HELP!! given sin(x)= -3/5, pi<x<3pi/2; cos(y)= 12/13, 3pi<y<2pi find exact value of:

Question

given sin(x)= -3/5, pi<x<3pi/2; cos(y)= 12/13, 3pi<y<2pi find exact value of:

 

a) sin(2x)

 

b) cos(y/2)

Answer 1

For (a), you can use sin(2x)=2sinxcosx. If you draw and "stretch" the unit circle, this is similar to being in the third quadrant in the Cartesian plane; both sine and cosine are negative. Cos x (adj/hyp) would be -4/5 from the 3,4,5 triangle Pythagorean Theorem. Thus, (a) is 2*(-3/5)*(-4/5) or 24/5.

 

For (b), you can apply the half-angle formula: sqrt((1+cosy)/2). (This would be negative if cos(y/2) is greater than zero). 

 

I am confused about the limits of (b), however. Is it (2pi,3pi)?