find exact value of trig problem?

Question

cos(sin-1(1/3)-tan-1(1/2))

i got it down to cos((2rt8)-3/6) but im not sure if i did it right, anyone willing to check and see if that is it, or maybe your able to simplify it more.

Answer 1

Let A = sin^(-1) (1/3)
Let B = tan^(-1) (1/2)

cos (A - B)
cos A cos B + sin A sin B
(√8/3)(2/√5) + (1/3)(1/√5)
(√8 + 1) / (3√5)
(√5)(√8 + 10) / 15
( (√5) (2)(√2) + 10 √5) / 15
(2√10 + 10√5) / 15
(2/15) (√10 + 10√5)

Answer 2

sin(A) = 1/3 so cos(A) = sqrt(8)/3

tan(B) = 1/2 so sin(B) = 1/sqrt(5) and cos(B) = 2/sqrt(5)

using Pythagoras and trig relations

also cos (A - B) = cos(A)cos(B) + sin(A)sin(B) =
sqrt(8) / 3 * 2/sqrt(5) + 1/3 * 1/sqrt(5) =
[ 2sqrt(8)+1 ] / [ 3sqrt(5) ]

Answer 3

tan^-1 (1/2) = 30 degrees = pi/6 radians
cos(sin^-1 (1/3) = cos (cos^-1(2sqrt(2)/3) = 2sqrt(2)/3
So answer is 2sqrt(2)/3 - pi/6

Answer 4

cos(sin-1(1/3)) = sqrt(8)/3
cos(tan-1(1/2)) = 2/sqrt(5)
sin(tan-1(1/2)) = 1/sqrt(5)

Therefore,
cos(sin-1(1/3)-tan-1(1/2))
= cos(sin-1(1/3)) cos(tan-1(1/2)) + sin(sin-1(1/3)) sin(tan-1(1/2))

Can you finish it now?