Question

how do i work this problem sin(2cos^-1 1/11)

Answer 1

$\sin\left(2\cos^{-1}\dfrac1{11}\right)=2\sin\left(\cos^{-1}\dfrac1{11}\right)\cos\left(\cos^{-1}\dfrac1{11}\right)$

When $0<x<\pi$, you have $\cos(\cos^{-1}x)=x$, so you can simplify the above slightly to get

$\sin\left(2\cos^{-1}\dfrac1{11}\right)=\dfrac2{11}\sin\left(\cos^{-1}\dfrac1{11}\right)$

Now picture a right triangle and pick one of the non-right angles. Let this angle's cosine be $\dfrac1{11}$. This means the leg adjacent to this angle must have length 1, while the hypotenuse must have length 11. By the Pythagorean theorem, the length of the remaining leg must be

$\sqrt{11^2-1^1}=\sqrt{121-1}=\sqrt{120}=2\sqrt{30}$

This means the sine of this angle is $\dfrac{2\sqrt{30}}{11}$, and so

$\sin\left(2\cos^{-1}\dfrac1{11}\right)=\dfrac{4\sqrt{30}}{121}$