How does WolframAlpha get an exact answer here? ${}{}$ I had a simple thing to compute with a calculator:
$$\sin\left(2\cos^{-1}\left(\frac{15}{17}\right)\right)$$
I got the decimal answer of about $0.83044983$, but when I typed it in WolframAlpha, it also gave an exact answer of $\frac{240}{289}$. How in the world would one get an exact answer here?
 A: 
\begin{align}
   \sin \theta &= \dfrac{8}{17} \\
   \cos \theta &= \dfrac{15}{17} \\
\hline
   \sin\left(2 \arccos \dfrac{15}{17} \right) 
   &= \sin(2 \theta) \\
   &= 2 \sin(\theta) \cos(\theta) \\
   &= \cdots
\end{align}
A: $ (8,15,17)$ are lengths of a Pythagorean triple right triangle. A narrow right triangle of these side lengths can be drawn if needed.
$$\sin(2\cos^{-1}\frac{15}{17}) = \sin(2\sin^{-1}\frac{8}{17}) = 2 \cdot \frac{8}{17}\cdot \frac{15}{17} =\frac{240}{289}.$$
A: Here are three relevant formulas:


*

*$\sin 2x = 2 \sin x \cos x$.

*$\cos \cos^{-1} x = x.$

*$\sin \cos^{-1} x = \sqrt{1 - x^2}$ after drawing an appropriate right triangle.
Combining these three to get the desired conclusion is left to the interested reader.
A: Call
$$
u = \cos^{-1}\frac{15}{17}
$$
Therefore
$$
\cos u = \frac{15}{17}
$$
and
$$
\sin u = \sqrt{1 - \cos^2 u} = \sqrt{1 - \frac{15^2}{17^2}} = \frac{8}{17}
$$
With these two you just need to calculate
$$
\sin 2u = 2\sin u \cos u = 2\frac{15}{17}\frac{8}{17} = \color{blue}{\frac{240}{289}}
$$
A: Let $\cos^{-1}x=y\implies\cos y=x$
Using Principal values,  $0\le x\le\pi\implies\sin y\ge0$
and $\sin y=+\sqrt{1-\cos^2y}=?$
Finally, $\sin2(\cos^{-1}x)=\sin2y=2\sin y\cos y=?$
A: Note that $$\sin\left(2\cos^{-1}(a)\right)=2\sin\left(\cos^{-1}(a)\right)\cos\left(\cos^{-1}(a)\right)=2a\sin\left(\cos^{-1}(a)\right)$$ and use the fact that $$\sin\left(\cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)\right)=\frac{\text{opposite}}{\text{hypotenuse}}$$
A: Hint.....$$\sin2x=2\sin x\cos x$$
