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

Question

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!

Answer 1

You should draw a picture, you have to angles that add up to $90^\circ$ ($\frac{\pi}{2}$ radians).

• 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?

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(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.

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 $\sin(\alpha)=\cos(\beta)$, $\sin(\beta)=\cos(\alpha)$.

Answer 2

$$1-x=\sin\left(\frac\pi2-\arccos\frac13\right).$$

This can be further simplified.

Answer 3

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 $x=\frac23$.

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On the other side you will get

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

Answer 4

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.