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what's the quickest way to find $$\sin^6 (x) +\cos^6 (x)$$ $\text{ if } \sin(x)+\cos(x) =a?$

This is what I did.

from the given expression, it's obvious $$\sin(x)\cos(x) = \frac{a^2 -1}{2}$$

And

$$\sin^6 (x) +\cos^6 (x)= (\sin^3(x)+\cos^3(x))^2 - 2(\sin^3(x)\cos^3(x))$$ This can further be expanded and solved

I was wondering if there's any easier, quicker method of solving this

Feng
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math and physics forever
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  • As a general technique when you know $s=x+y$ and $p=xy$ you can use the recursive sequence based on $x^2-sx+p=0$ to calculate any $x^n+y^n$. See here https://math.stackexchange.com/a/3897102/399263 – zwim Sep 06 '23 at 11:34

3 Answers3

6

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ we obtain \begin{align*} \sin^6x+\cos^6x&=(\sin^2 x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\ &=\sin^4x-\sin^2x\cos^2x+\cos^4x\\ &=(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x\\ &=(\sin^2x+\cos^2x)^2-3(\sin x\cos x)^2\\ &=1-3\left(\frac{a^2-1}2\right)^2=\frac{-3a^4+6a^2+1}4. \end{align*}

Feng
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3

I don't know if this is any quicker than your way. Both are pretty quick! $$1=(\cos^2 x+ \sin^2 x)^3=\cos^6 x + \sin ^6 x+3\cos^2 x \sin^2 x(\cos^2 x+ \sin^2 x)$$ $$=\cos^6 x +\sin ^6 x+3\left(\frac{a^2-1}{2}\right)^2$$ $$\cos^6 x +\sin ^6 x=1-3\left(\frac{a^2-1}{2}\right)^2$$

Blitzer
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$\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^x)^3=(\sin^2x+\cos^2x)(\sin^2x+\cos^2x)^2-3*\sin^2x\cos^2x)$

kmitov
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