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?
$\endgroup$ 43 Answers
$\begingroup$(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:
$$ \sin (4\theta) = 4\sin \theta \cos \theta \left( \cos^2 \theta - \sin^2 \theta \right)$$
$$ \sin (4\theta) = 4\sin \theta \cos^3 \theta - 4\sin^3 \theta \cos \theta$$
$\endgroup$ $\begingroup$The trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$ or the more general one $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ can be proved from the definitions of the functions $\sin$ and $\cos$ if that's what you are wondering. Classically, they are defined geometrically with a perpendicular triangle or a unit circle in a coordinate plane. However, to rigorously define it, $\sin$ and $\cos$ are defined in terms of integrals (See a good calculus books by Spivak, Apostol, Courant etc). After the sum of angles identity is proved from the definition, you just have to use it twice along with the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$. But for your assignment question, I don't think you have to prove the double angle identities.
$\endgroup$ $\begingroup$That's a trig identity.
So...
$\sin{4\theta} = 2\sin{2\theta}cos{2\theta}$
Can you take it from there?
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