4}\cos 140°+\frac{1}{4}\times \frac{1}{2}$…………………….(As, $\cos \left(-60°\right)=\frac{1}{2}$)

$=\frac{1}{4}\left(\cos 40°+\cos 140°\right)+\frac{1}{8}$

We know that $\left(2\cos a\cos b=\cos \left(a+b\right)+\cos \left(a-b\right)\right)$,

So, $90°+50°=140°,90°-50°=40°$

$=\frac{1}{4}\left(2\cos 90°.\cos \left(-50°\right)\right)+\frac{1}{8}$

$=\frac{1}{4}\left(2\left(0\right).\cos \left(-50°\right)\right)+\frac{1}{8}$…………………………($\cos 90°=0$)

$=0+\frac{1}{8}$

$=\frac{1}{8}$.

Therefore, it is proved that $\cos 20°\times \cos 40°\times \cos 80°=\frac{1}{8}$.