Question

Find the exact values of the six trigonometric functions at “a” given cos(2a) = - 4/5 and a is

in the 2nd quadrant.

Answer 1

If a is in the second quadrant, then cos(a) < 0 and sin(a) > 0.

Recall the double angle identity for cosine:

cos(2a) = 2 cos²(a) - 1 = 1 - 2 sin²(a)

It follows that

2 cos²(a) - 1 = -4/5 ==> cos²(a) = 1/10 ==> cos(a) = -1/√10

1 - 2 sin²(a) = -4/5 ==> sin²(a) = 9/10 ==> sin(a) = 3/√10

Then we find

1/cos(a) = sec(a) = -√10

1/sin(a) = csc(a) = √10/3

sin(a)/cos(a) = tan(a) = -3

1/tan(a) = cot(a) = -1/3