# What is sin(inverse tangent(12/5)?

Sep 30, 2015

$\sin \left(\arctan \left(\frac{12}{5}\right)\right) = \pm \frac{12}{13}$

#### Explanation:

$\arctan \left(\frac{12}{5}\right)$ implies one of the reference triangles pictured below
(with the hypotenuse caculated using the Pythagorean Theorem).

Since $s = \left(\text{opposite")/("hypotenuse}\right)$

$\sin \left(\arctan \left(\frac{12}{5}\right)\right) = \textcolor{red}{\frac{12}{13}} \text{ or } \textcolor{b l u e}{- \frac{12}{13}}$

Sep 30, 2015

$\sin \left(\arctan \left(\frac{12}{5}\right)\right) = \frac{12}{13}$

#### Explanation:

From the trig identity ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, we divide both sides by ${\sin}^{2} \left(\theta\right)$

$1 + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \left(\theta\right) = \frac{1}{\sin} ^ 2 \left(\theta\right)$

Since $\sin \frac{\theta}{\cos} \left(\theta\right) = \tan \left(\theta\right)$ we can rewrite the second term

$1 + \frac{1}{\tan} ^ 2 \left(\theta\right) = \frac{1}{\sin} ^ 2 \left(\theta\right)$

Taking the least common multiple,

$\frac{{\tan}^{2} \left(\theta\right) + 1}{{\tan}^{2} \left(\theta\right)} = \frac{1}{\sin} ^ 2 \left(\theta\right)$

Inverting both sides

${\tan}^{2} \frac{\theta}{{\tan}^{2} \left(\theta\right) + 1} = {\sin}^{2} \left(\theta\right)$

Subsituting $\theta = \arctan \left(\frac{12}{5}\right)$

${\left(\frac{12}{5}\right)}^{2} / \left({\left(\frac{12}{5}\right)}^{2} + 1\right) = {\sin}^{2} \left(\arctan \left(\frac{12}{5}\right)\right)$

${\sin}^{2} \left(\arctan \left(\frac{12}{5}\right)\right) = \frac{144}{25} \cdot \frac{25}{169} = \frac{144}{169}$

Taking the root

$\sin \left(\arctan \left(\frac{12}{5}\right)\right) = \pm \frac{12}{13}$

To pick the sign we look at the range of the arctangent, it only takes arguments on the first and fourth quadrants, during which the cosine is always positive. If, for $\frac{12}{5}$ the cosine is positive and the tangent is positive, then the sine must be positive too.

$\sin \left(\arctan \left(\frac{12}{5}\right)\right) = \frac{12}{13}$

Also, protip, you can use either the function with a "^-1" or put an arc before it to notate the inverse trig functions, but usually there's a lot less headache for everybody involved if you use the arc notation. (There's no grounds for mistaking it for other functions).