# \cos^{-1} \frac{3}{\sqrt{10}}+\cos^{-1} \frac{2}{\sqrt5}=? Let \cos^{-1} \frac{3}{\sqrt{10}}=\alpha, \cos^{-1} \frac{2}{\sqrt5}=\beta then, \cos \alpha=\frac{3}{\sqrt{10}},

$$\displaystyle{{\cos}^{{-{1}}}{\frac{{{3}}}{{\sqrt{{{10}}}}}}}+{{\cos}^{{-{1}}}{\frac{{{2}}}{{\sqrt{{5}}}}}}=?$$
Let $$\displaystyle{{\cos}^{{-{1}}}{\frac{{{3}}}{{\sqrt{{{10}}}}}}}=\alpha,{{\cos}^{{-{1}}}{\frac{{{2}}}{{\sqrt{{5}}}}}}=\beta$$ then, $$\displaystyle{\cos{\alpha}}={\frac{{{3}}}{{\sqrt{{{10}}}}}},{\cos{\beta}}={\frac{{{2}}}{{\sqrt{{5}}}}}$$
Therefore
$$\displaystyle{\cos{\alpha}}={\frac{{{3}\cdot{2}}}{{{2}\sqrt{{2}}\sqrt{{5}}}}}={\frac{{{3}}}{{{2}\sqrt{{2}}}}}\cdot{\cos{\beta}}$$

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sonorous9n
Use trig identity:$$\displaystyle{{\sin}^{{2}}\theta}+{{\cos}^{{2}}\theta}={1}$$
$$\displaystyle\Rightarrow{\sin{\alpha}}=\sqrt{{{1}-{{\cos}^{{2}}\alpha}}}=\sqrt{{{1}-{\left({\frac{{{3}}}{{\sqrt{{{10}}}}}}\right)}^{{2}}}}={\frac{{{1}}}{{\sqrt{{{10}}}}}}\forall\ {0}\leq\alpha\leq{\frac{{\pi}}{{{2}}}}$$
$$\displaystyle\Rightarrow{\sin{\beta}}=\sqrt{{{1}-{{\cos}^{{2}}\beta}}}=\sqrt{{{1}-{\left({\frac{{{2}}}{{\sqrt{{{5}}}}}}\right)}^{{2}}}}={\frac{{{1}}}{{\sqrt{{{5}}}}}}\forall\ {0}\leq\beta\leq{\frac{{\pi}}{{{2}}}}$$
Now, use trig identity
$$\displaystyle{\cos{{\left(\alpha+\beta\right)}}}={\cos{\alpha}}{\cos{\beta}}-{\sin{\alpha}}{\sin{\beta}}$$
$$\displaystyle{\cos{{\left(\alpha+\beta\right)}}}={\frac{{{3}}}{{\sqrt{{{10}}}}}}{\frac{{{2}}}{{\sqrt{{5}}}}}-{\frac{{{1}}}{{\sqrt{{{10}}}}}}{\frac{{{1}}}{{\sqrt{{5}}}}}={\frac{{{1}}}{{\sqrt{{2}}}}}$$
$$\displaystyle\Rightarrow\alpha+\beta={{\cos}^{{-{1}}}{\frac{{{1}}}{{\sqrt{{2}}}}}}={\frac{{\pi}}{{{4}}}}\ \ \ {\left(\therefore{\cos{{\left(\alpha+\beta\right)}}}\in{\left[-{1},{1}\right]}\right)}$$
$$\displaystyle\therefore{{\cos}^{{-{1}}}{\frac{{{3}}}{{\sqrt{{{10}}}}}}}+{{\cos}^{{-{1}}}{\frac{{{2}}}{{\sqrt{{5}}}}}}={\frac{{\pi}}{{{4}}}}$$
Or alternatively use trig identity
$$\displaystyle{\sin{{\left(\alpha+\beta\right)}}}={\sin{\alpha}}{\cos{\beta}}+{\cos{\alpha}}{\sin{\beta}}$$
$$\displaystyle{\sin{{\left(\alpha+\beta\right)}}}={\frac{{{1}}}{{\sqrt{{{10}}}}}}{\frac{{{2}}}{{\sqrt{{5}}}}}+{\frac{{{3}}}{{\sqrt{{{10}}}}}}{\frac{{{1}}}{{\sqrt{{5}}}}}={\frac{{{1}}}{{\sqrt{{2}}}}}$$
$$\displaystyle\Rightarrow\alpha+\beta={{\sin}^{{-{1}}}{\frac{{{1}}}{{\sqrt{{2}}}}}}={\frac{{\pi}}{{{4}}}}\ \ \ {\left(∵{\sin{{\left(\alpha+\beta\right)}}}\in{\left[-{1},{1}\right]}\right)}$$
$$\displaystyle\therefore{{\cos}^{{-{1}}}{\frac{{{3}}}{{\sqrt{{{10}}}}}}}+{{\cos}^{{-{1}}}{\frac{{{2}}}{{\sqrt{{5}}}}}}={\frac{{\pi}}{{{4}}}}$$
###### Not exactly what you’re looking for?
Kirsten Davis
Hint : apply this formula:
$$\displaystyle{{\cos}^{{-{1}}}{x}}+{{\cos}^{{-{1}}}{y}}={{\cos}^{{-{1}}}{\left[{x}{y}-\sqrt{{{\left({1}-{x}^{{2}}\right)}{\left({1}-{y}^{{2}}\right)}}}\right]}}$$
Put $$\displaystyle{x}={\frac{{{2}}}{{\sqrt{{{10}}}}}}$$ and $$\displaystyle{y}={\frac{{{2}}}{{\sqrt{{5}}}}}$$
Vasquez

Like Proof for the formula of sum of arcsine functions $$\arcsin x+\arcsin y$$
using $$\cos(A+B)$$ and the definition of principal values
$$\displaystyle{{\cos}^{{-{1}}}{x}}+{{\cos}^{{-{1}}}{y}}={\left\lbrace\begin{matrix}{{\cos}^{{-{1}}}{\left({x}{y}-\sqrt{{{\left({1}-{x}^{2}\right)}{\left({1}-{y}^{2}\right)}}}\right)}}&\ \text{ if }\ {{\cos}^{{-{1}}}{x}}+{{\cos}^{{-{1}}}{y}}\leq\pi\\{2}\pi-{{\cos}^{{-{1}}}{\left({x}{y}-\sqrt{{{\left({1}-{x}^{2}\right)}{\left({1}-{y}^{2}\right)}}}\right)}}&\ \text{ otherwise }\ \end{matrix}\right.}$$
Now

$$\begin{array}{}\cos^{-1}x+\cos^{-1}y \leq \pi \\will\ happen \\\Leftrightarrow \cos^{-1}x \leq \pi-\cos^{-1} y=\cos^{-1}(-y) \\\Leftrightarrow \frac{\pi}{2}-\sin^{-1}x \leq \frac{\pi}{2} -\sin^{-1}(-y) \\\Leftrightarrow \sin^{-1} x \geq \sin^{-1}(-y) \\\Leftrightarrow x \geq -y \end{array}$$