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So I need to use the fact that: $$\cos(4x) + i\sin(4x) = \left(\cos(x) + i\sin(x)\right)^4$$ to derive identities for $\cos(4x)$ and $\sin(4x)$ in terms of $\cos(x)$ and $\sin(x)$. I'm not sure how to go about this, could I please get some help.

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2 Answers 2

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Hint:

Try replacing $x$ by $-x$, and then adding the expressions. Using the fact that $\sin(-x) = -\sin(x)$ and that $\cos(-x) = \cos(x)$:

$$\cos(4x) + i \sin(4x) = \left(\cos(x) + i\sin(x)\right)^4$$ $$\cos(4x) - i \sin(4x) = \left(\cos(x) - i\sin(x)\right)^4$$ So: $$2\cos(4x) = \left(\cos(x) + i\sin(x)\right)^4 + \left(\cos(x) - i\sin(x)\right)^4 $$ Can you take it from there?

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$$(\cos x +i \sin x)^4=\cos 4x+i \sin 4x$$ $$(\cos x +i \sin x)^4=\cos^4x+4\cos ^3x \cdot (i \sin x)+6\cos ^2x \cdot (i \sin x)^2+4\cos x \cdot (i \sin x)^3+(i \sin x)^4=$$ $$=\cos^4x-6 \cos ^2x \sin^2x+\sin^4x+$$ $$+i(4 \cos ^3 x \sin x-4\cos x\sin^3x)$$ Then $$\cos 4x=\cos^4x-6 \cos ^2x \sin^2x+\sin^4x$$ and $$\sin 4x=4 \cos ^3 x \sin x-4\cos x\sin^3x$$

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  • $\begingroup$ Much faster than my answer :) $\endgroup$
    – nbubis
    Apr 10, 2016 at 8:19
  • $\begingroup$ Thank you so much for this, I was way off $\endgroup$
    – Drew
    Apr 10, 2016 at 21:33

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