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.

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- Apr 14th 2013, 02:52 PMbmorestudenthelp 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 PMtopsquarkRe: help w/ trig equation
- Apr 14th 2013, 05:09 PMHallsofIvyRe: help w/ trig equation
Are you sure you didn't mean $\displaystyle z= a cos(\theta)+ b sin(\theta)$? That's a much more interesting question. As topsquark says, your equation is just a matter of combining those two "$\displaystyle cos(\theta)$ terms.

To solve $\displaystyle z= a cos(\theta)+ b sin(\theta)$, 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 $\displaystyle a^2+ b^2= sin^2(A)+ cos^2(A)$ which is not, in general, true. We can fix that factoring $\displaystyle \sqrt{a^2+ b^2}$ out. That is, we write the equation as

$\displaystyle z= \sqrt{a^2+ b^2}\left(\frac{a}{\sqrt{a^2+ b^2}}cos(\theta)+ \frac{b}{\sqrt{a^2+b^2}} sin(\theta)\right)$ so that

$\displaystyle \frac{z}{\sqrt{a^2+b^2}}= \left(\frac{a}{\sqrt{a^2+ b^2}}cos(\theta)+ \frac{b}{\sqrt{a^2+b^2}} sin(\theta)\right)$

So that now $\displaystyle \left(\frac{a}{\sqrt{a^2+ b^2}}\right)^2+ \left(\frac{b}{\sqrt{a^2+ b^2}\right)^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1$

Now, we**can**let $\displaystyle \frac{a}{\sqrt{a^2+ b^2}}= sin(\phi)$ and $\displaystyle \frac{b}{\sqrt{a^2+b^2}}= cos(\phi)$ so that the equation can be written

$\displaystyle sin(\phi)cos(\theta)+ cos(\phi)sin(\theta)= sin(\theta+ \phi)= \frac{z}{\sqrt{a^2+ b^2}}$

then $\displaystyle \theta+ \phi= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)$

$\displaystyle \theta= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)- \phi= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)- arcsin(\frac{a}{\sqrt{a^2+ b^2}})$ - Apr 14th 2013, 08:26 PMbmorestudentRe: 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!