I was asked to find a formula for $\cos(5x)$ in terms of $\sin(x)$ and $\cos(x)$.

I tried to use the formula $\cos(5x) + i\sin(5x) = (\cos(x)+i\sin(x))^5 $ and what I get is $16i\sin^5(x) - 20i\sin^3(x) + 5i\sin(x) + \cos(x) + 16 \sin^4(x) \cos(x) - 12 \sin^2(x) \cos(x)$

But how do I deal with the $i\sin(5x)$? Because I only can use $\sin(x)$ and $\cos(x)$ to express $\cos(5x)$. Thank you for your help!


Your way is right. But I think it should be

$$(\cos x + i\sin x)^5 = \cos^5 x + 5i\cos^4x\sin x - 10\cos^3x\sin^2 x - 10i\cos^2x\sin^3x + 5\cos x\sin^4x +i\sin^5x$$

So, you have $$\cos 5x = \cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$ $$\sin 5x = \sin^5x-10\cos^2x\sin^3x+5\cos^4 x\sin x$$

  • $\begingroup$ Why all the imaginary units disappeared in the final equality? Is that because cos5x must be a real number so we can assume all the imaginary part equal to 0? $\endgroup$ – Parting Feb 1 '17 at 3:21

Hint: $\cos$ and $\sin$ are always real. So you can compare like terms to find what $\cos 5x$ is.

  • $\begingroup$ Every complex equation gives us tow real equation. Real parts of tow sides of equation are equal to each other and imaginary parts also are equal. So in above answer real part is $cos 5x$ and imaginary part is $sin 5x$ $\endgroup$ – Ali Feb 2 '17 at 12:07

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