I can't understand why, after evaluating this, I get √3/2. I need to evaluate cos(sin^-1(-3/5). sin^-1 is the Inverse of sin, or arcsin. I end up with cos(4/5), after performing the pythagorean theorem getting x=4. When I draw out cos(4/5) "x/r" with -3 as y, I end up with √3/2 with my knowledge of right triangles on the unit circle. I think I am wrong here, would anyone else evaluate this a different way?

Question

I can't understand why, after evaluating this, I get √3/2.  I need to evaluate cos(sin^-1(-3/5).  sin^-1 is the Inverse of sin, or arcsin.  I end up with cos(4/5), after performing the pythagorean theorem getting x=4.  When I draw out cos(4/5) "x/r"  with -3 as y, I end up with √3/2 with my knowledge of right triangles on the unit circle.  I think I am wrong here, would anyone else evaluate this a different way?

anonymous · Answer

cos(sin^-1(-3/5)
Where is the missing parentheses ?  There are two open and on closed.

anonymous · Answer

Sorry, "There are two open and one closed."

anonymous · Answer

$$\cos \left(\sin ^{-1}\left(-\frac{3}{5}\right)\right)=\frac{4}{5} $$

anonymous · Answer

$$\sqrt{25-9}=4 $$

anonymous · Answer

In a 5,4,3 right triangle one of the Cosines is 4/5.

anonymous · Answer

My calculator was terribly faulty.  I was being lazy.  Thank you.  It just had me really confused.