A) First, we must reconstruct the right triangle. We can safely assume this is a 3-4-5 right triangle based of the inverse cosine value. Then we must use our trigonometric identities. Finally we find our angles & evaluate the trigonometric ratio for each angle.
Sin 2x = 2*sin x * cos x
Sin 2(arccos -3/5) = 2 * sin (arccos -3/5) * cos (arccos -3/5) = 2(4/5)(-3/5) = -24/25
cos 2x = 1 - 2sin2x
cos 2(sin-1 24/25) = 1 - 2sin2(sin-1 24/25) = 1 - 2(24/25)2 = 1 - 1152/625 = -527/625
Tan 2x = (2*tan x)/(1 - tan2x)
Tan 2(tan-1 3/7) = (2*tan (tan-1 3/7))/(1 - tan2(tan-1 3/7)) = (2 * 3/7)/(1 - 9/49) = (6/7)/(40/49) = 21/20
B) For these questions, we must first find the reference angle. Then we must ascertain which quadrant the angle resides in. Lastly, we can find the angle.
sin (tan-1 -√3) The reference angle is 60 degrees (π/3) & the quadrant is 2 since tangent is negative there.
sin (180º - 60º) = sin 120º = √3/2
cos (sin-1 1) The reference angle is 90 degrees in quadrant 1
cos 90º = 0
tan (cos-1 0) The reference angle is 90º in quadrant 1
tan 90º = ∞
