SOLUTION: the question is : Find the exact value of cos(u/2) if sinu= -12/13 (3pi/2<u<2pi) I got -2&#8730;13/13 does anyone know if this is correct?

Algebra ->  Trigonometry-basics -> SOLUTION: the question is : Find the exact value of cos(u/2) if sinu= -12/13 (3pi/2<u<2pi) I got -2&#8730;13/13 does anyone know if this is correct?      Log On


   

Question 950267: the question is : Find the exact value of cos(u/2) if sinu= -12/13 (3pi/2 I got -2√13/13 does anyone know if this is correct?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
: Find the exact value of cos(u/2) if sinu= -12/13 (3pi/2 ***
Identity: cos%28x%2F2%29=sqrt%28%281%2Bcosx%29%2F2%29
sinu=-12/13 (Q4)
cosu=5/13 (working with (5-12-13) reference right triangle in quadrant IV)
cos%28u%2F2%29=-sqrt%28%281%2B%285%2F13%29%29%2F2%29=-sqrt%28%2818%2F13%29%2F2%29%29=-sqrt%28%289%2F13%29%29=-3%2Fsqrt%2813%29=-3sqrt%2813%29%2F3
Check:
sinu=-12/13
u=292.62
u/2=146.31
cos (u/2)=cos(146.31)≈-0.8320
exact value as computed=-3√3/13≈-0.8320