Question

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)
Prove:
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Answer 1

Given:

`\cos^4 \alpha+\sin^4\alpha=(1)/(4)(3+\cos 4 \alpha)`

To prove:

The given statement.

Proof:

We have,

`\cos^4 \alpha+\sin^4\alpha=(1)/(4)(3+\cos 4 \alpha)`

`LHS=\cos^4 \alpha+\sin^4\alpha`

`LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2`

`LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha`
`[\because a^2+b^2=(a+b)^2-2ab]`

`LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha`
`[\because \cos^2 \alpha+\sin^2\alpha=1]`

`LHS=1-2\cos^2 \alpha+2\cos^4 \alpha`

Now,

`RHS=(1)/(4)(3+\cos 4 \alpha)`

`RHS=(1)/(4)[3+(2\cos^2 2\alpha-1)]`
`[\because \cos 2\theta=2\cos^2\theta -1]`

`RHS=(1)/(4)[2+2\cos^2 2\alpha]`

`RHS=(1)/(4)[2+2(2\cos^2 \alpha-1)^2]`
`[\because \cos 2\theta=2\cos^2\theta -1]`

`RHS=(1)/(4)[2+2(4\cos^4 \alpha-4\cos \alpha+1)]`
`[\because (a-b)^2=a^2-2ab+b^2]`

`RHS=(1)/(4)[2+8\cos^4 \alpha-8\cos \alpha+2]`

`RHS=(1)/(4)[4+8\cos^4 \alpha-8\cos \alpha]`

`RHS=1+2\cos^4 \alpha-2\cos \alpha`

`RHS=1-2\cos^2 \alpha+2\cos^4 \alpha`

`LHS=RHS`

Hence proved.