Trig. troubles...?

a)

sin(2tan^-1(-3/4))

let 2tan^-1(-3/4) = x

tan^-1(-3/4) = x/2

so tan(x/2) = -3/4

tan(x/2) = sinx /(1+cosx)

so sinx /(1+cosx) = -3/4

4sinx = -3 - 3cosx

3cosx = -3 - 4sinx

9cos^2(x) = 9 + 16sin^2(x) + 24sinx

9 - 9sin^2(x) = 9 + 16sin^2(x) + 24sinx

25sin^2(x) = 24sinx

25sinx = 24

sinx = 24/25

so sin(2tan^-1(-3/4)) = sinx = 24/25

b)

cos[sin^-1(-1/√5) - tan^-1(-2)]

let sin^-1(-1/√5) = A

sinA = -1/√5 ------cosA = sqrt(1-sin^2(A)) = sqrt(1-1/5) = sqrt(4/5)
cosA = 2/√5

let tan^-1(-2) = B

tanB = -2 : sinB/cosB = -2 : opp.side = 2 and adj.side = 1

hypotenuse = sqrt(4+1) = sqrt(5)

sinB = opp.side/hypotenuse = -2/√5 : cosB = 1/√5

so cosA = 2/√5, cosB = 1/√5, sinA = -1/√5, sinB = -2/√5

cos[sin^-1(-1/√5) - tan^-1(-2)] = cos(A-B)

cosAcosB + sinAsinB = (2/√5)(1/√5) +( -1/√5)(-2/√5)

=>2/5 + 2/5 = 4/5

Some sketches of a unit circle on graph paper will be of considerable help in dealing with this.

a. -24/25

b. 4/5