Trig Identity Proof: a cos(t) + b sin(t) = sqrt{a^2+b^2} sin(t) + arctan(a/b)

[old_dog_learns_old_tricks]

In solving a problem on MIT OCW I had to find an identity I don't think I've ever seen before:

\(\displaystyle a\cos{t}+b\sin{t} = \sqrt{a^2+b^2} \sin {(t + \arctan ({\dfrac{a}{b}})} )\)

Using this identity solves my problem beautifully but I'm afraid I have no idea why it works. I've spent the last hour drawing triangles and I can't make out why this identity is true. Can anyone point me to a proof, or give me an idea how such a proof would go? Does this thing even have a name?

Edit: Fixed tex.

 

[Deleted member 4993]

In solving a problem on MIT OCW I had to find an identity I don't think I've ever seen before:

\(\displaystyle a\cos{t}+b\sin{t} = \sqrt{a^2+b^2} \sin {t + \arctan ({\dfrac{a}{b}}} )\)

Using this identity solves my problem beautifully but I'm afraid I have no idea why it works. I've spent the last hour drawing triangles and I can't make out why this identity is true. Can anyone point me to a proof, or give me an idea how such a proof would go? Does this thing even have a name?

b*sin(t) + a*cos(t) = √(a^2+b^2)[a/√(a^2+b^2) * cos(t) + b/√(a^2+b^2) * sin(t)] ..... and continue...

 

[old_dog_learns_old_tricks]

b*sin(t) + a*cos(t) = √(a^2+b^2)[a/√(a^2+b^2) * cos(t) + b/√(a^2+b^2) * sin(t)] ..... and continue...

First of all, thank you and I appreciate the response. But I'm not sure how this helps prove the identity. All we've done here is multiply both numerator and denominator by \(\displaystyle \sqrt{a^2+b^2}\). Since neither a nor b are given, I don't see how we reduce this to \(\displaystyle \sqrt{a^2+b^2}\sin{(t+\arctan{\dfrac{a}{b}})}\).

Here's where I got this from, if anybody cares... http://www.myphysicslab.com/trig_identity1.html

 

[old_dog_learns_old_tricks]

Aha, I found it

Here's the proof I'm looking for: http://pages.pacificcoast.net/~cazelais/252/lc-trig.pdf

Thanks!

 

[Deleted member 4993]

First of all, thank you and I appreciate the response. But I'm not sure how this helps prove the identity. All we've done here is multiply both numerator and denominator by \(\displaystyle \sqrt{a^2+b^2}\). Since neither a nor b are given, I don't see how we reduce this to \(\displaystyle \sqrt{a^2+b^2}\sin{(t+\arctan{\dfrac{a}{b}})}\).

Now do you see how you would reduce it to the given form after multiplying and dividing by √(a^2+b^2) ?