Question

Prove that, ${cos20}^{\circ }cos{40}^{\circ }{cos60}^{\circ }cos{80}^o=\frac{1}{16}$

Answer 1

We have,

$L.H.S.$

$\cos {20}^o\cos {40}^o\cos {60}^o\cos {80}^o$

$\Rightarrow \cos {20}^o\cos {40}^o\frac{1}{2}\cos {80}^o\therefore \cos {60}^o=\frac{1}{2}$

$\Rightarrow \frac{1}{2}\cos {20}^o\cos {40}^o\cos {80}^o$

Multiply and dived by 2 and we get,

$\Rightarrow \frac{1}{4}\left(2\cos {20}^o\cos {40}^o\right)\cos {80}^o$

$\Rightarrow \frac{1}{4}(\cos ({20}^o+{40}^o)+\cos ({20}^o-{40}^o))\cos {80}^o$

$\Rightarrow \frac{1}{4}(\cos {60}^o+\cos {20}^o)\cos {80}^o$

$\Rightarrow \frac{1}{4}(\frac{1}{2}+\cos {20}^o)\cos {80}^o$

$\Rightarrow \frac{1}{8}(1+2\cos {20}^o)\cos {80}^o$

$\Rightarrow \frac{1}{8}(\cos {80}^o+2\cos {20}^o\cos {80}^o)$

$\Rightarrow \frac{1}{8}(\cos {80}^o+(\cos ({20}^o+{80}^o)+\cos ({20}^o-{80}^o)))$

$\Rightarrow \frac{1}{8}(\cos {80}^o+(\cos {100}^o+\cos (-{60}^o)))$

$\Rightarrow \frac{1}{8}(\cos {80}^o+(\cos {100}^o+\cos {60}^o))$

$\Rightarrow \frac{1}{8}(\cos {80}^o+(\cos {100}^o+\frac{1}{2}))$

$\Rightarrow \frac{1}{8}(\cos {80}^o+\cos {100}^o+\frac{1}{2})$

$\Rightarrow \frac{1}{8}(\cos {80}^o+\cos {100}^o)+\frac{1}{16}$

$\Rightarrow \frac{1}{8}\left(2\cos \left(\frac{{80}^o+{100}^o}{2}\right)\cos \left(\frac{{80}^o-{100}^o}{2}\right)\right)+\frac{1}{16}$

$\Rightarrow \frac{1}{8}\left(2\cos {90}^o\cos {20}^o\right)+\frac{1}{16}$

$\Rightarrow \frac{1}{8}(2\times 0\cos {20}^o)+\frac{1}{16}$

$\Rightarrow \frac{1}{16}$

R.H.S.