What is cos[sin^-1(-3/5)]

cos[sin^(-1)(-3/5)]=cos(-0.6435011087933)=4/5

What is cos[sin^-1(-3/5)]

$$\boxed{~\cos{[\arcsin{(x)}]} = \pm \sqrt{1-x^2}~}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{1-\left(-\dfrac 35 \right)^2}\\\\
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{1- \dfrac{9}{25} }\\\\
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{\dfrac{16}{25} }\\\\
\mathbf{ \cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} } & \mahtbf{=} &
\mathbf{ \pm \dfrac{4}{5} }\\\\
\end{array}
$}}$$

I think it has to be the 4th quadrant because of the restrictions on inverse sine

So cos must be positive.

But since the hypotenuse is 5 and the opp is 3 the adj must be 4 ( just like Heureka said)    so it is +4/5

Let me see if I can explain the discrepencies between jdh3010's and  heureka's answers.....

Since the sine inverse returns an angle value in either the 4th or 1st quadrants, jdh3010's answer of (4/5) would be correct.....since this is a 4th quad angle .....(Wolframalpha agrrees with this result)

However, the angle could also plausibly fall into the 3rd quadrant since the sin would negative there, as well....and the cosine is also negative in this quadrant.......so heureka's  answer covers both possibilities

We would actually have to know the angle value to make the proper determination...........

No it can't be in the 3rd quad

Inverse sine is confined to -pi/2 to pi/2

I'll agree, Melody...that, in general, we would have to consider it to be a fouth quad angle...but.....we would have to admit that, if it was a third quad angle, the sine would be negative and thus, as heureka's second answer indicates, the cosine would be negative there, as well........

$$\small{\text{$
\begin{array}{lrcl}
1. & \sin{(A)} &=& -\frac 35 \quad \Rightarrow \quad A = \arcsin{(-\frac 35 )} = -36.87\ensurement{^{\circ}} ~ (IV.)\\\\
2. & \sin{(180\ensurement{^{\circ}}-A)} &=& -\frac 35 \quad \Rightarrow \quad A = 180\ensurement{^{\circ}}- \arcsin{(-\frac 35 )} = 180\ensurement{^{\circ}} + 36.87\ensurement{^{\circ}} = 216.87\ensurement{^{\circ}} ~(III.)\\\\
\end{array}
$}}$$

And each sqrt has a plus and a minus

I do not believe that your point 2 is relevant.

Arc sine must be between -pi/2 and pi/2 

It is only neg in the 4th quad.

Cos(4th quad angle) is positive.

WolframAlpha agrees with me (according to Chris)