## How might you solve this trig identity

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### How might you solve this trig identity

I have a trigonometric identity where I must solve for theta2 (theta2) but I'm tearing my hair out because I don't know to approach it. Could someone please assist me?

There are four variables:

x
L1
theta1
theta2

Each of the four variables are any number greater than or equal to zero and the angles theta1 and theta2 can only be between 0 and pi (or 0 and 180 degrees)
And I must solve for theta2. Please assume that I'm working in radians and not in degrees. Here's the equation.

xsine(theta2) = L1sine(theta2 - theta1)

So in this equation, there are two sine functions and each function has a coefficient. On the right side of the identity, two angles are subtracted from one another, the result of that number is put into sine, and then multiplied by L1, which, as stated, is a variable. The left is self-explanatory.

I have a second equation that is very similar and will most likely end up employing similar techniques. This one uses L2 instead of L1 as a variable but the same rules from the previous variables apply.

(sine (theta2 - theta1))/ x = sine theta1 / L2

I'm the kind of guy that likes to know the "why" in answers so any detail at all would be greatly appreciated. If the process is complicated however, I'd be fine with just an answer to the problems. I understand that these may possibly be difficult so I'll try to help but I really don't know that much trig. Thank you for your replies in advance!
korinkite

Posts: 2
Joined: Sun Apr 03, 2011 9:05 pm

### Re: How might you solve this trig identity

korinkite wrote:...I must solve for theta2....

xsine(theta2) = L1sine(theta2 - theta1)

I would use the angle-difference identity on the right-hand side. This will get the $\theta_2$ away from the $\theta_1$.

Then I would get the sines with $\theta_2$ together on the left-hand side, and then factor, so the sine is by itself. Then divide the factor to the right-hand side, and divide the cosine with the $\theta_2$ to the left-hand side. This can be redone as a tangent on the left-hand side. Then take the inverse tangent to get $\theta_2$ by itself.

maggiemagnet

Posts: 302
Joined: Mon Dec 08, 2008 12:32 am

### Re: How might you solve this trig identity

Okay.......

*hours of head scratching later* (just kidding!)

Here's what I received as an answer. Could you please let me know if I have it correctly? I got this for the first one:

theta2 = tan-1 ( (L1 sin(theta2) / (-x + L1 cos(theta1)) )

and, strangely (even though the two equations are really similar and shouldn't yield too different results) I simplified the second to be:

theta2 = theta1 + sin-1 ( (xsin(theta1)) / L2 )

Judging that these two equations came out drastically different, chances are I screwed up on at least one of them. Could someone confirm or deny my results?
korinkite

Posts: 2
Joined: Sun Apr 03, 2011 9:05 pm

### Re: How might you solve this trig identity

korinkite wrote:Here's what I received as an answer. Could you please let me know if I have it correctly? I got this for the first one:

theta2 = tan-1 ( (L1 sin(theta2) / (-x + L1 cos(theta1)) )

I don't get quite the same thing. How did you get your denominator on the right-hand side?

korinkite wrote:I simplified the second to be:

theta2 = theta1 + sin-1 ( (xsin(theta1)) / L2 )