Question

show that cos 20 . cos 40 .cos 60 .cos 80 = 1/16​

Answer 1

your question is solved above

Answer 2

$\mathcal\red{\underline{Question:-}}$

Show that cos 20° cos 40° cos 60° cos 80° = $\frac{1}{16}$

$\mathcal\red{\underline{Solution:-}}$

cos 20° cos 40° cos 60° cos 80°

= cos 60° ( cos 20° cos 60° ) cos 80°

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

{ °•° cos 60° = $\frac{1}{2}$}

= $\frac{1}{4}$ [{ cos ( 40° + 20° ) + cos ( 40° - 20° )} cos 80° ]

{ °•° 2 cos A cos B = cos ( A+B ) cos (A-B) ]

= $\frac{1}{4}$ [( cos 60° + cos 20° ) cos 80° ]

= $\frac{1}{4}$ [( 1/2 + cos 20° ) cos 80° ]

= $\frac{1}{8}$ ( cos 80° + 2 cos 80° cos 20° )

= $\frac{1}{8}$[ cos 80° + { cos ( 80° + 20° ) cos ( 80° - 20° )}]

{ °•° 2 cos A cos B = cos ( A+B ) cos ( A-B )]

= $\frac{1}{8}$ [ cos 80° + cos 100° + cos 60° ]

= $\frac{1}{8}$[ cos 80° + cos ( 180° - 100° ) + cos 60° ]

= $\frac{1}{8}$ [ cos 80° - cos 80° + cos 60° ]

= $\frac{1}{8} \: {[}\:\frac{1}{2}$ ]

= $\frac{1}{8}×\frac{1}{2}$

= $\frac{1}{16}$

$\mathcal\red{Thank\:You}$❤