please prove this answer, step by step.. $$\cos A - \cos 3A = 4 \sin^2A \cos A$$ I had just finished the left side $= -2 \sin 2A \sin A$ but then I have no idea to prove it..
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We use the identity $$\cos p-\cos q=-2\sin\left(\frac{p+q}{2}\right)\sin\left(\frac{p-q}{2}\right)$$ so with the double angle identity we have $$\cos A-\cos3A=\color{red}+2\sin(2A)\sin A=2\times 2\sin A\cos A\times \sin A=4\sin^2 A\cos A$$
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sorry.. but I don't understand why cos A - cos 3A can be + 2 sin (2A) sin A? – user163618 Jul 10 '14 at 08:08
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Apply the identity with $p=A$ and $q=3A$ you get $$-2\sin(2A)\sin(-A)=2\sin(2A)\sin A$$ – Jul 10 '14 at 08:10
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but sorry, why the identity is -2 sin and when it was applied to that problem it became +2? – user163618 Jul 10 '14 at 10:46
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As I explained because there are two negative signs: the negative sign of the identity and the negative sign from $\sin(-A)=-\sin A$. – Jul 10 '14 at 10:48
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ooo.. okay okay I understand.. thanks a lot ya! – user163618 Jul 10 '14 at 12:08
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$$\cos(A)-\cos(3A)=\cos(A)-(\cos(2A)\cos(A)-\sin(2A)\sin(A))$$
$$=\underbrace{\cos(A)-\cos^{3}(A)}_{=\cos(A)(1-\cos^{2}(A))}+\cos(A)\sin^{2}(A)+2\sin^{2}(A)\cos(A)$$
$$=\cos(A)\sin^{2}(A)+3\sin^{2}(A)\cos(A)=4\sin^{2}(A)\cos(A)$$
user71352
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$\cos(2A)\cos(A)=(\cos^{2}(A)-\sin^{2}(A))\cos(A)=\cos^{3}(A)-\sin^{2}(A)\cos(A)$ – user71352 Jul 12 '14 at 21:33
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$$\cos3A=4\cos^3A-3\cos A\implies\cos A-\cos3A=4\cos A(1-\cos^2A)$$
lab bhattacharjee
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@user163618, http://math.stackexchange.com/questions/175903/prove-cos-3x-4-cos3x-3-cos-x – lab bhattacharjee Jul 10 '14 at 12:14
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(for $A \in \mathbb{R})$ $$ \cos A - \cos 3A = \mathfrak{Re} (e^{iA} - e^{3iA}) \\ = \mathfrak{Re} \left(-2ie^{2iA}\frac{(e^{iA}-e^{-iA})}{2i} \right) \\ = \mathfrak{Im} \left(2e^{2iA} \right) \sin A \\ =2 \sin 2A \sin A \\ =4 \sin^2A \cos A $$
David Holden
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