Question

Prove that: 16.cos24.cos48.cos96.cos192=1

Answer 1

Answer:

Using the identity 2cosAcosB=cos(A+B)+cos(A−B) gives

2cos24∘cos96∘=cos120∘+cos72∘=sin18∘−12,

2cos48∘cos192∘=cos240∘+cos144∘=−cos36∘−12.

Therefore

16cos24∘cos48∘cos96∘cos192∘=(1+2cos36∘)(1–2sin18∘)

=1+2cos36∘−2sin18∘−4cos36∘sin18∘…(1)

From

sin72∘⋅sin36∘=(2sin36∘cos36∘)(2sin18∘cos18∘)

and

sin72∘=cos18∘

we get 4sin18∘cos36∘=1…(2)

Using the identity 2cosAsinB=sin(A+B)−sin(A−B) with A=36∘ and B=18∘ gives

2cos36∘sin18∘=sin54∘−sin18∘=cos36∘−sin18∘…(3)

Putting together eqns. (1), (2), (3), we get

16cos24∘cos48∘cos96∘cos192∘=1. ■

Step-by-step explanation: