Inverse Functions & Trig. Equations
posted by Claire .
7. Evaluate sin[sin^1(1/2)].
a. 1/2
b. ð /6
c. 1/2
d.  ð /6
8. Evaluate cos[tan^1(5/12)].
a. 12/13
b. 12/5
c. 5/12
d. 5/13
9. Evaluate cos[sin^1(3/5)].
a. 4/5
b. 5/4
c. 4/5
d. 3/4

I don't know at what level you have studied inverse trig functions, but Sal Khan has hundreds of very simple videos on Youtube.
http://www.youtube.com/watch?v=JGU74wbZMLg
Let's look at your #9
first of all sin^1 (3/5) to me means ..
give me the angle Ø so that sinØ = 3/5
and since sin (of angle) = opposite/hypotenuse, I see the famous 345 triangle.
But since the sine is negative, I could sketch that triangle in quadrants III or IV,
(I will pick IV since Ø is the smallest angle)
so now we want
cos(the angle of that triangle) = adj/hyp = 4/5
Of course you could just blindly let your calculator do the work for you ...
on mine :
2nd F
sin
(3/5)
=
cos
=
and I got 0.8 which is 4/5
Now do #2 the same way.
BTW, wouldn't #1 be 1/2 by the
"Just look at it" Theorem ?