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Consider we have been given that

$$\arcsin \left(1-x\right)+\arccos \left(\frac{1}{3}\right)=\frac{\pi }{2}$$

How would you solve this trigonometric equation? In other words, Is there any difference between

$$\sin \left(1-x\right)+\cos \left(\frac{1}{3}\right)=\frac{\pi }{2}$$

Regards!

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closed as off-topic by José Carlos Santos, cansomeonehelpmeout, Chris Custer, max_zorn, Namaste Jul 26 '18 at 0:25

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    $\begingroup$ It is as if you were asking for the difference between $x^2+e^x=2$ and $\tan\sqrt x=x+5$. $\endgroup$ – Yves Daoust Jul 25 '18 at 10:45
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You should draw a picture, you have to angles that add up to $90^\circ$ ($\frac{\pi}{2}$ radians). enter image description here

  • Where in the picture is $\arcsin(1-x)$?
  • Where in the picture is $\arccos(\frac{1}{3})$?
  • Can you find a relationship between the two?

(Edit)

Since OP has already accepted an answer, I'll add some more details. $\arcsin(1-x)$ and $\arccos(\frac{1}{3})$ are just two angles that add up to $90^\circ$, and it doesn't matter which angle is which in the picture.

enter image description here

Geogebra won't let me write in LaTeX, so let $$\alpha=\arcsin(1-x)$$ $$\beta=\arccos(\frac{1}{3})$$ Then from the figure we see that $\color{red}{\frac{1}{3}}$ is the $\color{red}{red}$ line, and $\color{green}{1-x}$ is the $\color{green}{green}$ line. These are equal in length, so you have to solve $$1-x=\frac{1}{3}$$ I'll leave the rest to you!

Notice that this proves a more general result:

Let $\alpha+\beta=90^\circ$, then $\sin(\alpha)=\cos(\beta)$, $\sin(\beta)=\cos(\alpha)$.

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    $\begingroup$ Wish I could upvote this more than once ~ $\endgroup$ – DanielV Jul 25 '18 at 11:13
  • $\begingroup$ I couldn't find where they are. $\endgroup$ – Cargobob Jul 25 '18 at 11:22
  • $\begingroup$ @Cargobob Ok, I'll try to help you some more. When looking at $\arcsin(1-x)$ and $1-x$, which of the two is a length, and which of the two is an angle? $\endgroup$ – cansomeonehelpmeout Jul 25 '18 at 11:26
  • $\begingroup$ @DanielV Divines' bless your kind heart! $\endgroup$ – cansomeonehelpmeout Jul 25 '18 at 11:55
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    $\begingroup$ What a great answer this is! $\endgroup$ – Cargobob Jul 25 '18 at 12:07
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$$1-x=\sin\left(\frac\pi2-\arccos\frac13\right).$$

This can be further simplified.

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  • $\begingroup$ How did you find that? Perhaps there would be an identitiy. $\endgroup$ – Cargobob Jul 25 '18 at 10:48
  • $\begingroup$ @Cargobob ! It comes from the definition of $sin$ and $arcsin$. They are reversed functions, for example, $sin(\frac{\pi}{2})=1$ and $arcsin(1)=\frac{\pi}{2}$ $\endgroup$ – yW0K5o Jul 30 '18 at 20:37
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By using the identity

$$\arcsin(x)+\arccos(x)~=~\frac{\pi}2 $$

The equation becomes

$$\begin{align} \arcsin(1-x)+\left[\frac{\pi}2-\arcsin\left(\frac13\right)\right]~&=~\frac{\pi}2\\ \arcsin(1-x)~&=~\arcsin\left(\frac13\right)\\ 1-x~&=~\frac13 \end{align}$$

and therefore you get $x=\frac23$.


On the other side you will get

$$\begin{align} \sin(1-x)+\cos\left(\frac13\right)~=~\frac{\pi}2\\ \sin(1-x)~=~\frac{\pi}2-\cos\left(\frac13\right)\\ 1-x~=~\arcsin\left(\frac{\pi}2-\cos\left(\frac13\right)\right)\\ x~=~1-\arcsin\left(\frac{\pi}2-\cos\left(\frac13\right)\right)\\ \end{align}$$

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Take $\cos$ of both sides, expand and then use https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Inverse_trigonometric_functions, solve the quadratic to get $x=1\pm1/3$, from which only $2/3$ qualifies as a real solution.

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