I got this question from my teacher:
$\sin {6x}=\dots$
Try to make this one from this: $\sin(3x+3x)$, then according to the formula ended up like this:
$$2\sin{(2x+x)}\cos{(2x+x)}$$ $$2((\sin(2x)\cos(x)+\cos(2x)\sin(x))\cdot(\cos(2x)\cos(x)+\sin(2x)\sin(x)))$$
Then I just have no idea how to continue. Should I use the formula again or what?
$\endgroup$ 22 Answers
$\begingroup$Since you are willing to learn DeMoivre's formula, I'll use it.
For any $n$, we have $\cos nx + i\sin nx = (\cos x + i\sin x)^n$.
Now, lets set $n = 6$ and use the Binomial theorem:
$\cos 6x + i\sin 6x = (\cos x + i\sin x)^6 = \displaystyle\sum_{k = 0}^{6}\dbinom{6}{k}(\cos x)^k(i\sin x)^{6-k}$
If we write out all the terms, and use the fact that $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$, we get:
$\cos 6x + i\sin 6x = \cos^6 x + 6i\cos^5 x\sin x - 15\cos^4 x \sin^2 x - 20i\cos^3 x\sin^3 x + 15\cos^2 x\sin^4 x +6i\cos x\sin^5 x - \sin^6 x$
Now, set the real parts and imaginary parts equal:
$\cos 6x = \cos^6 x - 15\cos^4 x \sin^2 x + 15\cos^2 x\sin^4 x - \sin^6 x$
$\sin 6x = 6\cos^5 x\sin x - 20\cos^3 x\sin^3 x +6\cos x\sin^5 x$
As you can see, this gets us both the identities for $\cos 6x$ and $\sin 6x$ simultaneously, but it does require you to know a bit about complex numbers, and the binomial theorem.
If you don't know these, you can continue what you have already done by using the following:
$\sin 2x = \sin(x+x) = \sin x \cos x + \cos x \sin x = 2\sin x \cos x$
$\cos 2x = \cos(x+x) = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$
$\endgroup$ 4 $\begingroup$You can manipulate with those two formulas $$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$ $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$ for example $$\sin(6x)=\sin(3x+3x)=\sin(2\cdot3x)=2\sin(3x)\cos(3x)$$ $$\sin(3x)=\sin(x+2x)=\sin(x)\cos(2x)+\sin(2x)\cos(x)$$ $$\cos(3x)=\cos(x+2x)=\cos(x)\cos(2x)-\sin(2x)\sin(x)$$ $$\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x)=2\sin(x)\cos(x)$$ $$\cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x)=\cos^2(x)-\sin^2(x)$$
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