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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?

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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, $\sin\left(\arcsin\left(\frac45\right)+\arccos\left(\frac1{\sqrt{50}}\right)\right)=\frac{3\pi}4$.

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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.

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  • $\begingroup$ So even some formulae (for example, the ones I have used above) are not helpful sometimes? $\endgroup$
    – user36339
    Jun 28 '17 at 8:35
  • $\begingroup$ @Tug'tekin Now I see that my solution based on your $\arccos$, which is helpful if we'll end this solution. $\endgroup$ Jun 28 '17 at 9:42
  • $\begingroup$ 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. $\endgroup$
    – user36339
    Jun 30 '17 at 18:33

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