Question

Find the exact value of cos[cot^-1 (-√3) + sin^-1 (-1/2)].

I'm having trouble with inverses. Please help by showing work.

Answer 1

cos[cot^-1 (-√3) + sin^-1 (-1/2)]

let's take it in parts

cot^-1 (-√3)
is the angle so that cotØ = -√3
or tanØ = -1/√3
I know tan30° = +1/√3
so Ø = 180-30 = 150° or -30°
(usually we take the smallest positive angle)

sin^-1 (-1/2)
= 180+45 = 225°

cos[cot^-1 (-√3) + sin^-1 (-1/2)]
= cos(150 + 225)
= cos 375°
= cos(360 + 15)
= cos15
= cos(45-30)
= cos45cos30 + sin45sin30
= (√2/2)(√3/2) + (√2/2)(1/2)
= (√6 + √2)/4

Answer 2

just for ease of readability, let's say

x = cot^-1 (-√3)
y = sin^-1 (-1/2)

The principal values of these inverse trig functions will be in QIV, so draw the triangles there. Then you can see that

sinx = -1/2
cosx = √3/2

siny = -1/2
cosy = √3/2

cos(x+y) = cosx cosy - sinx siny
= √3/2 * √3/2 - 1/2 * 1/2
= 3/4 - 1/4
= 1/2

Or, you could just recognize that
x = y = -π/6
so cos(x+y) = cos(-π/3) = 1/2

Answer 3

go with Steve's answer,

I forgot that we could pin-point the quadrant (although mine had the more-fun calculations)

Answer 4

Thanks so much! I'm starting to get it. :)