# How do you find sin if tan is 4?

May 20, 2018

See below

#### Explanation:

We know that ${\sin}^{2} x + {\cos}^{2} x = 1$

Dividing by ${\cos}^{2} x$ last identity we have

${\tan}^{2} x + 1 = {\sec}^{2} x = \frac{1}{\cos} ^ 2 x$

Then $16 + 1 = \frac{1}{\cos} ^ 2 x$ then ${\cos}^{2} x = \frac{1}{17}$

And applying first identity

${\sin}^{2} x + \frac{1}{17} = 1$

${\sin}^{2} x = 1 - \frac{1}{17} = \frac{16}{17}$

$\sin x = \frac{\sqrt{16}}{\sqrt{17}} = \frac{4}{\sqrt{17}}$

May 20, 2018

$\sin x = \pm \frac{4}{\sqrt{17}} = \frac{4 \sqrt{17}}{17}$

#### Explanation:

Use trig identity:
${\sin}^{2} x = \frac{1}{1 + {\cot}^{2} x}$
In this case tan x = 4 --> $\cot x = \frac{1}{4}$
${\sin}^{2} x = \frac{1}{1 + \frac{1}{16}} = \frac{16}{17}$
$\sin x = \pm \frac{4}{\sqrt{17}}$
tan x = 4 --> x could be in Quadrant 1 or Quadrant 3, therefor
sin x could be positive or negative.