Using de Moivre's formula to find an expression for $\sin 3x$ in terms of $\sin x$ and $\cos x$

Question

I was asked to use de Moivre's formula to find an expression for $\sin 3x$ in terms of $\sin x$ and $\cos x$.

De Moivre's formula is this: $$\cos nx+i\sin nx=(\cos x+i\sin x)^n$$ I plugged $3$ in for $n$ and got the following: $$\cos 3x+i\sin 3x = (\cos x+i\sin x)^3$$ At this point, I am stuck. I don't know how to take this and solve for $\sin 3x$. I could do: $\sin 3x=\frac{(\cos x+i\sin x)^3-\cos 3x}{i}$, but I do not think the answer is supposed to have an imaginary number in it.
If I expand my numerator I get:$$\frac{\cos^3x-3\sin^2x\cos x+i(3\cos^2x\sin x-\sin^3x)-\cos 3x}{i}$$ Simplifying this gets me: $$\frac{-3\sin^2x\cos x}i+3\cos^2x\sin x-\sin^3x$$ This still leaves me with an imaginary number. How do I get rid of it? And, are my other steps correct?

Answer 1

$$(\cos x+i\sin x)^3=\cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x$$ $$=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x$$ $$=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x)$$

Since this is $\cos 3x +i\sin3x$, we can conclude that $\cos3x=\cos^3x-3\cos x\sin^2x$ and $\sin3x=3\cos^2x\sin x-\sin^3x$. You can simplify with the Pythagorean identities as you desire.

Answer 2

• Expand the right hand side, i.e. the term $( ... )^3$;
• Gather the imaginary part in the right hand side
• Identify the imaginary parts.

Answer 3

Let Z is a coplex number, Z=r(cos x +i sin x), where i=√-1

Whe have from de moivre's theorem for any n Z^n = cos nx +i sin nx , Hence (cos nx+i sin nx)^3 = cos 3x +i sin nx

By expanding and equting we get that Sin3x =3cos^2 x*sinx - sin^3 x