It is known that sin α = 13 and α is the angle of the second quadrant. (1) find the value of sin (α − π 6); (2) find the value of COS 2 α

It is known that sin α = 13 and α is the angle of the second quadrant. (1) find the value of sin (α − π 6); (2) find the value of COS 2 α

(1) ∵ sin α = 13, and α is the angle of the second quadrant ∵ cos α = - 1 − sin2 α = - 223 ∵ sin (α − π 6) = sin α· cos π 6-cos α· sin π 6 = 3 + 226 (2) Cos2 α = 1-2sin2 α = 1-29 = 79

2sin^2 a·cos^2 a=(1/2)sin^2 a
How did it come about?

This uses the double angle formula sin2a = 2sinacosa
∴ 2sin²a*cos²a
=(1/2)*(2sinacosa)²
=(1/2)*sin²2a

If θ is the second quadrant angle and sin θ / 2-cos θ / 2 = √ (1-sin θ), then what quadrant angle is θ / 2?

Drawing method, using y = x and y = - x bisection coordinate system, we can get eight parts. From the first quadrant, we can get 1-8 parts anticlockwise, sin θ / 2-cos θ / 2 > 0, which shows that θ / 2 is in 2,3,4,5 four parts, and θ is the second quadrant, that is 90 ° + 2pi

Let cos (a-b / 2) = - 1 / 9, sin (A / 2-B) = 2 / 3, and a be the second quadrant angle, and B be the first quadrant angle, then find the value of COS (a + b) / 2