# Simplify the expression: sin(2cos^(-1)6x) Thoughts? =)

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`sin(2cos^(-1)(6x))`

Using the double angle formula

`sin(2theta)=2cos(theta)sin(theta) `

we get

`sin(2cos^(-1)(6x))=2cos(cos^(-1)(6x))sin(cos^(-1)(6x))`

Now `sin(theta)=sqrt(1-(cos(theta))^2)` by the pythagorean theorem

`sin(cos^(-1)(6x))=sqrt(1-(cos(cos^(-1)(6x)))^2)`

And using `cos(cos^(-1)(a))=a` by definition we get

`2cos(cos^(-1)(6x))sin(cos^(-1)(6x))=2(6x)(sqrt(1-(6x)^2))`

We get by simplifying

`sin(2cos^(-1)(6x))=12xsqrt(1-36x^2)` which is our answer

Let cos^-1(6x) = arccos 6x

sin(2cos^(-1)(6x)) = sin(2arccos (6x))

We'll use the following identity:

sin (2alpha) = 2sin (alpha)*cos(alpha)

We'll put alpha = arccos 6x

sin(2arccos (6x)) = 2sin (arccos 6x)*cos(arccos 6x)

But sin(arccos alpha) = sqrt(1 - alpha^2) and cos(arccos alpha) = alpha

sin (arccos 6x) = sqrt(1 - 36x^2)

cos (arccos 6x) = 6x

sin(2arccos (6x)) = 2*6x*sqrt(1 - 36x^2)

sin(2arccos (6x)) = 12xsqrt(1 - 36x^2)

**Therefore, the requested value is sin(2arccos (6x)) = 12xsqrt(1 - 36x^2).**