Compound angle formula to determine expression for Sin3x in terms of sinx and cosx?

a) Write 3x as the sum of two terms involving x.

b) Substitute the expression from part a) into sin3x.

c) Use an appropriate compound angle formula to expand the expression from part b).

d) Use an appropriate double angle formula to expand any terms involving 2x in the expanded expression from part c). Simply as much as possible.

e) Select a value for x, and show that your expression in part d) holds true for that value. (use pi/3 for x)

**What I've got so far**

a) (2x+x)

b) sin(2x+x)

c) sin2xcosx + cos2xsinx

d) I'm a bit unsure of this one.

e) After I solve d) I sub in pi/3, remembering that sin pi = 0 and sin pi/3 = sqrt(3) / 2.

So pretty much D) is the only one im having trouble with, can anyone enlighten me on this? Thanks in advance!

Re: Compound angle formula to determine expression for Sin3x in terms of sinx and cos

I'm bumping this thread because I'm working on the exact same problem.

I don't understand this last part: e) Select a value for x, and show that your expression in part d) holds true for that value. (use pi/3 for x)

What is the point of subbing pi/3 into the final equation form of d) What is supposed to hold true?

Any help would be great I'm trying to finish off my studies and this very last part of the very last problem is giving me trouble.

Re: Compound angle formula to determine expression for Sin3x in terms of sinx and cos

Update: Took an hour break and looked at it again, I got it :D Thanks anyway.

Re: Compound angle formula to determine expression for Sin3x in terms of sinx and cos

In case someone else reads this, your question is a valid one, so to help them out, you should show your full solution:

On the LHS, you have: $\displaystyle \sin(3x)$ On the RHS, you have $\displaystyle 3\sin x - 4 \sin^3 x$

Plugging in $\displaystyle \dfrac{\pi}{3}$ as suggested, you have:

LHS: $\displaystyle \sin\left( 3\dfrac{\pi}{3} \right) = \sin(\pi) = 0$

RHS: $\displaystyle 3\sin\left( \dfrac{\pi}{3} \right) - 4\sin^3\left( \dfrac{\pi}{3} \right) = 3\dfrac{\sqrt{3}}{2} - 4\left( \dfrac{\sqrt{3}}{2} \right)^3 = 0$

So, the two sides give the same answer, demonstrating that this solution is correct for $\displaystyle \dfrac{\pi}{3}$.