So I need to simplify a problem down to ten characters, and I got this far and now I'm stuck:

2cos3t(3sint-4sin³t) +1

I don't really know what to do with the 2cos3t... I feel like distributing it in would just make the problem even messier...

Here's the original problem:
2cos(3t)(sin(2t)cos(t)+cos2t* sqr(1-cos²t))+1

Here's what I was able to do:
I turned the (sin2tcost)
into 2sintcost(cost)
then into 2sintcos²t,
then into 2sint(1-sin²t),
then into 2sint-2sin³t.

Then I turned cos2t*sqr(1-cos²t)
into 1-2sin²t*sqrt(sin²t),
then into 1-2sin²t(sint),
then into sint-2sin³t.

Then I added the two sides together to get 3sint-4sin³t

Right now we're using reciprocal id, quotient id, pythagorean id, cofunction id, even/odd id, sum&difference formulas, half-angle formulas, double angle formulas, sum-to-product formulas


Any help would be fantastic! Thanks!

Hi dftba28!

I think you should have gone the other way.

In your initial expression you have something that looks a lot like: sin2t cost + cos2t sint.
This happens to be part of the sum formula: sin(2t + t)=sin2t cos t + cos2t sint.

So instead of making the angles smaller, I think you should make them bigger.

Hello, dftba28!

ILikeSerena has great advice . . . did you follow it?

$\displaystyle \text{Simplify: }\:2\cos 3t(\sin2t\cos t+\cos2t\sqrt{1-\cos^2t})+1$

$\displaystyle \text{We have: }\:2\cos 3t(\sin2t\cos t+\cos2t\underbrace{\sqrt{1-\cos^2t}}_{\text{This is }\sin t})+1$

. . . . . . . $\displaystyle =\;2\cos3t\underbrace{(\sin2t\cos t + \cos2t\sin t)}_{\text{This is }\sin3t} + 1$

. . . . . . . $\displaystyle =\;\underbrace{2\cos3t\sin3t}_{\text{This is }\sin6t} + 1$

. . . . . . . $\displaystyle =\;\sin6t + 1$

Why all the trouble. we know that sin 3A = 3sinA-4sin^3A
Thus we get 2cos3t(3sint-4sin3t) +1 = 2 cos 3t sin 3t + 1 = sin 6t + 1 [Because sin 2A = 2 sinA cosA]