periodic functionstrigonometry
I have an assignment question that says "Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers."
I'm told that $\sin 2\theta = 2 \sin\theta \cos\theta$ but I don't know how this was found.
I used Wolfram Alpha to get the answer but this is what I could get :
$$
4\cos^3\theta\sin\theta- 4\cos\theta \sin^3\theta
$$
How can I solve this problem?
The equation has no real solutions.
For every $\theta\in\mathbb R$, we have $\sin^2\theta\in[0,1]$ and $\cos\theta\in[-1,1]$. This means that $\sin^2\theta+\cos\theta=2$ is only possible if $\sin^2\theta=1$ and $\cos\theta=1$. But if $\sin^2\theta=1$ we immediately have $\cos^2\theta=1-\sin^2\theta=0$, so $\cos\theta$ would have to be equal to $0$. This means the equation has no real solutions.
Proof without words: $\displaystyle\cot(\theta/2)=\frac{\color{green}{\sin(\theta)}}{\color{red}{1-\cos(\theta)}}$
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Best Answer
(Note that this method is pretty explanatory and slow, you can do it faster).
Let $u = 2\theta$, then we have:
$$ \sin 4\theta = \sin 2u $$
We know that:
$$ \sin 2u = 2\sin u\cos u$$
Now put $u = 2\theta$ back in:
$$ \sin (2 \cdot 2\theta) = 2\sin 2\theta \cos 2\theta $$ $$ \sin (4\theta) = 2\sin 2\theta \cos 2\theta $$
We know that $\sin 2\theta = 2\sin\theta\cos\theta$, so: $$ \sin (4\theta) = 4\sin \theta \cos \theta \cos 2\theta $$
Still, we must get rid of that pesky $\cos 2\theta$. You should know the other double angle sum formula for $\cos$:
$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$
So:
$$ \sin (4\theta) = 4\sin \theta \cos \theta \left( \cos^2 \theta - \sin^2 \theta \right)$$
$$ \sin (4\theta) = 4\sin \theta \cos^3 \theta - 4\sin^3 \theta \cos \theta$$