How do you find sin if tan is 4?

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How do you find sin if tan is 4?

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Answer 1 · 2018-05-20T18:59:15

See below

Explanation:

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

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

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

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

And applying first identity

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

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

$sin x = \frac{\sqrt{16}}{\sqrt{17}} = \frac{4}{\sqrt{17}}$ $sinx=sqrt16/sqrt17=4/sqrt17$

Answer 2 · 2018-05-20T21:56:03

$sin x = \pm \frac{4}{\sqrt{17}} = \frac{4 \sqrt{17}}{17}$ $sin x = +- 4/sqrt17 = (4sqrt17)/17$

Explanation:

Use trig identity:
${sin}^{2} x = \frac{1}{1 + {cot}^{2} x}$ $sin^2 x = 1/(1 + cot^2 x)$
In this case tan x = 4 --> $cot x = \frac{1}{4}$ $cot x = 1/4$
${sin}^{2} x = \frac{1}{1 + \frac{1}{16}} = \frac{16}{17}$ $sin^2 x = 1/(1 + 1/16) = 16/17$
$sin x = \pm \frac{4}{\sqrt{17}}$ $sin x = +- 4/sqrt17$
tan x = 4 --> x could be in Quadrant 1 or Quadrant 3, therefor
sin x could be positive or negative.

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How do you find sin if tan is 4?

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