# Trigonometric Identity involving sin()+cos()

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• erobz
In summary: Hence, the same argument goes through.In summary, the 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)## is valid for all values of ##a## and ##b##, and can be used to simplify equations involving trigonometric functions. However, it is important to be mindful of the signs of ##a## and ##b## when applying this identity, as it can affect the overall result.

#### erobz

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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?

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
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
erobz said:
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
I read the identity from my Mechanical Measurements Text:

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

erobz said:
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
FactChecker said:
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
PeroK said:
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
erobz said:
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
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
It is really common for people to be sloppy when they say ##\phi = tan^{-1}(b/a)##. As others have said, there are sign issues depending on which quadrant you are in. When things seem wrong always look at a sketch in the complex plane first. This is why all (good) programming libraries have the atan2 function. Also, beware, you'll see this mistake again, from yourself or others.

FactChecker and erobz