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The following is a known trigonometric identity,

$$a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,$$

where

$$c=\sqrt {a^2+b^2+2ab\cos \alpha },\,$$

and

$$\beta =\arctan \left(\frac {b\sin \alpha }{a+b\cos \alpha }\right) + \begin{cases} 0 & \text{if } a+b \cos \alpha \geq 0,\\ \pi & \text{if } a+b\cos \alpha <0.\end{cases}$$

Is there an identity for the following ?

$$a\sin 2x + b\sin(x+\alpha )$$

Alternatively, can you find the values of $x$ that satisfy the following equation?

$$a\sin 2x - b\sin(x+\alpha )=0$$

Michael Hardy
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Heisenberg
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  • I'd try to look at the proof of the first identity and then apply the same idea. – TZakrevskiy Feb 12 '14 at 23:05
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    There are probably lots of identities saying that particular expression is equal to something, but you won't get a sine wave like the first one, as you'll see if you draw the graph. – Michael Hardy Feb 13 '14 at 00:40
  • I remark with some amusement that the easiest way to see that the first left-hand side is a sinusoidal wave, while the later expressions aren't, is to take their second derivatives (with respect to $x$) and compare them with the negatives of the original expressions. – Greg Martin Feb 13 '14 at 01:52

1 Answers1

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As for the last question: \begin{align*} a\sin2x-b\sin(x+\alpha)&=0 \\ 2a\sin x\cos x - b(\sin x\cos\alpha + \cos x\sin\alpha)&=0\\ -b\cos\alpha\sin x &= \cos x \cdot(b\sin\alpha-2a\sin x) \\ \frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} &= \cos x \\ \bigg( \frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} \bigg)^2 &= \cos^2 x = 1-\sin^2x \end{align*} gives you an equation of degree $4$ in $\sin x$ that you can try to solve your favorite way. (You'll need to check any solutions against the original equation, since squaring both sides will introduce extraneous solutions.)

Greg Martin
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