How do I solve sin(cos^-1(6x))?

Question

How do I solve sin(cos^-1(6x))?

Answer 1

use the identity sin(arccos(a)) = sqrt(1-a^2)

sqrt(1-(6x)^2)

Answer 2

sin[arccos(6x)] =

let [arccos(6x)] = y

hence: 6x = cosy

thus, rewriting siny in terms of cosy, you get:

siny = ±√(1 - cos²x)

now note that, due to arccosine function range (that is [0, π]), y ends

either in the 1st or in the 2nd quadrant and siny is therefore positive;

thus, taking the plus sign, you get:

siny = √(1 - cos²x)

then plugging in cosy = 6x, you get:

siny = √(1 - cos²x) = √[1 - (6x)²] = √(1 - 36x²)

thus, in conclusion, being [arccos(6x)] = y:

sin[arccos(6x)] = siny = √(1 - 36x²)

I hope it helps...

Bye!