Math help... Need it ASAP?

Question

find the exact value:

1. tan[2arccos(-4/5)]

2. cos [arcsin(2/3) + 2arcsin(-1/3)]

Solve the equations

3. sin^2 x +5cos x+2=0 , 0≤x≤2π

4. Find all the solutions to 4sin2θ-3cosθ=0

please show the complete solutions.

thanks... :P

Answer 1

1) arccos(-4/5) = pi - arccos(4/5) so given expr = tan(2pi - arccos(4/5)) = - tan[2arccos(4/5)]  = - tan2T (say) = -{2tanT/(1-tan^2T)} and

tan(arccosm) = [rt(1-m^2)]/m

tan(arccos(4/5)) = 3/5. [Draw a rt triangle 3,4,5 to prove this]

So give expr = (6/5)/[1 - (9/25)] Now simplify.

2) Use cos(A-B) formula as arcsin(-x) = -arcsin(x) and use cos(arcsina)

= rt(1-a^2)

3) Use sin^2x = 1- cos^2x and form a quadratic in cos x and solve.

4) It is 8sinTcosT - 3cosT = 0

so, cosT ( 8sinT -3) = 0 so cosT =0 or sinT = 3/8.

this means T = 3pi/2 or pi/2. or T = arcsin(3/8) [ pl read Theta for T]