Multiple trigonometric functions?

Question

I have a question about solving equations with trigonometric functions. To my understanding, one can use inverse functions like this:

$$\sin \theta = a, \quad \arcsin a = \theta$$

But I do not know how to solve something with more than one function, like this one:

$$xy\sin \theta = zxy\cos \theta $$

I was able to simplify it to:

$$\sin \theta = z * \cos \theta $$

How can I rewrite this to be equal to theta?

Sorry if I missed anything; this is for an intro calculus class. Thanks.

By the way, I was unable to get something from these: Multiple trigonometric functions, How can I solve for a single variable which occurs in multiple trigonometric functions in an equation?

Answer 1

In general, there is no way to solve transcendental equations except by numerical techniques.

In this case, however, we have a shorthand: $\tan \theta = \frac{\sin \theta}{\cos \theta}$. This allows us to divide both sides of your final equation by $\cos \theta$ to arrive at:

$$\tan \theta = z \quad \Rightarrow \quad \theta = \arctan z$$

This is really the same equation, except $\tan$ and its inverse are used so much in daily calculations that even our desk calculators have them-- they already have the numerical approximations built in to solve the equation to a good degree of accuracy.

Answer 2

Even if we didn't have $\tan$, we could've squared both sides and used $\sin^2\theta=1-\cos^2\theta$, i.e. the pythagorean theorem, which would give

$$1-\cos^2\theta=z^2\cos^2\theta\\1=(1+z^2)\cos^2\theta\\\cos^2\theta=\frac1{1+z^2}\\\cos\theta=\pm\sqrt{\frac1{1+z^2}}\\\theta=\arccos\left(\pm\sqrt{\frac1{1+z^2}}\right)$$

Which is the alternate solution.