/sci/ - Science & Math

>>15089675
The point of an infinite series is to closely
approximate the trig function as a form of a
polynomial of some degree (the more terms/higher
the degree of polynomial, the better the graph).
This is not needed here...unless you like playing
with potentially infinitely many terms.

By the property you have used, you recovered
the same sine term as what the equation was
equal to. In that case, expand and group that
sine term. You will then have:
[eqn] {1\over 2}\sin(3\theta) = {1 \over 2}\sin(\theta)+\cos(2\theta)\sin(\theta) [/eqn]
Swap out [math] \cos(2\theta) = 1-2\sin^2(\theta) [/math] .
Then, you'll have:
[eqn] {1\over 2}\sin(3\theta) = {1 \over 2}\sin(\theta)+\sin(\theta)-2\sin^3(\theta) [/eqn]
Complete the rest and you got it.

Pic related is the infinite series of sine and
cosine for reference. You may probably see why
switching one or all parts to infinite series could
be troublesome in proving the identity.