# How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$?

How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?

One reasonable approach seems to be using $(\cos^2x+\sin^2x)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.

Are there other solutions, simpler approaches?

I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $\int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx$ but again it is transformed to a different from then required here.

Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.

Well, it's very easy if you know some basic formulas from algebra.

$a^3+b^3$

$=(a+b)(a^2+b^2−ab)$

$=(a+b)((a+b)^2−2ab−ab)$

$=(a+b)^ 3−3ab(a+b)$

Now substitute $a$ and $b$ as

$a=\sin^2x,\;b=\cos^2x$

and use the identity

$\sin^2x+\cos^2x=1$

You will get your answer which is

$\sin^6x+\cos^6x=1−3\sin^2x\cos^2x$

• Nice approach, but I do by simple operations – Fawad Nov 16 '16 at 14:33

$$(c^2+s^2)^3=c^6+3c^4s^2+3c^2s^4+s^6=c^6+s^6+3c^2s^2(c^2+s^2).$$

Hint. Use the identity: $A^3+B^3=(A+B)(A^2-AB+B^2)=(A+B)((A+B)^2-3AB)$.

use $$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Factoring from $u^3+v^3=(u+v)(u^2-uv+v^2)$: \begin{align} \cos^6x+\sin^6x&=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\\ &=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x-\cos^2x\sin^2x\\ &=1-3\cos^2x\sin^2x \end{align}

$$\cos^6x+\sin^6x=(\cos^2x)^3+(\sin^2x)^3=$$ $$=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)=$$ $$=\cos^4x-\cos^2x\sin^2x+\sin^4x=$$ $$=((\cos^2x)^2+(\sin^2x)^2+2\cos^2x\sin^2x)-3\cos^2x\sin^2x=$$ $$=(\cos^2x+\sin^2x)^2-3\cos^2x\sin^2x=$$ $$=1-3\cos^2x\sin^2x$$

Hint:

Apply $a^3+b^3 = (a+b)(a^2-ab+b^2)$ where you take $a = \cos^2 x$ and $b=\sin^2 x$.

If $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$

Now write Identity as $$\sin^2 x+\cos^2 x+(-1) = 0\;,\text{ Then }\sin^6 x+\cos^6 +(-1)^3 = -3\sin^2 x\cos^2 x$$

• I suppose justifying the initial statement would be desirable... – Wojowu Nov 21 '16 at 11:18

Here's a slick way:

$(\cos^2(\theta)+\sin^2(\theta))^3=1$

Use Pascal's Triangle to get

$\cos^6(\theta)+3 \cos^4(\theta) \sin^2(\theta)+3 \cos^2(\theta) \sin^4(\theta)+\sin^6(\theta)=1$

$\cos^6(\theta)+\sin^6(\theta)=1-3 \cos^2(\theta) \sin^4(\theta)-3 \cos^4(\theta) \sin^2(\theta)$

$\cos^6(\theta)+\sin^6(\theta)=1-3\cos^2(\theta) \sin^2(\theta)(\cos^2(\theta)+\sin^2(\theta))$

$\cos^6(\theta)+\sin^6(\theta)=1-3\cos^2(\theta) \sin^2(\theta)$

$$(\cos^2 x+\sin^2x)^3=1^3=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x$$

$$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$ so $$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x$$

Any symmetric polynomial in $X$ and $Y$ can be expressed as a polynomial in $S=X+Y$ and $P=XY$. If $X=\cos^2x$ and $Y=\sin^2x$, then $S=\cos^2x+\sin^2x=1$, so a symmetric polynomial expression in $\cos^2x$ and $\sin^2x$ can be written as a polynomial in $P=\cos^2x\sin^2x$.

If the symmetric polynomial is also homogeneous, all terms in the new expression are homogeneous as well.

In this case, $P$ appears at most with degree $1$, because $P$ counts for degree $2$ in $X$ and $Y$.

Hence $X^3+Y^3=a+b(X+Y)^3+c(X+Y)XY$; evaluating at $X=Y=0$ entails $a=0$. With $X=1$ and $Y=0$ we get $b=1$; with $X=1$ and $Y=1$ we get $$1+1=8+2c$$ so $c=-3$; therefore $$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x$$

As another example, $X^4+Y^4=a(X+Y)^4+bXY(X+Y)^2+c(XY)^2$. Evaluating at $X=1$ and $Y=0$ gives $1=a$; evaluating at $X=1$ and $Y=-1$ gives $2=c$; evaluating at $X=Y=1$ gives $2=16a+4b+c$, so $b=-4$.

Therefore $$\cos^8x+\sin^8x=1-4\cos^2x\sin^2x+2\cos^4x\sin^4x$$