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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?

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    $\begingroup$ Hint: Divide both sides $\cos\theta$ and you get? $\endgroup$ – Simply Beautiful Art Jul 26 '17 at 20:55
  • $\begingroup$ @SimplyBeautifulArt oops tan theta $\endgroup$ – red_kb Jul 26 '17 at 21:01
  • $\begingroup$ :D $~{}{}{}{}~$ $\endgroup$ – Simply Beautiful Art Jul 26 '17 at 21:03
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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.

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  • $\begingroup$ Actually, any polynomial involving $\sin$ and $\cos$ can be solved using the method I describe in my answer. Basically, convert the $\sin$ into $\cos$ or the $\cos$ into $\sin$ via the pythagorean theorem and solve the rest using algebra. $\endgroup$ – Simply Beautiful Art Jul 26 '17 at 21:16
  • $\begingroup$ @SimplyBeautifulArt - Nice! Thanks for your answer. In the end, we have to use $\arcsin$ or $\arccos$ anyway though, which are transcendental and require numerical techniques or approximations to solve. But at least all desk calculators can solve them! $\endgroup$ – Myridium Jul 26 '17 at 21:17
  • $\begingroup$ Oh, that's what you meant. I was thinking more along the lines of "impossible to algebraically solve in terms of elementary functions", for example, it is not possible to algebraically solve $x=\cos x$ in the manners we've presented. $\endgroup$ – Simply Beautiful Art Jul 26 '17 at 21:19
  • $\begingroup$ That is what I meant, except I don't consider $\cos$ or $\sin$ as being elementary functions. It's an arbitrary choice what we call elementary functions after all. $\endgroup$ – Myridium Jul 26 '17 at 21:20
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    $\begingroup$ Personally, my definition of an elementary function involves algebraic functions, exponential functions, and algebraic combinations of the two as well as their inverses. Following from Euler's formula, one finds trig functions fall under this definition as well. $\endgroup$ – Simply Beautiful Art Jul 26 '17 at 21:22
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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.

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