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So this sprung up in a lecture note on quantum harmonic oscillation.

The equation is

$a \cos\left( \omega t \right)+b\sin\left( \omega t \right)=A \cos\left( \omega t+\phi \right)$

I must have forgotten how the right hand side follows from the left hand side.

Some hints would be helpful

JimmyK4542
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Mathematicing
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3 Answers3

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Let $A = \sqrt{a^2+b^2}$ and $\phi$ be an angle such that $\cos\phi = \dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi = \dfrac{-b}{\sqrt{a^2+b^2}}$. Then, by using the identity $\cos(x+y) = \cos x\cos y-\sin x\sin y$, we have:

\begin{align}a\cos(\omega t)+b\sin(\omega t) &= \sqrt{a^2+b^2}\left[\dfrac{a}{\sqrt{a^2+b^2}}\cos(\omega t)-\dfrac{-b}{\sqrt{a^2+b^2}}\sin(\omega t)\right] \\ &=A\left[\cos\phi\cos(\omega t)-\sin\phi\sin(\omega t)\right] \\ &= A\cos(\omega t + \phi).\end{align}

JimmyK4542
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2

Try breaking down $A\cos(\omega t+\phi)$

$$\begin{array}{lll}A\cos(\omega t+\phi) &=& A(\cos \omega t\cos\phi-\sin\omega t\sin\phi)\\ &=&(A\cos\phi)\cos \omega t-(A\sin\phi)\sin\omega t\\ \end{array}$$

Try to show that for arbitrary constants $a$ and $b$ that there exists constants $A$ and $\phi$ such that $$a = A\cos\phi$$ and $$b=A\sin\phi$$

What is $a^2+b^2$?

John Joy
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A good general way to prove trigonometric identities is to switch to complex representation using Euler's formula. It sometimes involves a little bit of algebra, but it's more algorithmic and requires less magical insight. We can express

$$ \cos\omega t = \frac{e^{i\omega t}+e^{-i\omega t}}{2} $$

and

$$ \sin\omega t = \frac{e^{i\omega t}-e^{-i\omega t}}{2i} $$

Then we desire to prove that

$$ \frac{a}{2}\left( e^{i\omega t}+e^{-i\omega t} \right) + \frac{b}{2i}\left( e^{i\omega t}-e^{-i\omega t} \right) = \frac{A}{2}\left( e^{i(\omega t + \phi)}+e^{-i(\omega t + \phi)} \right) $$

Rearranging, we have

$$ \left( \frac{a}{2} + \frac{b}{2i} \right)e^{i\omega t} + \left( \frac{a}{2} - \frac{b}{2i} \right)e^{-i\omega t} = \frac{A}{2}e^{i\phi}e^{i\omega t} + \frac{A}{2}e^{-i\phi}e^{-i\omega t} $$

Now, comparing terms on the left and right hand sides, we see that we will have equality if the following relations are satisfied:

$$ Ae^{i\phi} = a - bi $$ $$ Ae^{-i\phi} = a + bi $$

or, written out,

$$ A\cos\phi + iA\sin\phi = a - bi $$ $$ A\cos\phi - iA\sin\phi = a + bi $$

We may equate the real and imaginary parts of the foregoing expressions. Both furnish the following:

$$ A\cos\phi = a $$ $$ A\sin\phi = -b $$

Squaring both sides of these equations and adding yields:

$$ A^2(\cos^2\phi + \sin^2\phi) = a^2 + b^2 $$ $$ A = \sqrt{a^2 + b^2} $$

Then, dividing them we have

$$ \frac{\sin\phi}{\cos\phi} = \tan\phi = -\frac{b}{a} $$

which are the desired relations.

pr0gramR
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