1. ## Trig Equations

$Cosx*Cos3x - Sinx*Sin3x = 0$

I have tried sing the addition formula, and I end up with a lot of work. I'm trying it for the third time, and nothing seems to cancel.

Is there another way to do this?

2. Maybe use some triple angle formula? Look here.

Double & Triple Angle formulas

3. I would suggest try playing around with the Sum and Difference and Product-to-Sum Formulas.

$\displaystyle sin(u\pm v)=sin(u)cos(v)\pm cos(u)sin(v), \ cos(u\pm v)=cos(u)cos(v)\mp sin(u)sin(v),$ $\ sin(u)sin(v)=\frac{1}{2}[cos(u-v)-cos(u+v), \ cos(u)cos(v)]=\frac{1}{2}[cos(u-v)+cos(u+v)]$

4. Originally Posted by dwsmith
I would suggest try playing around with the Sum and Difference and Product-to-Sum Formulas.

$\displaystyle sin(u\pm v)=sin(u)cos(v)\pm cos(u)sin(v), \ cos(u\pm v)=cos(u)cos(v)\mp sin(u)sin(v),$ $\ sin(u)sin(v)=\frac{1}{2}[cos(u-v)-cos(u+v), \ cos(u)cos(v)]=\frac{1}{2}[cos(u-v)+cos(u+v)]$
I've tried using the Product-to-sum formula, but it only makes things more confusing (to me). My teacher said that we should be able to solve this only using the addition/subtraction formulas.

and this is what I have gotten so far

[Cosx (cos^2x - sin^2x) cosx - (2sinxcosx) sinx] - [sinx (2sinxcosx) cosx + (cos^2x - sin^2x) sinx]

I don't know if that is right, I'm still playing around with what what I have.

5. $\cos(x)\cos(3x)-\sin(x)\sin(3x)=cos(4x)=0$

$x=\dfrac{\pi}{8}$

6. Originally Posted by Plato
$\cos(x)\cos(3x)-\sin(x)\sin(3x)=cos(4x)=0$

$x=\dfrac{\pi}{8}$
I should have seen that.

Thank you so much, you have saved me so much work and time