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egoroff_w [7]
3 years ago
11

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

Mathematics
1 answer:
asambeis [7]3 years ago
8 0

Given:

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

To prove:

The given statement.

Proof:

We have,

\cos^4 \alpha+\sin^4\alpha=\dfrac{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=\dfrac{1}{4}(3+\cos 4 \alpha)

RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]

RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]        [\because (a-b)^2=a^2-2ab+b^2]

RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]

RHS=\dfrac{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.

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