Question

sin 20 sin 40 sin 80 =1/8​

Answer 1

Answer:

The Actual Question is P.T.

$cos20^0\;cos40^0\;cos80^0=\frac{1}{8}$

Because

$sin20^0sin40^0sin80^0=0.2165063......\neq\frac{1}{8}$

Step-by-step explanation:$cos20^0\;cos40^0\;cos80^0\\\\=\frac{cos20^0}{2}(2cos40^0cos80^0)\\\\=\frac{cos20^0}{2}(cos120^0+cos40^0)\\\\=\frac{2cos20^0cos40^0}{4}+\frac{cos20^0cos120^0}{2}\\\\=\frac{cos20^0+cos60^0}{4}-\frac{cos20^0}{4}\\\\=\frac{cos60^0}{4}\\\\=\frac{1}{8}$

HOPE IT HELPS

Answer 2

Step-by-step explanation:

We know formula

sin theta. sin(60-theta).sin(60+theta) = sin3theta /4

sin 20 . sin 40 . sin 80. = sin3(20)/4

= √3/2/4

= √3/8

Hope it helps uh!