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How can I show that this equation is true for all $x \in \mathbb{R}$? $$\sin^6x + \cos^6x = 1 - 3\sin^2x \ \cos^2x$$

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2 Answers 2

up vote 2 down vote accepted

Hint: $\cos^2\varphi+\sin^2\varphi=1$.

Full proof

$$ \begin{align*} \sin^2x+\cos^2x=1 &\Leftrightarrow (\sin^2x+\cos^2x)^3=1^3=1\\ &\Leftrightarrow \sin^6 +3\cdot \sin^4x\cdot \cos^4x+3\cdot \sin^2x \cdot \cos^4x+\cos^6x =1\\ &\Leftrightarrow \sin^6x+\cos^6x=1-(3\cdot \sin^4x\cdot \cos^4x+3\cdot \sin^2x \cdot \cos^4x)\\ &\Leftrightarrow \sin^6x+\cos^6x=1-3\cdot \sin^2x\cdot \cos^2x(\sin^2x+\cos^2x) \\ &\Leftrightarrow \sin^6x + \cos^6x = 1 - 3\cdot \sin^2x \cdot \cos^2x \end{align*} $$

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I already knew that, but I don't know how to use it. –  Paul Manta Feb 12 '12 at 18:57
    
Do you know what does $(a+b)^3$ equal to? –  Lazar Ljubenović Feb 12 '12 at 18:58
    
Substitute $1-\sin^2 x$ for $\cos^2 x$ and $(1 - \sin^2 x)^3$ for $\cos^6 x$, and expand... –  Robert Israel Feb 12 '12 at 19:00
    
I added full proof since you seem to be having trouble. –  Lazar Ljubenović Feb 12 '12 at 19:09

$$\begin{align*}\sin^6x + \cos^6x&= (\sin^2x)^3 + (\cos^2x)^3\\&=(\sin^2x + \cos^2x)(\sin^4x -\sin^2x \cdot \cos^2x +\cos^4x) \\&=(\sin^2x+\cos^2x)^2-3 \cdot \sin^2x \cdot \cos^2x\\&=1-3 \cdot \sin^2x \cdot \cos^2x\end{align*}$$

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I am unsure if you don't know this $\TeX$ construct. Please go through the edit. –  user21436 Feb 12 '12 at 21:06

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