9
$\begingroup$

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 ?

$\endgroup$
14
$\begingroup$

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 $$\cos^6 \theta - \sin^6 \theta = \bigg({1 + x \over 2}\bigg)^3 - \bigg({1 - x \over 2}\bigg)^3$$ $$ = {3 \over 4} x + {1 \over 4} x^3$$

$\endgroup$
  • $\begingroup$ 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 . $\endgroup$ – A Googler Aug 11 '13 at 5:13
  • $\begingroup$ @AGoogler you saying you never knew half angle equations? :) $\endgroup$ – Kaster Aug 11 '13 at 5:50
  • $\begingroup$ @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 . $\endgroup$ – A Googler Aug 11 '13 at 7:28
  • $\begingroup$ @AGoogler OK, what I meant is $\cos 2\theta = x$, so $\cos^2 \theta = \frac {1+\cos 2\theta}2 = \frac {1+x}2$; $\sin^2 \theta = \frac {1-\cos 2\theta}2 = \frac {1-x}2$ $\endgroup$ – Kaster Aug 11 '13 at 7:40
5
$\begingroup$

$$ \cos^6\theta-\sin^6\theta = \left ( \cos^2 \theta\right )^3 - \left (\sin^2 \theta \right )^3 = \\ = \left( \cos^2 \theta - \sin^2 \theta\right ) \left(\cos^4 \theta + \sin^2\theta \cos^2\theta + \sin^4\theta \right ) = \\ = x \left ( \cos^4 \theta - 2\cos^2\theta\sin^2\theta + \sin^4 \theta + 3 \cos^2\theta\sin^2\theta\right ) = \\ = x \left( \left(\cos^2\theta - \sin^2 \theta \right )^2 + \frac 34 \sin^22\theta\right ) = x \left( x^2 + \frac 34 \left( 1-\cos^2 2\theta\right )\right ) = \\ = x \left(x^2 + \frac 34 (1-x^2) \right ) = \frac x4 \left(x^2+3 \right ) $$

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – A Googler Aug 11 '13 at 7:36
  • $\begingroup$ @AGoogler, I don't really know actually. Never thought about it. It's just the way I was taught in high school. $\endgroup$ – Kaster Aug 11 '13 at 7:44
4
$\begingroup$

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 $ab$.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

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

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

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

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.