# Find $\arcsin(\frac{4}{5})+\arccos(\frac{1}{\sqrt{50}})$

I can't solve this problem due to the irrational number. Used 2 methods: (1) with Pythagorean theorem I got: \begin{align} \arccos(\frac{1}{\sqrt{50}})&=\theta\\\frac{1}{\sqrt{50}}&=\cos(\theta)\\ \frac{7}{\sqrt{50}}&=\sin(\theta)\\\arcsin(\frac{7}{\sqrt{50}})&=\theta\end{align}So got $$\arcsin(\frac{7}{\sqrt{50}})+\arcsin(\frac{4}{5})=\arccos(\sqrt{1-\frac{7}{\sqrt{50}}}\cdot\sqrt{1-\frac{16}{25}}-\frac{7}{\sqrt{50}}\cdot\frac{4}{5})=...$$The calculations didn't give the desired result (which mustn't include an irrational number) since it has got much more complicated. The second one was by using the formula \begin{align}\arccos(\frac{1}{\sqrt{50}})&=\arcsin(\sqrt{1-\frac{1}{50}})\\&=\arcsin(\frac{7}{\sqrt{50}}),\\\end{align}Which also includes irrationality, although there shouldn't be in an answer. Any other way around?

• Jun 28 '17 at 7:06

## 2 Answers

We have\begin{multline*}\sin\left(\arcsin\left(\frac45\right)+\arccos\left(\frac1{\sqrt{50}}\right)\right)=\\=\sin\left(\arcsin\left(\frac45\right)\right)\cos\left(\arccos\left(\frac1{\sqrt{50}}\right)\right)+\\+\cos\left(\arcsin\left(\frac45\right)\right)\sin\left(\arccos\left(\frac1{\sqrt{50}}\right)\right)=\\=\frac45\times\frac1{\sqrt{50}}+\frac35\times\frac7{\sqrt{50}}=\frac1{\sqrt2}.\end{multline*}Since $\arcsin\left(\frac45\right),\arccos\left(\frac1{\sqrt{50}}\right)\in\left(0,\frac\pi2\right)$, $\arcsin\left(\frac45\right)+\arccos\left(\frac1{\sqrt{50}}\right)\in(0,\pi)$. So, $\arcsin\left(\frac45\right)+\arccos\left(\frac1{\sqrt{50}}\right)=\frac\pi4$ or $\arcsin\left(\frac45\right)+\arccos\left(\frac1{\sqrt{50}}\right)=\frac{3\pi}4$. But $\frac45>\frac1{\sqrt{2}}\Longrightarrow\arcsin\left(\frac45\right)>\arcsin\left(\frac1{\sqrt{2}}\right)=\frac\pi4$. Therefore, $\sin\left(\arcsin\left(\frac45\right)+\arccos\left(\frac1{\sqrt{50}}\right)\right)=\frac{3\pi}4$.

It's obvious that $$0<\arcsin\frac{4}{5}+\arccos\frac{1}{\sqrt{50}}<\pi.$$ But $$\cos\left(\arcsin\frac{4}{5}+\arccos\frac{1}{\sqrt{50}}\right)=\frac{3}{5}\cdot\frac{1}{\sqrt{50}}-\frac{4}{5}\cdot\frac{7}{\sqrt{50}}=-\frac{1}{\sqrt2},$$ which gives the answer: $\frac{3\pi}{4}$.

If we'll use $\sin$ then we'll get two cases and it's a bit of harder, I think.

• So even some formulae (for example, the ones I have used above) are not helpful sometimes? Jun 28 '17 at 8:35
• @Tug'tekin Now I see that my solution based on your $\arccos$, which is helpful if we'll end this solution. Jun 28 '17 at 9:42
• You substituted sine to the left side so I thought that I can also do it to like questions, but it turned out to be incorrect on the post math.stackexchange.com/questions/2342039/…, so could you explain when to use this method? After that your use of that here also has become unclear to me so could you explain why you used this method. Jun 30 '17 at 18:33