Question

prove that cos20.cos40.cos60.cos80=1/16​

Answer 1

Answer:

This is the full answer.....

Answer 2

Consider the given equation

Cos 20° · Cos40° · Cos60° . Cos80° = 1/16

LHS = Cos 20° · Cos40° · Cos60° . Cos80°

We know that Cos60° = 1/2

LHS = Cos 20° · Cos40° · 1/2 . Cos80°

Multiply and divide the equation by 2

LHS = 1/4 (2. Cos 20° · Cos40° · Cos80°)

We know the formula 2 cosa cosb= cos(a+b) + cos(a-b)

LHS = 1/4 [Cos(20+80)+ Cos(20-80)] . Cos40

LHS = 1/4 [Cos(-60)+ Cos(100)] Cos40

LHS = 1/4 [1/2 + cos100] Cos40

LHS = 1/8 Cos40+ 1/4 (Cos40 . Cos100)

Multiply and divide the equation by 2

LHS = 2/2 (1/8 Cos 40) + 1/8(2 Cos40 Cos100)

We know the formula

2cosa cosb= cos(a+b) cos(a-b)

LHS = 1/8 Cos40+ 1/8 [Cos 140 + Cos (-60)]

LHS = 1/8 Cos 40+ 1/8 Cos 140 + 1/16

Since Cos 60= 1/2

LHS = 1/8 (Cos 40 + Cos 140) + 1/16

LHS = 1/8 [2 Cos 90 Cos (-50)] + 1/16

LHS = Cos 90

Cos 90 = 0

LHS = 1/16

LHS = RHS