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Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$

I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$

so $$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$ and $\cos2\alpha\sin\alpha$ can be expressed in three ways: $$(\cos^2\alpha-\sin^2\alpha)\sin\alpha =\sin\alpha\cos^2\alpha-\sin^3\alpha$$ $$(2\cos^2\alpha -1)\sin\alpha = 2\cos^2\alpha\sin\alpha - \sin\alpha$$ $$(1-2\sin^2\alpha)\sin\alpha = \sin\alpha - 2\sin^3\alpha$$ I tried adding these, but nothing came close to the required answer. So then I tried calculating $\sin4\alpha$ (from the required answer): $$\sin4\alpha=2\sin(2\alpha)\cdot\cos(2\alpha)$$ $$\sin4\alpha=2\cdot2\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$ $$\sin4\alpha=4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$ so $$\sin4\alpha\cos\alpha= 4\sin\alpha\cos^2\alpha(\cos^2\alpha-\sin^2\alpha)$$ Still looking at the answer, I calculated $\cos4\alpha$ $$\cos4\alpha = 1- \sin^2(2\alpha)$$ If $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ then $$\sin^22\alpha= (2\sin\alpha\cos\alpha)^2$$ and $$\cos4\alpha = 1 - 4\sin^2\alpha\cos^2\alpha$$ and $$\cos4\alpha\sin\alpha=\sin\alpha-4\sin^3\alpha\cos^2\alpha$$ I have tried subtracting my values for $\sin4\alpha\cos\alpha$ and $\cos4\alpha\sin\alpha$, but I have not come close to a solution.

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    $\begingroup$ the formulas for cos4α and sin4α you are trying to use are not correct. and answers are pretty comprehensive. I just wanted to point out that cos4α is not equal to cos2α*cos2α. $\endgroup$
    – Omega
    Jul 28, 2013 at 8:23

4 Answers 4

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use $$\sin(A+B) = \sin A\cos B + \cos A\sin B $$on LHS and $$\sin(A-B) =\sin A\cos B - \cos A\sin B$$ on RHS

so $$\sin(3\alpha) = \sin(3\alpha)$$

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  • $\begingroup$ Hi @yanbo, may I know why you removed kahan academy links from my answer ? $\endgroup$ Jul 27, 2013 at 13:30
  • $\begingroup$ I didn't realise I did that. Terribly sorry. Gave you a +1 though, hope that makes it up. $\endgroup$
    – BlackAdder
    Jul 27, 2013 at 13:34
  • $\begingroup$ Ya Ya no problem .I added the links so that he can know the proof also . $\endgroup$ Jul 27, 2013 at 13:37
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    $\begingroup$ A = $2\alpha$ and B= $\alpha$ so A+B = $3\alpha$ $\endgroup$ Jul 27, 2013 at 14:37
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    $\begingroup$ In RHS A -> $4\alpha$ and B -> $\alpha$ so A-B = $4\alpha - \alpha $ = $3\alpha$ $\endgroup$ Jul 27, 2013 at 14:40
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You're making this way harder than it needs to be. One is $\sin{(2 \alpha + \alpha)}$. The other is $\sin{(4 \alpha-\alpha)}$.

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Just remember that $\sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$ and apply the r.h.s. of each of these equalities to your example.

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Hint

Take the real part of this complex number

$$-ie^{i\alpha}e^{2i\alpha}=-ie^{-i\alpha}e^{4i\alpha}$$

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