Trigonometric simplification: -a*sin(wt) + w*cos(wt)

[SubtleAphex]

I'm trying to figure out the steps required to do the following simplification:

This equation:
b/(a^2+ ω^2 )*(-a*sin(ω*t)+ω*cos(ω*t))

can be simplified to the following:
(b/√(a^2+ ω^2 ))*sin(ω*t+ θ)
θ= tan^(-1) (- ω/a)

I can numerically verify that this is true but I am having trouble figuring out the steps to do this simplification. Any help would be greatly appreciated. See attached .doc file for a cleaner presentation of the equations.

 

[mathman]

I'm trying to figure out the steps required to do the following simplification:

This equation:
b/(a^2+ ω^2 )*(-a*sin(ω*t)+ω*cos(ω*t))

can be simplified to the following:
(b/√(a^2+ ω^2 ))*sin(ω*t+ θ)
θ= tan^(-1) (- ω/a)

I can numerically verify that this is true but I am having trouble figuring out the steps to do this simplification. Any help would be greatly appreciated. See attached .doc file for a cleaner presentation of the equations.

For this problem use the sin sum formula sin(ω*t+θ)=cos(θ)sin(ω*t)+sin(θ)cos(ω*t), where xcos(θ)=-a and xsin(θ)=ω. You then need to to solve for x and θ.

 

[pmsrw3]

The other way to solve it, and many similar problems, is to write it in terms of complex exponentials. This isn't really any simpler, but it eliminates the need to memorize a million trig identities.

 

[LCKurtz]

I'm trying to figure out the steps required to do the following simplification:

This equation:
b/(a^2+ ω^2 )*(-a*sin(ω*t)+ω*cos(ω*t))

can be simplified to the following:
(b/√(a^2+ ω^2 ))*sin(ω*t+ θ)
θ= tan^(-1) (- ω/a)

I can numerically verify that this is true but I am having trouble figuring out the steps to do this simplification. Any help would be greatly appreciated. See attached .doc file for a cleaner presentation of the equations.

Write it like this:

$$\frac{b}{a^2+\omega^2}\left(-a\sin(\omega t)+\omega\sin(\omega t)\right)<br /> =<br /> \frac b {\sqrt{a^2+\omega^2}}\left(\frac{-a}{\sqrt{a^2+\omega^2}}\sin(\omega t)<br /> +\frac{\omega}{\sqrt{a^2+\omega^2}}\cos(\omega t)\right)$$

Now, if you draw an angle in the second quadrant whose tangent is -ω/a you will see that the above expression becomes

$$\frac b {\sqrt{a^2+\omega^2}}\left(cos(\theta)\sin(\omega t)<br /> +\sin(\theta)\cos(\omega t)\right)=\frac b {\sqrt{a^2+\omega^2}}\sin(\omega t+\theta)$$

My only quarrel with the formula is that θ, being in the second quadrant, is not the principal value, which may not matter.