# help w/ trig equation

• Apr 14th 2013, 02:52 PM
bmorestudent
help w/ trig equation
would like to solve the equation for theta:

z = b*cos(theta) + a*cos(theta)

been a long time since i took trig, any help would be appreciated.
• Apr 14th 2013, 04:12 PM
topsquark
Re: help w/ trig equation
Quote:

Originally Posted by bmorestudent
z = b*cos(theta) + a*cos(theta)

The first step is to factor:
$z = b~cos(\theta) + a~cos(\theta) = (b + a)~cos(\theta)$

Can you finish it from there?

-Dan
• Apr 14th 2013, 05:09 PM
HallsofIvy
Re: help w/ trig equation
Are you sure you didn't mean $z= a cos(\theta)+ b sin(\theta)$? That's a much more interesting question. As topsquark says, your equation is just a matter of combining those two " $cos(\theta)$ terms.

To solve $z= a cos(\theta)+ b sin(\theta)$, use the identity sin(A+B)= sin(A) cos(B)+ cos(A) sin(B). That would be the same as the previous equation if a= sin(A) and b= cos(A). The problem with that is that we would have to have $a^2+ b^2= sin^2(A)+ cos^2(A)$ which is not, in general, true. We can fix that factoring $\sqrt{a^2+ b^2}$ out. That is, we write the equation as
$z= \sqrt{a^2+ b^2}\left(\frac{a}{\sqrt{a^2+ b^2}}cos(\theta)+ \frac{b}{\sqrt{a^2+b^2}} sin(\theta)\right)$ so that
$\frac{z}{\sqrt{a^2+b^2}}= \left(\frac{a}{\sqrt{a^2+ b^2}}cos(\theta)+ \frac{b}{\sqrt{a^2+b^2}} sin(\theta)\right)$
So that now $\left(\frac{a}{\sqrt{a^2+ b^2}}\right)^2+ \left(\frac{b}{\sqrt{a^2+ b^2}\right)^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1$

Now, we can let $\frac{a}{\sqrt{a^2+ b^2}}= sin(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}}= cos(\phi)$ so that the equation can be written
$sin(\phi)cos(\theta)+ cos(\phi)sin(\theta)= sin(\theta+ \phi)= \frac{z}{\sqrt{a^2+ b^2}}$
then $\theta+ \phi= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)$
$\theta= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)- \phi= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)- arcsin(\frac{a}{\sqrt{a^2+ b^2}})$
• Apr 14th 2013, 08:26 PM
bmorestudent
Re: help w/ trig equation
thanks for the quick replies! yes, i meant to type b*SIN(theta), not cos. glad you caught that mistake and answered it hallsofivy, many thanks!