# Trigonometric Identity involving sin()+cos()

• B
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I'm trying to use the following trigonometric identity:

$$a \cos ( \omega t ) + b \sin ( \omega t ) = \sqrt{a^2+b^2} \cos ( \omega t - \phi )$$ Where ##\phi = \tan^{-1} \left( \frac{b}{a} \right)## for the following equation:

$$x(t) = -\frac{g}{ \omega^2} \cos ( \omega t) + \frac{v_o}{ \omega } \sin ( \omega t ) + \frac{g}{ \omega^2}$$

When I apply the identity I get:

##a = -\frac{g}{ \omega^2}##
##b = \frac{v_o}{ \omega }##
##\phi = \tan^{-1} \left( \frac{-v_o \omega}{g} \right)##

$$X(t) = \sqrt{\left( \frac{g}{ \omega^2} \right)^2+\left( \frac{v_o}{ \omega } \right)^2} \cos \left( \omega t - \tan^{-1} \left( \frac{-v_o \omega}{g} \right) \right) + \frac{g}{ \omega^2}$$

However, on a plot they are not matching up...What am I doing wrong?

Homework Helper
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I suspect it may be related to the fact that
##\phi = \tan^{-1} \left( \frac{b}{a} \right)## can't distinguish ##\frac{b}{a}## from ##\frac{-b}{-a}##.
Try https://en.wikipedia.org/wiki/Atan2
or add ##\pi## to the result of ##\tan^{-1}## if the x-component is negative.

erobz and topsquark
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You can see from your original equation that CORRECTION the right-hand side is the same no matter what the signs of ##a## and ##b## are, whereas the left-hand side is different when a sign of ##a## or ##b## changes. Your plot of the right-hand side is correct for a positive ##a##, but your example has a negative ##a=-g/\omega^2##.
CORRECTION: I overlooked that ##\phi## of the right-hand side depends on the signs of ##a## and ##b##.

Last edited:
erobz, PeroK and topsquark
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I'm trying to use the following trigonometric identity:

$$a \cos ( \omega t ) + b \sin ( \omega t ) = \sqrt{a^2+b^2} \cos ( \omega t - \phi )$$ Where ##\phi = \tan^{-1} \left( \frac{b}{a} \right)## for the following equation:
This is not quite right. According to Wikipedia, it should be:
$$a \cos ( \omega t ) + b \sin ( \omega t ) = sgn(a) \sqrt{a^2+b^2} \cos ( \omega t - \phi )$$ Where ##\phi = \tan^{-1} \left( \frac{b}{a} \right)##.

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations

topsquark and erobz
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I read the identity from my Mechanical Measurements Text:

Perhaps they are assuming something about ##A,B##, but I don't see it anywhere.

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I read the identity from my Mechanical Measurements Text:

Perhaps they are assuming something about ##A,B##, but I don't see it anywhere.

View attachment 315809
If ##A## is taken to be the amplitude (##y(0) = A##), then tacitly ##A > 0##.

topsquark
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You can see from your original equation that CORRECTION the right-hand side is the same no matter what the signs of ##a## and ##b## are, whereas the left-hand side is different when a sign of ##a## or ##b## changes. Your plot of the right-hand side is correct for a positive ##a##, but your example has a negative ##a=-g/\omega^2##.
CORRECTION: I overlooked that ##\phi## of the right-hand side depends on the signs of ##a## and ##b##.
You were right if you looked at the case where ##t = 0##, as ##\cos## is positive on the range of ##\tan^{-1}##, which is ##(-\frac \pi 2, \frac \pi 2)##:
$$a = \sqrt{a^2 + b^2}\cos(-\phi) = \sqrt{a^2 + b^2}\cos(\phi)> 0$$

topsquark
Gold Member
If ##A## is taken to be the amplitude (##y(0) = A##), then tacitly ##A > 0##.
But wouldn't the amplitude be ##\sqrt{ A^2 + B^2}## ?

topsquark and PeroK
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But wouldn't the amplitude be ##\sqrt{ A^2 + B^2}## ?
That's true. The important thing is to remember the mathematical pitfall you uncovered here. It's possible the book's author overlooked it.

topsquark and erobz
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First, assume ##a > 0## and let ##\phi = \tan^{-1}\big (\frac b a \big )##. Note that ##-\frac \pi 2 < \phi < \frac \pi 2##. Hence: ##\cos \phi > 0## and ##\sin \phi## has the same sign as ##b##. We have:
$$\cos(x - \phi) = \cos x \cos \phi + \sin x \sin \phi$$Where
$$\cos \phi = \sqrt{\cos^2 \phi} = \frac 1 {\sqrt{\sec^2 \phi}} = \frac 1 {\sqrt{1 + \tan^2 \phi}} = \frac 1 {\sqrt{1 + \frac{b^2}{a^2}}} = \frac{a}{\sqrt{a^2 + b^2}}$$And:
$$\sin \phi = sgn(b) \sqrt{\sin^2 \phi} = sgn(b) \sqrt{1 - \cos^2 \phi} = sgn(b)\sqrt{1 - \frac{1}{\sec^2 \phi}}$$$$= sgn(b)\sqrt{\frac{b^2}{a^2 + b^2}} = \frac{b}{\sqrt{a^2+b^2}}$$Hence, for ##a > 0## we have:
$$\cos(x - \phi) = \frac 1 {\sqrt{a^2 + b^2}}\big(a\cos x + b\sin x\big)$$And$$a\cos x + b\sin x = \sqrt{a^2 + b^2}\cos(x - \phi)$$Finally, if ##a < 0##, then:
$$a\cos x + b\sin x = -\big((-a)\cos x + (-b)\sin x \big ) = -\sqrt{a^2 + b^2}\cos(x - \phi)$$Where ##\phi = \tan^{-1}\big(\frac{-b}{-a} \big) = \tan^{-1}\big(\frac{b}{a} \big)##

erobz and topsquark