SOLUTION: Give sin(a) =8/9, pi/2< a < pi, Find the exact value of sin(a/2) I tried solving it but I got stuck in some part. Here is my step for solving it. Do pythagoream theorem to

Algebra ->  Trigonometry-basics -> SOLUTION: Give sin(a) =8/9, pi/2< a < pi, Find the exact value of sin(a/2) I tried solving it but I got stuck in some part. Here is my step for solving it. Do pythagoream theorem to      Log On


   

Question 1007340: Give sin(a) =8/9,
pi/2< a < pi,
Find the exact value of sin(a/2)
I tried solving it but I got stuck in some part.
Here is my step for solving it.
Do pythagoream theorem to find cos (a)
sin(a)=8/9 = y/r
y=8, r=9
r^2=x^2+y^2
9^2=x^2+8^2
81=x^2+64
sqrt(17)=x
cost=x/r=sqrt(17)/9
use double angle formula, sin(x/2)= +/- sqrt((1-costx)/2)
sin(x/2)= +/- sqrt((1-sqrt((17)/9))/2)
sin(x/2)= +/- sqrt((9/9-sqrt(17)/9))/2), I am stuck here.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If sin%28a%29+=+8%2F9, then cos%28a%29+=+%28sqrt%2817%29%29%2F%289%29. So you are correct there.

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Start with pi%2F2+%3C+a+%3C+pi.

Multiply every side by 1%2F2 to get pi%2F4+%3C+a%2F2+%3C+pi%2F2

The angle a%2F2 is larger than pi%2F4 but smaller than pi%2F2. This places angle a%2F2 in quadrant 1. So sin%28a%2F2%29 is positive. So we only use the "plus" version of the formula (NOT the minus, NOT the plus/minus)


sin%28a%2F2%29+=+sqrt%28%281-cos%28a%29%29%2F2%29

sin%28a%2F2%29+=+sqrt%28%281-%28sqrt%2817%29%29%2F%289%29%29%2F2%29

sin%28a%2F2%29+=+sqrt%28%289%2F9-%28sqrt%2817%29%29%2F%289%29%29%2F2%29

sin%28a%2F2%29+=+sqrt%28%28%289-sqrt%2817%29%29%2F%289%29%29%2F2%29

sin%28a%2F2%29+=+sqrt%28%28%289-sqrt%2817%29%29%2F%289%29%29%2A%281%2F2%29%29

sin%28a%2F2%29+=+sqrt%28%289-sqrt%2817%29%29%2F%2818%29%29 Exact value

sin%28a%2F2%29+=+0.52051760426961 Approximate value (use a calculator)