How to solve $\cos(\sin^{-1}(-3/5))$? I'm stuck with question $$\cos\Bigl(\sin^{-1}\Bigl(-\frac35\Bigr)\Bigr)$$ 
I looked for the answer in the book and it is $\frac45$ 
I tried solving it using the formula $\sin^2x+\cos^2x=1$
and I also got $\frac45$ as the answer but I got it by inputting the value of $\cos x$ in the given question in place of$$ \sin^{-1}\Bigl(-\frac35\Bigr)$$
But I cant see any logic there.
Please explain it to me.
 A: Hint:
$$\cos(\arcsin(x))=\sqrt{1-x^2},$$ for all $x\in (-\pi/2,\pi/2)$.
A: You can do this purely with (trigonometric) formulas, but here's a more geometric approach.
Since $\sin^{-1}(-a)=-\sin^{-1}(a)$ and $\cos(-a)=\cos(a)$, you have:
$$\cos\left(\sin^{-1}\left(-\tfrac{3}{5}\right)\right)=\cos\left(\sin^{-1}\left(\tfrac{3}{5}\right)\right)$$
Now imagine (or draw!) a right triangle with sides 3, 4 and 5:


*

*$\sin^{-1}\left(\tfrac{3}{5}\right)$ corresponds to the angle with opposite side 3;

*the cosine of this angle is the ratio of the adjacent side (4) and the hypotenuse (5).

A: Actually, for better understanding, you can translate this to English. For this particular expression
$$\cos\Bigl(\sin^{-1}\Bigl(-\frac35\Bigr)\Bigr)$$
what it asks is 

What is the cosine of an angle whose sine value is $-\frac{3}{5}$?

Now, there are two things to be careful here. First, we need to find an angle whose sine value is $-\frac{3}{5}$. Second, we are asked to find the cosine value of this angle.
From here, I refer to StackTD's answer since I find solving it geometrically more beneficial.
A: $\displaystyle \sin^{-1}\left(-\dfrac 35 \right)$ is the angle with its sine ratio equal to  $-\dfrac35$.  Note that angles in the first and the second quadrants have positive sine ratios and angles in the third and the fourth quadrants have negative sine ratios. So, $\displaystyle \sin^{-1}\left(-\dfrac 35 \right)$ must be an angle in the third or the fourth quadrant. We don't want $\sin^{-1}$ to have multiple values and therefore we restrict its range to $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}2\right]$, with $\sin^{-1}$ of a positive number not greater than $1$ has the value in $\displaystyle \left(0, \frac{\pi}2\right]$  and that of a negative number not less than $-1$ has the value in $\displaystyle \left[-\frac{\pi}{2}, 0\right)$ (Also, we have $\sin^{-1}0=0$). This is the way we define the inverse sine function.
So if $\displaystyle \theta = \sin^{-1}\left(-\dfrac 35 \right)$, then $\theta$ is an angle in $\displaystyle \left[-\frac{\pi}{2}, 0\right)$ such that $\displaystyle \sin\theta=-\dfrac35$. $\theta$ is an angle in the fourth quadrant and hence $\cos\theta\ge0$.
As $\sin^2\theta+\cos^2\theta=1$, $\cos^2\theta=\displaystyle 1-\left(-\dfrac35\right)^2=\frac{16}{25}$. $\cos\theta\ge0$ implies that $\cos\theta=\dfrac45$.
A: Let $\sin^{-1}\left(-\frac 35\right)=\alpha$ so that $\sin \alpha =-\frac 35$
Then $\sin^2\alpha +\cos ^2 \alpha =1$ and $\cos^2 \alpha = \frac {16}{25}$
I think this is essentially want you did - or similar to it, and all you need then is some care about which square root you want - which in reality may depend on context, but when no context is stated there is a convention for the principal value which others have explained.
I just wanted to highlight that giving a name like $\alpha$ to a complicated expression can sometimes help you to see what is going on. It is often useful to use a name other than commonly used variable names like $x$ or $n$, because his avoids using the same symbol to mean different things.
A: To find $\cos(\arcsin(-3/5))$, since $\sin(x)=-3/5 $, $x=\arcsin(-3/5)$. Then
$\cos(x)=\pm \sqrt{1-(3/5)^2 }$, since  $\cos^2 x+\sin^2 x=1$, i.e. $\cos(x) = \pm 4/5$.
