SOLUTION: Can someone please help, I have a test tommorow and have no clue what to do on this problem. given sinx=1/4 and 0<x<pi/2, find the exact value of sin2x

Algebra ->  Trigonometry-basics -> SOLUTION: Can someone please help, I have a test tommorow and have no clue what to do on this problem. given sinx=1/4 and 0<x<pi/2, find the exact value of sin2x      Log On


   

Question 78790: Can someone please help, I have a test tommorow and have no clue what to do on this problem.
given sinx=1/4 and 0
Found 2 solutions by mathdoc314, Earlsdon:
Answer by mathdoc314(58) About Me  (Show Source):
You can put this solution on YOUR website!
This can be solved with a double-angle formula
+sin+2x+=+2+sin+x+cos+x+
+cos+2x+=+cos%5E2+%28x%29+-+sin%5E2+%28x%29+
(I remember it better if I see them both at once)
That x is in the first quadrant is useful to know because there you know that the cosine is positive. Therefore, since +sin%5E2+%28x%29+%2B+cos%5E2+%28x%29+=+1+, you can solve and find +cos+x+=+sqrt%281+-+%281%2F4%29%5E2%29+ = +sqrt%2815%2F16%29+ = +sqrt%2815%29%2F4+.
The double angle formula for sin%282x%29 tells me +sin%282x%29+=+2+%2A+%281%2F4%29+%2A+%28sqrt%2815%29%2F4%29+=+%0D%0A%28sqrt%2815%29%29%2F8+ exactly.

Answer by Earlsdon(6294) About Me  (Show Source):