# 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?

• Hint: $15^2+8^2=17^2$ Mar 19 '18 at 21:38
• @PrasunBiswas actually $8^2$ instead of $6^2$. I knew that, but still wasn't able to do it, because as it turns out i forgot about double angle formula Mar 19 '18 at 21:40
• Hint #2: $2\times\dfrac 8{17}\times\dfrac{15}{17}=\dfrac{240}{289}$ Mar 19 '18 at 21:41
• how did this simple question get that many upvotes? Aug 4 '18 at 17:45
• @James I have no idea lol, I was thinking the same thing Aug 8 '18 at 19:54

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.

• My students don't reliably remember the condition on your second equality, so I'd recommend including it. ", if $-1 \leq x \leq 1$." Mar 20 '18 at 15:13
• @EricTowers I might be one of those stupid students in this case, but if you allow complex numbers, doesn't that second equality hold? Or is there some multivalued function weirdness going on that you have to care about? Mar 20 '18 at 21:00
• Not "stupid" -- focusing on the "big", misremembering the "small". Since this Question is tagged "trigonometry", we're very likely talking about real-valued functions. The domain of real-valued arccosine is $[-1,1]$. (The usual extension to the complex plane has branch cuts along the rest of the real line. There is definitely some multivalued weirdness going on; cosine isn't injective, so this is unavoidable.) In fact, in Trig, any time a student uses the shortcut you write, they are expected to explicitly write that they have verified that $x$ is in the domain of arccosine. Mar 21 '18 at 0:01
• @EricTowers: Arccos is multivalued, but for all possible values $y$ of $\cos^{-1}x$ you have $\cos y=x$, by definition. Thus, in complex numbers $\cos\cos^{-1}x=x$ is always true. The converse is not - you could have $\cos^{-1}\cos x\neq x$ if you chose the wrong branch. Whether students in a trig class should first show that $-1\le x\le1$, is more a question of pedagogy than mathematics. Mar 21 '18 at 10:18
• The same is with other inverse functions - for any branch of $\sqrt{x}$, you always have $\left(\sqrt{x}\right)^2=x$, but $\sqrt{x^2}$ can be either $x$ or $-x$ depending on $x$ and the branch. Mar 21 '18 at 10:22

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

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

\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}

• Nice idea.͏͏͏͏͏͏ Mar 21 '18 at 20:01
• I love it! Draw the triangle! Mar 23 '18 at 19:00
• I didn't notice this answer before, but it deserves more points, I like that idea. If I could give two answers best answer, you would be the best answer too Apr 18 '18 at 14:27

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

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

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

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

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