if sinA=2/5, and A is in the second quadrant, find cosA, cos2A, and sin2A

Question

I have a feeling cos2A and sin2A are the same but I'm not exactly sure

Answer 1

2nd quadrant, sine is positive as given,

cosine is negative

sin A = 2/5 --> opposite = 2 and hypotenuse=5;

pythagorean says adjacent= sqrt(25-4)=sqrt(21)

sin A = 2/5

cos A = -sqrt(21)/5

cos 2A = cos^2 - sin^2

= (21/25) - 4/25

= 17/25

sin 2A = 2 sin A cos A

= 2 ( 2/5)(-sqrt(21)/5)

= -4 sqrt(21) / 25