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Problem : If $\sin\theta +\sin^2\theta +\sin^3\theta=1$ Then prove $\cos^6\theta -4\cos^4\theta +8\cos^2\theta =4$

My working :

As $\sin\theta +\sin^2\theta +\sin^3\theta=1 \Rightarrow \sin\theta +\sin^3\theta = \cos^2\theta$

Now the given equation : $\cos^6\theta -4\cos^4\theta +8\cos^2\theta$ can be written as

$(\sin\theta +\sin^3\theta)^3-4(\sin\theta+\sin^3\theta )^2+8(\sin\theta +\sin^3\theta)$

= $\sin^3\theta +\sin^6\theta +3\sin^5\theta +3\sin^7\theta -4\sin^2\theta -4\sin^6\theta -8\sin^4\theta + 8\sin\theta + 8\sin^3\theta$

But I think this is not the right way of doing this...Please suggest other alternative.. Thanks...

Jonas Kgomo
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Sachin
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  • Please check your expansion. There should be a $\sin^9 \theta$ involved (I believe your second term is wrong). – Calvin Lin Jun 03 '13 at 15:08

2 Answers2

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Let $x = \sin \theta$. We are given that $x^3 + x^2 + x - 1 = 0$.

We want to show that $(1-x^2)^3 -4(1-x^2)^2 +8(1-x^2) = 4$. Expanding and comparing terms, this is equivalent to

$$ x^6 + x^4 + 3x^2 - 1 = 0. $$

This is true because $$x^6 + x^4 + 3x^2 - 1 = (x^3 + x^2 + x -1 ) ( x^3 - x^2 + x +1).$$

Calvin Lin
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  • That was great @Calvin...genius...Thanks a lot.... – Sachin Jun 03 '13 at 15:23
  • @sultan questions like this generally follow the above method, unless the original polynomial equation has some special form which allows you to guess at the roots. – Calvin Lin Jun 03 '13 at 15:26
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Clearly, we need to eliminate $\sin\theta$ and let us derive the proposition instead of verifying it.

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

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

$\implies \sin\theta(2 -\cos^2\theta) =\cos^2\theta$ using $\cos^2\theta=1-\sin^2\theta$

Squaring we get, $\sin^2\theta(2 -\cos^2\theta)^2 =\cos^4\theta$

or, $(1-\cos^2\theta)(2 -\cos^2\theta)^2 =\cos^4\theta$

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

lab bhattacharjee
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