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Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.

This is my proof.

Triangle

Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha})=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.

Is my proof valid? Is there a more general way to prove it?

For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?

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    This is called Harmonic Addition Theorem, usually used for the trigonometric Fourier compact form. it's proof can be found here: http://mathworld.wolfram.com/HarmonicAdditionTheorem.html – 2018-12-14

3 Answers 3

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Let $a = r \cos \alpha$ and $b = r \sin \alpha$. Then,

$ a\cos x + b\sin x\\ = r\cos \alpha \cos x + r\sin \alpha \sin x\\ = r \cos (x - \alpha) $

since we know that $\cos \theta \cos \phi + \sin \theta \sin \phi = \cos (\theta - \phi)$. Now, we already have $a = r \cos \alpha$ and $b = r \sin \alpha$. Squaring and adding both of these,

$ a^2 + b^2 = r^2(\cos^2 \alpha + \sin^2 \alpha) = r^2 \\ \implies r = \sqrt{a^2 + b^2} $

which gives us our desired result.

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The proof is only valid for $0

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    The condition $0 – 2017-01-06
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    @MyGlasses It's the assumption implicit in the diagram. – 2017-01-06
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You are largely correct. However you could prove the first part in a very simple way:

We can write $$P= a\cos x+b\sin x =\sqrt {a^2+b^2} [\frac {a}{\sqrt {a^2+b^2}}\cos x+\frac{b}{\sqrt {a^2+b^2}}\sin x] $$ Now we can take $\frac{a}{\sqrt {a^2+b^2}} $ as $\cos \alpha $ giving us $$P=\sqrt {a^2+b^2}[\cos \alpha \cos x+\sin \alpha \sin x]=\sqrt{a^2+b^2}\cos (x-\alpha) $$ And also $$\alpha =\arccos \frac {a}{\sqrt {a^2+b^2}} $$ which I think maybe a small typo in your calculation.

You need not prove the first part with the help of a triangle but your approach is fine. Hope it helps.

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    Typos have been fixed. I have one question. In my proof, I assumed that $x$ and $\alpha$ were complementary. But in your proof, it seems that this is not necessary. Does the identity hold for any $x$ and $\alpha$? – 2017-01-06
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    @W.Zhu It is obvious that $0 – 2017-01-06
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    Is it necessary that $x+\alpha=\frac\pi2$? – 2017-01-06
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    Yes it is necessary. – 2017-01-06
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    Having carefully read your proof again, now I see. Your step of taking $\frac{a}{\sqrt{a^2+b^2}}$ as $\cos\alpha$ has the implicit assumption that $x+\alpha=\frac\pi2$. Maybe the second part of the question has a simpler answer: $\alpha=\frac\pi2-x$. – 2017-01-06