SOLUTION: if sin alpha = 1/ square root 5 , 0< alpha < pie by 2,. find the exact values of a) cos alpha b) sin2 alpha c) cos 2 alpha

Algebra ->  Trigonometry-basics -> SOLUTION: if sin alpha = 1/ square root 5 , 0< alpha < pie by 2,. find the exact values of a) cos alpha b) sin2 alpha c) cos 2 alpha      Log On


   

Question 818904: if sin alpha = 1/ square root 5 , 0< alpha < pie by 2,. find the exact values of
a) cos alpha
b) sin2 alpha
c) cos 2 alpha
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sin%28alpha%29+=+1%2Fsqrt%285%29
a) cos%28alpha%29
cos%5E2%28alpha%29+=+1+-+sin%5E2%28alpha%29
cos%5E2%28alpha%29+=+1+-+%281%2Fsqrt%285%29%29%5E2
cos%5E2%28alpha%29+=+1+-+1%2F5
cos%5E2%28alpha%29+=+4%2F5
Now we will use a square root of each side. Usually one needs to use a + when finding the square root of both sides of the equation. But since 0+%3C+alpha+%3C+pi%2F2 alpha terminates in the 1st quadrant where all Trig functions are positive. So we only need the positive square root:
cos%28alpha%29+=+sqrt%284%2F5%29
cos%28alpha%29+=+sqrt%284%29%2Fsqrt%285%29
cos%28alpha%29+=+2%2Fsqrt%285%29

b) sin%282alpha%29
Using sin(2x) = 2sin(x)cos(x) on this:
2sin%28alpha%29cos%28alpha%29
Now just substitute in sin%28alpha%29 (which was given) and cos%28alpha%29 which we found in part a. Then simplify.

c) cos%282alpha%29
There are 3 variations to cos(2x). Any of them may be used. I'm going to use cos%282x%29+=+2cos%5E2%28x%29-1:
2cos%5E2%28alpha%29-1
Now just substitute cos%28alpha%29 which we found in part a. Then simplify.