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Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .

Here is what I did: $\cos^4 \theta −\sin^4 \theta = x$.

($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$

Thus ($\cos^2 \theta −\sin^2 \theta)=x$ , so $\cos 2\theta=x$ .

Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $

So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?

A Googler
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5 Answers5

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You have $$\cos^2 \theta + \sin^2 \theta = 1$$ $$\cos^2 \theta - \sin^2 \theta = x$$ Adding and subtracting the two equations gives $$\cos^2 \theta = {1 + x \over 2}$$ $$\sin^2 \theta = {1 - x \over 2}$$ Substituting you have $$\cos^6 \theta - \sin^6 \theta = \bigg({1 + x \over 2}\bigg)^3 - \bigg({1 - x \over 2}\bigg)^3$$ $$ = {3 \over 4} x + {1 \over 4} x^3$$

Zarrax
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  • Very elegant solution. Even though I had those two equations , I never thought of solving them simultaneously to obtain cos^2 theta & sin^2 theta . Thanks a lot . – A Googler Aug 11 '13 at 05:13
  • @AGoogler you saying you never knew half angle equations? :) – Kaster Aug 11 '13 at 05:50
  • @Kaster Nope I do not mean that. I'm saying that even though I obtained the two equations (cos^2 - sin^2 =x and cos^2+sin^2=1 ) , I did not think of solving them to get cos^2 and sin^2 . – A Googler Aug 11 '13 at 07:28
  • @AGoogler OK, what I meant is $\cos 2\theta = x$, so $\cos^2 \theta = \frac {1+\cos 2\theta}2 = \frac {1+x}2$; $\sin^2 \theta = \frac {1-\cos 2\theta}2 = \frac {1-x}2$ – Kaster Aug 11 '13 at 07:40
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$$ \cos^6\theta-\sin^6\theta = \left ( \cos^2 \theta\right )^3 - \left (\sin^2 \theta \right )^3 = \\ = \left( \cos^2 \theta - \sin^2 \theta\right ) \left(\cos^4 \theta + \sin^2\theta \cos^2\theta + \sin^4\theta \right ) = \\ = x \left ( \cos^4 \theta - 2\cos^2\theta\sin^2\theta + \sin^4 \theta + 3 \cos^2\theta\sin^2\theta\right ) = \\ = x \left( \left(\cos^2\theta - \sin^2 \theta \right )^2 + \frac 34 \sin^22\theta\right ) = x \left( x^2 + \frac 34 \left( 1-\cos^2 2\theta\right )\right ) = \\ = x \left(x^2 + \frac 34 (1-x^2) \right ) = \frac x4 \left(x^2+3 \right ) $$

Kaster
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  • I like the way you substituted ( -2cos^2sin^2 + 3cos^2sin^2 ) for (+cos^2sin^2) . How do you guys manipulate equations like this with ease ? – A Googler Aug 11 '13 at 07:36
  • @AGoogler, I don't really know actually. Never thought about it. It's just the way I was taught in high school. – Kaster Aug 11 '13 at 07:44
4

For typing ease, let $a=\cos^2\theta$ and $b=\sin^2\theta$. Thus $a+b=1$. We are told that $a^2-b^2=x$, or equivalently that $a-b=x$.

We want $a^3-b^3$. It will be enough to find $a^2+ab+b^2$. Note the identity $$4(a^2+ab+b^2)=3(a+b)^2+(a-b)^2.$$

Remark: For other problems of a similar character, it might be preferable to extract $ab$ from the identity $4ab=(a+b)^2-(a-b)^2$, and use standard techniques for expressing symmetric functions of $a$ and $b$ in terms of the elementary symmetric functions $a+b$ and $ab$.

André Nicolas
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It would be easier if you decompose $\cos^6 \theta -\sin^6 \theta $ into $(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta +\cos^2 \theta \sin^2 \theta+\sin^4 \theta)$

Cheong
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As already found $\cos^2\theta-\sin^2\theta=x,\implies \cos2\theta=x$

Using $a^3-b^3=(a-b)^3+3ab(a-b),$

$$\cos^6\theta-\sin^6\theta=(\cos^2\theta)^3-(\sin^2\theta)^3$$

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

$$=x^3+3x\cos^2\theta\sin^2\theta$$

$$=x^3+\frac{3x}4\sin^22\theta\text{ as }\sin2A=2\sin A\cos A$$

$$=x^3+\frac{3x}4(1-\cos^22\theta)$$

$$=x^3+\frac{3x}4(1-x^2)=\frac{4x^3+3x-3x^3}4=\frac{x(x^2+3)}4$$

lab bhattacharjee
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