# How do you find the value of sin(tan^-1 (7/24))?

Aug 28, 2016

$\frac{7}{25}$

#### Explanation:

Let $\theta$ be the angle whose tan is $\frac{7}{24}$
Draw the right angle triangle with the two shorter sides are 7and 24

Pythagoras gives the hypotenuse as 25

Sin $\theta$ =$\frac{7}{25}$

Aug 28, 2016

$\sin \left({\tan}^{-} 1 \left(\frac{7}{24}\right)\right) = \frac{7}{25}$.

#### Explanation:

Suppose that, ${\tan}^{-} 1 \left(\frac{7}{24}\right) = \theta$. Then, by Defn. of ${\tan}^{-} 1$ function, $\tan \theta = \frac{7}{24} , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

Since, tan theta >0, theta !in (-pi/2,0), &, so, theta in (0,pi/2).

We know that, ${\csc}^{2} \theta = 1 + {\cot}^{2} \theta = 1 + \frac{1}{\tan} ^ 2 \theta$.

$\therefore {\csc}^{2} \theta = 1 + {\left(\frac{24}{7}\right)}^{2} = \frac{{7}^{2} + {24}^{2}}{7} ^ 2 = {25}^{2} / {7}^{2}$

$\therefore \csc \theta = \pm \frac{24}{7} \text{, but, as,} \theta \in \left(0 , \frac{\pi}{2}\right) , \csc \theta = + \frac{25}{7} , \mathmr{and} , \sin \theta = \frac{7}{25}$.

Therefore, $\text{the reqd. value} = \sin \left({\tan}^{-} 1 \left(\frac{7}{24}\right)\right)$

$= \sin \theta = \frac{7}{25}$.

Alternatively , we can use the conversion formula :

${\tan}^{-} 1 x = {\sin}^{-} 1 \left(\frac{x}{\sqrt{1 + {x}^{2}}}\right) , w h e r e , x > 0$, to give,

${\tan}^{-} 1 \left(\frac{7}{24}\right) = {\sin}^{-} 1 \left(\frac{7}{25}\right)$, and hence,

$\sin \left({\tan}^{-} 1 \left(\frac{7}{24}\right)\right) = \sin \left({\sin}^{-} 1 \left(\frac{7}{25}\right)\right) = \frac{7}{25}$, as before!

Enjoy Maths.!