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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 $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.

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?

TShiong
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W. Zhu
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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 – Faroq AL-Tam Dec 14 '18 at 11:20

4 Answers4

2

You are largely correct. However you could prove the first part in a very simple way:

We can write $$P= a\cos x+b\sin x =\sqrt {a^2+b^2} \left[\frac {a}{\sqrt {a^2+b^2}}\cos x+\frac{b}{\sqrt {a^2+b^2}}\sin x \right] $$ 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.

Obsessive Integer
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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 $\alpha$? – W. Zhu Jan 06 '17 at 07:44
  • @W.Zhu It is obvious that $0 <x<\frac {\pi}{2} $, so the condition follows. Didn't explicitly state that as you got it right. –  Jan 06 '17 at 07:50
  • Is it necessary that $x+\alpha=\frac\pi2$? – W. Zhu Jan 06 '17 at 07:52
  • Yes it is necessary. –  Jan 06 '17 at 07:54
  • 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: $\alpha=\frac\pi2-x$. – W. Zhu Jan 06 '17 at 08:11
1

The proof is only valid for $0<x<\frac\pi2$. For a more general proof, let: $$a\cos x+b\sin x\ \equiv\ R\cos(x-\alpha)=R\cos\alpha\cos x + R\sin\alpha\sin x$$ So $a=R\cos\alpha$ and $b=R\sin\alpha$, from which you can readily find $R$ and $\alpha$.

George Law
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1

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.

Nilabro Saha
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1

We can prove this by using properties of complex numbers: $$ \begin{align}a\cos(x) + b\sin(x) &= a \cos(x) + b \cos\left(x-\frac{\pi}{2}\right)\\ &=a \operatorname{Re}\left\{e^{ix}\right\} + b \operatorname{Re}\left\{e^{i(x-\pi/2)}\right\}\\ &= \operatorname{Re}\left\{ae^{ix} + be^{i(x-\pi/2)}\right\}\\ &= \operatorname{Re}\left\{e^{ix}(a-bi) \right\} \\ &= \operatorname{Re}\left\{e^{ix}\sqrt{a^2+b^2}e^{i\cdot\operatorname{atan2}(-b,a)} \right\}\\ &= \operatorname{Re}\left\{\sqrt{a^2+b^2}e^{i(x+\operatorname{atan2}(-b,a))}\right\}\\ &= \sqrt{a^2+b^2} \cos(x+\operatorname{atan2}(-b,a)) \end{align} $$

where $\operatorname{atan2}(y_0,x_0)$ is the angle of the complex number $x_0+iy_0$.

This formula works for all real numbers $a$ and $b$ except when $a=b=0$.

Generalization

Following the same procedure, we can find a general formula for the sum of two sinusoids with same frequency but (possibly) different phases:

$$ \begin{align}a\cos(x+\alpha) + b \cos(x+\beta) &= \operatorname{Re}\left\{ae^{i(x+\alpha)} + be^{i(x+\beta)}\right\}\\ &= \operatorname{Re}\left\{e^{ix}\left(ae^{i\alpha}+be^{i\beta}\right) \right\}\\ &= \operatorname{Re}\left\{e^{ix}ce^{i\gamma} \right\}\\ &= \operatorname{Re}\left\{ce^{i(x+\gamma)} \right\}\\ &= c \cos(x+\gamma) \end{align} $$

where $ce^{i\gamma} = ae^{i\alpha}+be^{i\beta}$.

We have $$\begin{align}ce^{i\gamma}&=a (\cos(\alpha)+i\sin(\alpha)) +b (\cos(\beta)+i\sin(\beta))\\ &= (a\cos(\alpha)+b\cos(\beta)) + (a\sin(\alpha)+b\sin(\beta))i.\\ \end{align} $$

So $$c=\sqrt{(a\cos(\alpha)+b\cos(\beta))^2+(a\sin(\alpha)+b\sin(\beta))^2}$$ and $$\gamma=\operatorname{atan2}\big(a\sin(\alpha)+b\sin(\beta),a\cos(\alpha)+b\cos(\beta)\big).$$

Sidenotes:

  • This result is equivalent to phasor addition.
  • We can get similar formula involving $\sin$ by taking the imaginary part.
s114
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