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Let $$ \left\{ \begin{array}{c} \sin x = \frac{4}{5} \\ \cos x = \frac{3}{5} \end{array} \right. $$

Find all of the possible values for $x$.

My try: By dividing the equations we obtain $\tan x = \frac{4}{3}$ and then $$x = \arctan\frac{4}{3} + k\pi$$ But WolframAlpha gives $$x = 2k\pi + 2\arctan\frac{1}{2}$$

Using $\arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$, we get $$2\arctan\frac{1}{2} = \arctan\frac{4}{3}$$ but the answers are different still.

Why does this happen? And what is the correct answer?

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S.H.W
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2 Answers2

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Your original (set of) equations implies $\tan x=\frac{4}{3}$ but not the other way around. When you solve $\tan x=\frac{4}{3}$ you get the solutions of your original equation and the solutions $\sin x=\frac{-4}{5}$, $\cos x=\frac{-3}{5}$.

  • Thanks, so after getting $x = \arctan\frac{4}{3} + k\pi$ what is the next step? – S.H.W Jul 23 '19 at 10:24
  • If $k$ is even then it's the solution of your equation, if $k$ is odd it's the solution of the wrong one ($\sin\theta=-4/5$, $\cos\theta=-3/5$). –  Jul 23 '19 at 10:26
  • Think about the position of $\sin(x)$ and $\cos(x)$ in the unit circle. Clearly they are in the first quadrant, so any possible answers $x$ should satisfy $$2k\pi\leq x\leq 2k\pi+\frac{1}{2}\pi.$$ – Floris Claassens Jul 23 '19 at 10:27
  • @FlorisClaassens That's right, so in general for solving trigonometric equations putting the answer in the original equations is always necessary? – S.H.W Jul 23 '19 at 10:33
  • If you had just solved arccos($\frac{3}{5}$) and arcsin($\frac{4}{5}$) you can find the answers by taking the values they both have in common. In that case it is not necessary to go back to the original equations. If you use the arctan way, which I would advice it is necessary. In all cases of trigonometry I would definitely advise to always consider the original equations to see if the answers you got make sense. – Floris Claassens Jul 23 '19 at 10:38
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    @S.H.W In general for solving any equations it should always be possible to put the answer back in the original equation. If you can be absolutely sure that all your steps were "if and only if" steps then you may be able to skip that final check, but if you do not know how to prove that it is unnecessary, you should plug back into the final equation before writing your final answer. – David K Jul 23 '19 at 10:38
  • @FlorisClaassens Okay, thanks a lot. – S.H.W Jul 23 '19 at 10:58
  • @DavidK It was really a helpful advice, thanks. – S.H.W Jul 23 '19 at 11:00
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$\cos x=\dfrac35$

$\implies x=2n\pi\pm\arccos\dfrac35$

As for $x>0,\arccos x=\arcsin\sqrt{1-x^2}=\arctan\dfrac{\sqrt{1-x^2}}x$

$\implies x=2n\pi\pm\arcsin\dfrac45$ where $n$ is any integer

Now , $\sin x=\sin\left(2n\pi\pm\arcsin\dfrac45\right)=\pm\dfrac45$

But $\sin x=+\dfrac45$

$$\implies x=2n\pi+\arcsin\dfrac45=2n\pi+\arccos\dfrac35=2n\pi+\arctan\dfrac43$$

Further if $2y=\arctan\dfrac43,\dfrac43=\tan2y=\dfrac{2\tan y}{1-\tan^2y}$

Solve the quadratic equation in $\tan y$ to find the values to be $\dfrac12,-2$

But as $0<2y<\dfrac\pi2, y=\arctan\dfrac12$

which can be validated using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

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