# Trig Equations

• November 22nd 2010, 01:05 PM
>_<SHY_GUY>_<
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?
• November 22nd 2010, 01:13 PM
pickslides
Maybe use some triple angle formula? Look here.

Double & Triple Angle formulas
• November 22nd 2010, 01:13 PM
dwsmith
I would suggest try playing around with the Sum and Difference and Product-to-Sum Formulas.

$\displaystyle sin(u\pm v)=sin(u)cos(v)\pm cos(u)sin(v), \ cos(u\pm v)=cos(u)cos(v)\mp sin(u)sin(v),$ $\ sin(u)sin(v)=\frac{1}{2}[cos(u-v)-cos(u+v), \ cos(u)cos(v)]=\frac{1}{2}[cos(u-v)+cos(u+v)]$
• November 22nd 2010, 01:23 PM
>_<SHY_GUY>_<
Quote:

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.
• November 22nd 2010, 01:24 PM
Plato
$\cos(x)\cos(3x)-\sin(x)\sin(3x)=cos(4x)=0$

$x=\dfrac{\pi}{8}$
• November 22nd 2010, 01:27 PM
>_<SHY_GUY>_<
Quote:

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 :)