How does WolframAlpha get an exact answer here? ${}{}$

Question

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?

Answer 1

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.

Answer 2

Hint.....$$\sin2x=2\sin x\cos x$$

Answer 3

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

$$ \sin 2u = 2\sin u \cos u = 2\frac{15}{17}\frac{8}{17} = \color{blue}{\frac{240}{289}} $$

Answer 4

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, $\sin2(\cos^{-1}x)=\sin2y=2\sin y\cos y=?$

Answer 5

$ (8,15,17)$ are lengths of a Pythagorean triple right triangle. A narrow right triangle of these side lengths can be drawn if needed.

$$\sin(2\cos^{-1}\frac{15}{17}) = \sin(2\sin^{-1}\frac{8}{17}) = 2 \cdot \frac{8}{17}\cdot \frac{15}{17} =\frac{240}{289}.$$

Answer 6

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 $$\sin\left(\cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)\right)=\frac{\text{opposite}}{\text{hypotenuse}}$$

Answer 7

\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}