\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}},

killjoy1990xb9 2022-01-03 Answered
\(\displaystyle{{\cos}^{{-{1}}}{\frac{{{3}}}{{\sqrt{{{10}}}}}}}+{{\cos}^{{-{1}}}{\frac{{{2}}}{{\sqrt{{5}}}}}}=?\)
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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Expert Answer

sonorous9n
Answered 2022-01-04 Author has 5435 answers
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
\(\displaystyle{\sin{{\left(\alpha+\beta\right)}}}={\sin{\alpha}}{\cos{\beta}}+{\cos{\alpha}}{\sin{\beta}}\)
\(\displaystyle{\sin{{\left(\alpha+\beta\right)}}}={\frac{{{1}}}{{\sqrt{{{10}}}}}}{\frac{{{2}}}{{\sqrt{{5}}}}}+{\frac{{{3}}}{{\sqrt{{{10}}}}}}{\frac{{{1}}}{{\sqrt{{5}}}}}={\frac{{{1}}}{{\sqrt{{2}}}}}\)
\(\displaystyle\Rightarrow\alpha+\beta={{\sin}^{{-{1}}}{\frac{{{1}}}{{\sqrt{{2}}}}}}={\frac{{\pi}}{{{4}}}}\ \ \ {\left(∵{\sin{{\left(\alpha+\beta\right)}}}\in{\left[-{1},{1}\right]}\right)}\)
\(\displaystyle\therefore{{\cos}^{{-{1}}}{\frac{{{3}}}{{\sqrt{{{10}}}}}}}+{{\cos}^{{-{1}}}{\frac{{{2}}}{{\sqrt{{5}}}}}}={\frac{{\pi}}{{{4}}}}\)
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Kirsten Davis
Answered 2022-01-05 Author has 1051 answers
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 \(\displaystyle{y}={\frac{{{2}}}{{\sqrt{{5}}}}}\)
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Vasquez
Answered 2022-01-08 Author has 9499 answers

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

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

0

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