sina=3/5, pi/2<a<pi evaluate cos2a
cos2a = 1 - 2sin^2(a)
= 1 - 2*9/25
...
in this case, it doesn't really matter which quadrant a is in, as sin^2 is always positive.
extra credit: which quadrant is 2a in?
To evaluate cos(2a) given that sin(a) = 3/5 and π/2 < a < π, we need to use the double-angle identity for cosine.
The double-angle identity for cosine states that:
cos(2a) = cos²(a) - sin²(a)
To find cos(a), we can use the Pythagorean identity:
sin²(a) + cos²(a) = 1
Given that sin(a) = 3/5, we can substitute this value into the Pythagorean identity:
(3/5)² + cos²(a) = 1
9/25 + cos²(a) = 1
Now, we can solve for cos²(a):
cos²(a) = 1 - 9/25
cos²(a) = 16/25
Now that we know cos²(a), we can substitute it back into the double-angle identity for cosine to find cos(2a):
cos(2a) = cos²(a) - sin²(a)
= 16/25 - (3/5)²
= 16/25 - 9/25
= 7/25
Therefore, cos(2a) = 7/25.