Question

if sin theta + sin square theta + sin cube theta is equals to 1 prove that cos raised for 6 theta minus 4 cos raised for 4 theta + 8 cos square theta is equals to 4

Answer 1

Answer:

$cos^6\theta-4cos^4\theta+8cos^2\theta=4$

Step-by-step explanation:

Given:

$sin\theta+sin^2\theta+sin^3\theta=1$

$sin\theta+sin^3\theta=1-sin^2\theta\\\\sin\theta(1+sin^2\theta)=cos^2\theta\\\\sin\theta(1+1-cos^2\theta)=cos^2\theta\\\\sin\theta(2-cos^2\theta)=cos^2\theta$

squaring on both sides we get

$sin^2\theta(2-cos^2\theta)^2=cos^4\theta\\\\(1-cos^2\theta)(4+cos^4\theta-4cos^2\theta)=cos^4\theta\\\\4+cos^4\theta-4cos^2\theta-4cos^2\theta-cos^6\theta+4cos^4\theta=cos^4\theta$

Rearranging terms we get

$cos^6\theta-4cos^4\theta+8cos^2\theta=4$

Answer 2

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