B Trigonometric Identity involving sin()+cos()

[erobz]

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?

 

[robphy]

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.

 

[FactChecker]

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 $b$.

 

[PeroK]

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

 

[erobz]

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

Yeah, putting the factor of $-1$ out front fixes it.

 

[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.

 

[PeroK]

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 $A > 0$.

 

[PeroK]

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 $(-\frac \pi 2, \frac \pi 2)$:
$$a = \sqrt{a^2 + b^2}\cos(-\phi) = \sqrt{a^2 + b^2}\cos(\phi)> 0$$

 

[erobz]

If $A$ is taken to be the amplitude ($y(0) = A$), then tacitly $A > 0$.

But wouldn't the amplitude be $\sqrt{ A^2 + B^2}$ ?

 

[PeroK]

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.

 

[PeroK]

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 $\phi = \tan^{-1}\big(\frac{-b}{-a} \big) = \tan^{-1}\big(\frac{b}{a} \big)$

 

[DaveE]

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.