killjoy1990xb9

## Answered question

2022-01-03

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

### Answer & Explanation

sonorous9n

Beginner2022-01-04Added 34 answers

Use trig identity:${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$

Now, use trig identity
$\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{sin}\alpha \mathrm{sin}\beta$
$\mathrm{cos}\left(\alpha +\beta \right)=\frac{3}{\sqrt{10}}\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{10}}\frac{1}{\sqrt{5}}=\frac{1}{\sqrt{2}}$

$\therefore {\mathrm{cos}}^{-1}\frac{3}{\sqrt{10}}+{\mathrm{cos}}^{-1}\frac{2}{\sqrt{5}}=\frac{\pi }{4}$
Or alternatively use trig identity
$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{cos}\alpha \mathrm{sin}\beta$
$\mathrm{sin}\left(\alpha +\beta \right)=\frac{1}{\sqrt{10}}\frac{2}{\sqrt{5}}+\frac{3}{\sqrt{10}}\frac{1}{\sqrt{5}}=\frac{1}{\sqrt{2}}$

$\therefore {\mathrm{cos}}^{-1}\frac{3}{\sqrt{10}}+{\mathrm{cos}}^{-1}\frac{2}{\sqrt{5}}=\frac{\pi }{4}$

Kirsten Davis

Beginner2022-01-05Added 27 answers

Hint : apply this formula:
${\mathrm{cos}}^{-1}x+{\mathrm{cos}}^{-1}y={\mathrm{cos}}^{-1}\left[xy-\sqrt{\left(1-{x}^{2}\right)\left(1-{y}^{2}\right)}\right]$
Put $x=\frac{2}{\sqrt{10}}$ and $y=\frac{2}{\sqrt{5}}$

Vasquez

Expert2022-01-08Added 669 answers

Like Proof for the formula of sum of arcsine functions $\mathrm{arcsin}x+\mathrm{arcsin}y$
using $\mathrm{cos}\left(A+B\right)$ and the definition of principal values

Now

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