Can anyone please guide me step by step on: "use the maclaurin series for f(x)=sin^2 Hint: using the identity sin^2 x = 1/2 (1-cos(2x))"

Question

anonymous · Answer

Follow the hint that $\large \sin^2x=\frac{1}{2}(1−\cos(2x))$ along with the fact that 
$$\large \cos u=\sum_{n=0}^{\infty}\frac{(−1)^nu^{2n}}{(2n)!}\tag {1}$$and hence you get the power series for $\large \cos(2x)$ by letting $\large u=2x$ in $\large(1)$.

Can you take things from here? :-)

anonymous · Answer

Thanks. cos U you just put in be course the derivative of sin is cos? so you put that outside?…how does the 1 turn into (-1)?

anonymous · Answer

There's no differentiating being done here.  What I meant by what I wrote originally was that you can rewrite
$$\large \sin^2 x = \frac{1}{2}(1-\cos(2x)) = \frac{1}{2}\left(1-\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n}}{(2n)!} \right) = \ldots\tag{1}$$using the trig identity they provided and the fact that we know the Maclaurin series for $\large \cos x$.  All you need to do now is to write $\large (1)$ as a single power series. 

Does this clarify things? :-)

anonymous · Answer

yes i guess so. I have to try out some more of these types. My brain is not that sharp. but i think i got it
thanks