# What does sin(arc cot(5))+5sin(arc tan(2)) equal?

Jan 21, 2016

$\sin \left({\cot}^{-} 1 \left(5\right)\right) + 5 \sin \left({\tan}^{-} 1 \left(2\right)\right) = \frac{\sqrt{26} + 52 \sqrt{5}}{26}$

#### Explanation:

The trick is to evaluate $\sin \left({\cot}^{-} 1 \left(5\right)\right) + 5 \sin \left({\tan}^{-} 1 \left(2\right)\right)$ is
breaking down the problem and solving each part separately.

Let us start with evaluating ${\cot}^{-} 1 \left(5\right)$

Let $\theta = {\cot}^{-} 1 \left(5\right)$

Then $\cot \left(\theta\right) = 5$

$\cot \left(\theta\right) = \text{adjancent"/"opposite}$

If I have to show that in a right triangle. The hypotenuse would be $\sqrt{{1}^{2} + {5}^{2}}$
Hypotenuse $= \sqrt{26}$

$\sin \left(\theta\right) = \text{Opposite"/"hypotenuse}$
$\sin \left(\theta\right) = \frac{1}{\sqrt{26}}$

Rationalizing the denominator we get
$\sin \left(\theta\right) = \frac{\sqrt{26}}{26}$

Remember $\theta$ was cot^-1(5))

$\sin \left({\cot}^{-} 1 \left(5\right)\right) = \frac{\sqrt{26}}{26}$

Now let us work on the second part.

$5 \sin \left({\tan}^{-} 1 \left(2\right)\right)$
$\Theta = {\tan}^{-} 1 \left(2\right)$
$\tan \left(\Theta\right) = 2$
$\tan \left(\Theta\right) = \text{Opposite"/"Adjacent}$

Let us put that in a right triangle. Now to find the hypotenuse

Hypotenuse $= \sqrt{{2}^{2} + {1}^{2}}$
Hypotenuse $= \sqrt{5}$

$\sin \left(\Theta\right) = \text{Opposite"/"hypotenuse}$

$\sin \left(\Theta\right) = \frac{2}{\sqrt{5}}$

Rationalizing the denominator

$\sin \left(\Theta\right) = \frac{2 \sqrt{5}}{5}$

Remember $\Theta = {\tan}^{-} 1 \left(2\right)$

$\sin \left({\tan}^{-} 1 \left(2\right)\right) = \frac{2 \sqrt{5}}{5}$

$5 \sin \left({\tan}^{-} 1 \left(2\right)\right) = 5 \ast \frac{2 \sqrt{5}}{5}$

$5 \sin \left({\tan}^{-} 1 \left(2\right)\right) = 2 \sqrt{5}$

Now the final steps

$\sin \left({\cot}^{-} 1 \left(5\right)\right) + 5 \sin \left({\tan}^{-} 1 \left(2\right)\right) = \frac{\sqrt{26}}{26} + 2 \sqrt{5}$

$\sin \left({\cot}^{-} 1 \left(5\right)\right) + 5 \sin \left({\tan}^{-} 1 \left(2\right)\right) = \frac{\sqrt{26} + 52 \sqrt{5}}{26}$