By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
$\endgroup$ 29 Answers
$\begingroup$You have nice answers by expansion, so I present a solution without it.$$ \sin^6x + \cos^6x $$$$ = \sin^6x + (1-\sin^2 x)^3 $$$$ = \sin^6x + 1 - 3\sin^2 x+3\sin^4 x-\sin^6 x $$$$ = 1 - 3\sin^2 x + 3\sin^4 x $$$$ = 1 - 3\sin^2(1 - \sin^2 x) $$$$ = 1 - 3\sin^2 x\cos^2 x $$$$ = 1 - \frac{3}{4}\sin^22x $$Hence Proved.
$\endgroup$ $\begingroup$You can use $a^3+b^3=a^2-ab+b^2$ and proceed$$\begin{equation}\begin{aligned} &\sin^6 x+\cos^6 x\\[3ex] &=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\[2ex] &=\color{green}{\sin^4x}-\sin^2x\cos^2x+\color{green}{\cos^4x}\\[2ex] &=\color{green}{(\sin^2x+\cos^2x)^2}-3\sin^2x\cos^2x\\[2ex] &=1-\frac{3}{\color{red}4}(\color{red}4\sin^2x\cos^2x)&\text{($\because 4\sin^2x\cos^2x=\sin^22x$)}\\[2ex] &=\fbox{$1-\frac{3}{4}(\sin^22x)$}\\ \end{aligned}\end{equation}$$
$\endgroup$ $\begingroup$It should be
$(\cos^2x+\sin^2x)^3=1$
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x\color{red}{=1}$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x=1$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x=1$
$\cos^6x + \sin^6x = 1-3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = 1-\dfrac3{\color{red}4}(4\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = 1-\dfrac3{\color{black}4}(2\cos x\sin x)^2$
$\cos^6x + \sin^6x = 1-\dfrac34(\sin^22x)$
$\endgroup$ $\begingroup$Use the high-school formula$$a^3+b^3=(a+b)(a^2-ab+b^2).$$You'll obtain\begin{align} \cos^6x+\sin^6x&=(\cos^2+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\cr &=\cos^4x-\cos^2x\sin^2x+\sin^4x \end{align}Can you proceed?
$\endgroup$ $\begingroup$We have that
$$\cos^6x + \sin^6x =(\cos^2x +\sin^2x)(\cos^4x -\cos^2x\sin^2x +\sin^4x)=$$
$$=\cos^4x -\cos^2x\sin^2x +\sin^4x$$
and
$$\cos^4x -\cos^2x\sin^2x +\sin^4x=1-3\cos^2x\sin^2x=1-\frac34\sin^2(2x)$$
$\endgroup$ $\begingroup$You have $$(\cos^2(x)+\sin^2(x))^3=\cos^6(x)+\sin^6(x)+3\cos^4(x)\sin^2(x)+3\sin^4(x)\cos^2(x)$$$$=\cos^6(x)+\sin^6(x)+3\sin^2(x)\cos^2(x)(\cos^2(x)+\sin^2(x))$$$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}\cdot 4\sin^2(x)\cos^2(x)$$$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}(2\sin(x)\cos(x))^2$$$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}\sin^2(2x)$$
using the identity $\cos^2(x)+\sin^2(x)=1$ and the double angle formula $\sin(2x)=2\sin(x)\cos(x)$. Now the LHS is also equal to $1$.
$\endgroup$ 2 $\begingroup$Hint Use $$a^3+b^3=(a+b)\left({(a+b)}^2-3ab \right)$$
with
$a=\cos^2 x, b=\sin^2 x$.
$\endgroup$ $\begingroup$Following along your approach,
After expansion,$$(\cos^2x + \sin^2x)^3 = \cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$$$$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$$$$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x = 1$$Hence,$$\cos^6x + \sin^6x = 1 - 3\cos^2x\sin^2x$$$$ = 1 - \frac{3}{4}(4\sin^2x\cos^2x)$$$$ = 1 - \frac{3}{4}\sin^22x$$Hence Proved. There just were $2$ errors in your solution, that I have corrected here.
You start by writing
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
but that's not a "sentence". It's just noun without a verb. What about $\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$? what are you saying about it? Are you claiming it is equal to something? To what?
Then you go from this line
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
which is also a noun without a verb to this line
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
Which has a verb ("is equal to") and you moved the chunk "$(3\cos^2x\sin^2 x$" to the other side of the equation. But that assumes $\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$ had to have been equal to $0$ all along. Was $\cos^6x + (3\cos^2x\sin^2x) + \sin^6x= 0$?
I think you assumed that because you had the "noun" $\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$ just sitting there by itself with nothing else when you moved "$(3\cos^2x\sin^2 x$" to "the other side" there was no other side (because you never made an equation sentence in the first place) you didn't think it matter.
In essence what you did was equivalent to the following:
$ 7$
$10- 3$
$10=3 $
Which .... would be true if we assume that $7=0$. But wouldn't be true otherwise.
So lets do this with full sentences.
We start with $(\cos^2 x + \sin^2 x)^3$. Will what about it. It's equal to something but to what? Well we know $(\cos^2 + \sin^2 x) = 1$ so:
$\cos^2 x + \sin^2 x = 1$
$(\cos^2 x + \sin^2 x)^3 = 1^3$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x = 1$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x=1$
$\cos^6x + \sin^6x = 1-3\cos^2x\sin^2x$
Now we we can complete this but showing that $3\cos^2\sin^2 x = \frac 34 \sin 2x$.
And here made a second mistake. You figured that $sin^2 2x = 2\cos^2x\sin^2 x$
In actuality $\sin 2x = \sin(x+x) = \cos x\sin x + \sin x \cos x =$ \cos x \sin x$.
So $sin^2 2x = (2 \cos x \sin x) = 2^2 \cos^2 x \sin^2 x=4 \cos^2 sin^2 x$.
So....
$\cos^6x + \sin^6x = 1-3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = 1-\frac 34(4\cos^2x\sin^2x)$
$\endgroup$$\cos^6x + \sin^6x = 1 - \frac 34 \sin^2 x$.
