Expert: Sherman D. - 5/24/2008

Question
Prove the following identities
Cos2θ≡Cos^4θ-Sin^4θ
Cos3θ≡4Cos^3θ-3Cosθ
Sin3θ≡3Sinθ-4Cos^3θ
Cos4θ≡8Cos^4θ-8Cos^2θ+1
Sin4θ≡4(SinθCos^3θ-Sin^3θCosθ)

Answer
cos(2A) = cos(A)^4 - sin(A)^4

taking the right side and factoring it.

cos(A)^4 - sin(A)^4 =
(cos(A)^2 + sin(A)^2)(cos(A)^2 - sin(A)^2) =
1(cos(2A))

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cos(3A) = 4cos(A)^3 - 3cos(A)

cos(3A) =
cos(2A + A) =
cos(2A)cos(A) - sin(2A)sin(A)
(2cos(A)^2 - 1)cos(A) - 2cos(A)sin(A)^2
2cos(A)^3 - cos(A) - 2cos(A)(1 - cos(A)^2)
2cos(A)^3 - cos(A) - 2cos(A) + 2cos(A)^3
4cos(A)^3 - 3cos(A)

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sin(3A) = 3sin(A) - 4cos(A)^3

sin(2A + A) = 3sin(A) - 4cos(A)^3

sin(2A)cos(A) + sin(A)cos(2A)
2sin(A)cos(A)^2 + sin(A)(1 - 2sin(A)^2)
2sin(A)(1 - sin(A)^2) + sin(A) - 2sin(A)^3
2sin(A) - 2sin(A)^3 + sin(A) - 2sin(A)^3
3sin(A) - 4sin(A)^3

i've checked your problem, and sin(3A) doesn't equal 3sin(A) - 4cos(A)^3

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cos(4A) = 8cos(A)^4 - 8cos(A)^2 + 1

cos(4A) = cos(2(2A))

cos(2(2A)) =
2cos(2A)^2 - 1 =
2(2cos(A)^2 - 1)^2 - 1 =
2(4cos(A)^4 - 4cos(A)^2 + 1) - 1 =
8cos(A)^4 - 8cos(A)^2 + 2 - 1 =
8cos(A)^4 - 8cos(A)^2 + 1

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sin(4A) = 4(sin(A)cos(A)^3 - sin(A)^3cos(A))

sin(4A) = sin(2(2A))

2sin(2A)cos(2A) =
4(sin(A)cos(A))(cos(A)^2 - sin(A)^2) =
4(sin(A)cos(A)^3 - sin(A)^3cos(A))

you can find more info at http://math2.org/math/trig/identities.htm