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Prove the following equation.

\begin{eqnarray} \\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\ \end{eqnarray}

I can't prove it by many methods I use.

Please give me some hints.


Thank you for your attention

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It seems one sign in your equality is wrong. One actually has \begin{align} 1=\left(\cos^2x+\sin^2x\right)^3&=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x=\\ &=\cos^6x+3\sin^2x\cos^2x\underbrace{\left(\cos^2x+\sin^2x\right)}_{=1}+\sin^6x=\\ &=\cos^6x+3\sin^2x\cos^2x+\sin^6x. \end{align}

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    $\begingroup$ That was fast! Saw there was an issue, but you nailed it! +1 $\endgroup$ – Amzoti Oct 26 '13 at 3:59
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HINT:

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

Can you continue?

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  • $\begingroup$ @O.L. no problem, sir. $\endgroup$ – Shobhit Oct 26 '13 at 4:10
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It’s false as stated: try it with $x=\frac{\pi}4$. The middle sign is wrong. If you subsitute $u=\cos^2x$ and change the middle sign, it becomes

$$u^3+3u(1-u)+(1-u)^3=1\;,$$

which is easily verified.

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  • $\begingroup$ Today is the big day! $\endgroup$ – The Chaz 2.0 Oct 26 '13 at 4:25
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I'll start from your left hand side instead of R.H.S.

All terms contain integer powers of $\sin^2x$ and $\cos^2x$. Rewrite the l.h.s to

$$(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)+(\cos^2x)^3$$

Cubic? It feels like $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ but notice the lack of $a^2b$ and $ab^2$. To make up for the missing powers, I'm tempted to tweak the L.H.S by introducing a harmless multiplier $1=\sin^2 x+\cos^2 x$:

$$ \begin{align} L.H.S.&=(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)\cdot1+(\cos^2x)^3\\ &=(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)\cdot(\sin^2x+\cos^2x)+(\cos^2x)^3\\ &=\quad ?\quad \text{(Try to expand)} \\ &=(\sin^2 x + \cos^2x)^3=1 \end{align} $$

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Hint: Expand $(\cos^2x+\sin^2x)^3$ and apply Pythagorean theorem.

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An alternate approach

From algebra we know if $a+b+c=0$ then $ a^3+b^3+c^3 = 3.a.b.c$

$$Derivative \begin{cases}a+b+c = 0 & \text{(1)}\\ \Rightarrow a+b=-c \\ \Rightarrow (a+b)^3=(-c)^3 & \text{cubing both sides} \\ \Rightarrow a^3+b^3+3a^2b+3ab^2=(-c)^3 \\ \Rightarrow a^3+b^3+3ab(a+b)=(-c)^3 & \text{but }a+b = -c\text{ from (1)} \\ \Rightarrow a^3+b^3-3abc=-c^3 \\ \Rightarrow a^3+b^3+c^3=3abc & \text{ subtracting } c^3 \text{from both sides} \end{cases}$$

Now if $a=cos^2x$, $b=sin^2x$ and $c = -1$, we have $$a+b+c=cos^2x + sin^2x - 1 = 0$$

thus we have $$\Rightarrow (cos^2x)^3 + (sin^2x)^3 +(-1)^3 - 3 \cos ^2x .\sin ^2x . (-1) = 0 $$

$$\Rightarrow\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$$

Hence Provded

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  • $\begingroup$ Nice method~!!but why $a^3+b^3+c^3 = 3abc$ if $a+b+c=0$? Are there any references? $\endgroup$ – Casper Oct 28 '13 at 16:27
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    $\begingroup$ @CasperLi: I thought it was well known identity. For a proof I have added a spoiler $\endgroup$ – Abhijit Oct 28 '13 at 17:00
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    $\begingroup$ Nice~! Thank you. It is very useful. $\endgroup$ – Casper Oct 28 '13 at 17:05

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