Verified answer

(a) Consider (cos θ + i sin θ)^4; we will evaluate it in two different ways.

By De Moivre's Theorem, this equals cos(4θ) + i sin(4θ).

On the other hand, using the Binomial Theorem (or expanding it out directly), we obtain

cos^4(θ) + 4 cos^3(θ) * i sin θ + 6 cos^2(θ) (i sin θ)^2 + 4 cos θ (i sin θ)^3 + (i sin θ)^4

= cos^4(θ) + 4i cos^3(θ) sin θ - 6 cos^2(θ) sin^2(θ) - 4i cos θ sin^3(θ) + sin^4(θ)

= [cos^4(θ) - 6 cos^2(θ) sin^2(θ) + sin^4(θ)] + i [4 cos^3(θ) sin θ - 4 cos θ sin^3(θ)]

Equating real parts yields

cos(4θ) = cos^4(θ) - 6 cos^2(θ) sin^2(θ) + sin^4(θ)

............= (1 - sin^2(θ))^2 - 6 (1 - sin^2(θ)) sin^2(θ) + sin^4(θ)

............= 1 - 8 sin^2(θ) + 8 sin^4(θ).

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(b) Repeatedly apply double angle identities.

cos(4θ) = cos(2 * 2θ)

............= 1 - 2 sin^2(2θ)

............= 1 - 2 (2 sin θ cos θ)^2

............= 1 - 8 sin^2(θ) cos^2(θ)

............= 1 - 8 sin^2(θ) (1 - sin^2(θ))

............= 1 - 8 sin^2(θ) + 8 sin^4(θ).

I hope this helps!

2sin^2?+3cos?=0 2(a million-cos^2?)+3cos?=0 cos?=x 2(a million-x^2)+3x=0 -2x^2+3x+2=0 -2x^2+4x-x+2=0 -2x(x-2)-(x-2)=0 (-2x-a million)(x-2)=0 2x=a million x=2 it fairly is cos?=0.5 cos?=2 the different answer is misguided as a results of fact cos? would nicely be between -a million and a million! cos?=0.5 ?=60 tiers or PI/3