precalc hw

Question

Use a half-angle formula to rewrite the given expression so that it involves the sum or difference of only constants and sine and cosine functions to the first power.
(sin 8x)2(cos 8x)2

Answer 1

(sin8x)²(cos8x)² = [(sin8x)(cos8x)]² because Aⁿ ⋅ Bⁿ = (AB)ⁿ

Let 8x=X. We get

(sin8x)²(cos8x)² = [(sinX)(cosX)]² = [2sinXcosX/2]²

= 1/4 (2sinXcosX)² = 1/4 (sin2X)².................... (1) [since sin2A= 2sinAcosA]

Recall that cos2X = 1- 2 (sin X)²

Therefore, 2(sinX)² = 1 - cos2X,

Or, (sinX)² = 1/2(1 - cos2X) and therefore (sin2x)² = 1/2(1 - cos4X);

Substituting this in (1) above, we get

R.H.S. = 1/4(sin2X)² = 1/4 x 1/2(1 - cos4X)

Again replace X by 8x and we get the final answer as

(sin8x)²(cos8x)² = 1/8(1-cos32x).