Question

If sinθ+sin²θ+sin³θ=1, prove that cos⁶θ–4cos⁴θ+8cos²θ=4.

Answer 1

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Answer 2

We have:

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

To prove that: $\cos^6 \theta$ - 4$\cos^4 \theta$ + 8$\cos^2 \theta$ = 4.

Solution:

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

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

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

Using the trigonometric identity:

$\cos^2 \theta$ = 1 - $\sin^2 \theta$

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

⇒ $\sin \theta$ (2 - $\cos^2 \theta$) =

Squaring both sides, we get

 $[\sin \theta(2-\cos^2 \theta)]^2$ = $(\cos^2 \theta)^2$

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

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

⇒ 4 + 4$\cos^4 \theta$ - 4$\cos^2 \theta$ - 4

⇒ 4 + 4$\cos^4 \theta$ - 4$\cos^2 \theta$ - 4

⇒ 4 + 4$\cos^4 \theta$ - 4$\cos^2 \theta$ - 4

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

Thus, if $\sin \theta$ +  $\sin^2 \theta$ +  $\sin^3 \theta$ = 1, then $\cos^6 \theta$ - 4$\cos^4 \theta$ + 8$\cos^2 \theta$ = 4. proved.