# Prove that cos^6A+sin^6A = 1/4 (1 + 3cos^2A)

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### 2 Answers

You should notice that you may get `sin^6 A + cos^6 A` if you use the formula `(sin^3A + cos^3A)^2 = sin^6 A + cos^6 + 2(sinA*cosA)^3` `sin^3A + cos^3A = (sin A + cos A)(sin^2A - sinA*cosA + cos^2A)`

You should use fundamental formula of trigonometry such that:

`sin^2A + cos^2A = 1`

`sin^3A + cos^3A = (sin A + cos A)(1 - sinA*cosA)`

`(sin^3A + cos^3A)^2 = (sin A + cos A)^2(1 - sinA*cosA)^2`

`(sin^3A + cos^3A)^2 = (1 + 2sinA*cosA)(1 - sinA*cosA)(1 - sinA*cosA)`

`(sin^3A + cos^3A)^2 = (1 + sinA*cosA - 2(sinA*cosA)^2)(1 - sinA*cosA)`

`(sin^3A + cos^3A)^2 = (1 - sinA*cosA + sinA*cosA - (sinA*cosA)^2 - 2(sinA*cosA)^2 + 2(sinA*cosA)^3)`

`(sin^3A + cos^3A)^2 = (1- 3(sinA*cosA)^2 + 2(sinA*cosA)^3)`

`4(1- 3(sinA*cosA)^2 + 2(sinA*cosA)^3) != (1 + 3cos^2A)`

**Notice that the last line `4(1- 3(sinA*cosA)^2 + 2(sinA*cosA)^3) != (1 + 3cos^2A)` proves that the given expression is not an identity.**

**Sources:**

We use the identity (a+b)^3 = a^3 + b^3 +3ab(a+b)

(sinA)^6 +(cosA)^6 = [(sinA)^2 +(cosA)^2]^3 -3[(sinA)^2] [ (cosA)^2] [(sinA)^2 +(cosA)^2]

now put

(sinA)^2 +(cosA)^2 =1

and use 2(sinA)^2=1-cos2A

2(cosA)^2=1+cos2A