If $tan θ = \frac{2}{3}$ $tan theta=2/3$ and $cos θ >$ $cos theta >$ , how do you find $sin θ$ $sin theta$ ?

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If $tan θ = \frac{2}{3}$ $tan theta=2/3$ and $cos θ >$ $cos theta >$ , how do you find $sin θ$ $sin theta$ ?

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Answer 1 · 2018-05-25T21:59:40

$sin t = \frac{2 \sqrt{13}}{13}$ $sin t = (2sqrt13)/13$

Explanation:

$tan t = \frac{2}{3}$ $tan t = 2/3$ -->
$cot t = \frac{1}{tan t} = \frac{3}{2}$ $cot t = 1/(tan t) = 3/2$
${sin}^{2} t = \frac{1}{1 + {cot}^{2} t} = \frac{1}{1 + \frac{9}{4}} = \frac{4}{13}$ $sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 9/4) = 4/13$
$sin t = \pm \frac{2}{\sqrt{13}}$ $sin t = +- 2/sqrt13$
Because tan t > 0 and cos t > 0, therefor, t lies in Quadrant 1 -->
and sin t is positive
$sin t = \frac{2 \sqrt{13}}{13}$ $sin t = (2sqrt13)/13$

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If $tan θ = \frac{2}{3}$ $tan theta=2/3$ and $cos θ >$ $cos theta >$ , how do you find $sin θ$ $sin theta$ ?

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