If $tan (θ) = \frac{3}{4}$ $tan(theta) = 3/4$ and $sin (θ) < 0$ $sin(theta) < 0$ , how do you find $cos (θ)$ $cos(theta)$ ?

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If $tan (θ) = \frac{3}{4}$ $tan(theta) = 3/4$ and $sin (θ) < 0$ $sin(theta) < 0$ , how do you find $cos (θ)$ $cos(theta)$ ?

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Answer 1 · 2017-05-05T09:19:37

$- \frac{4}{5}$ $-4/5$

Explanation:

Since $tan (θ) = \frac{3}{4} > 0$ $tan(theta)=3/4>0$ and $sin (θ) < 0$ $sin(theta)<0$ , $θ$ $theta$ is in quadrant 3 (since $tan (θ) > 0$ $tan(theta)>0$ iff $θ$ $theta$ is in quadrant 1 or 3 and $sin (θ) < 0$ $sin(theta)<0$ iff $θ$ $theta$ is in quadrant 3 or 4).

$cos (θ)$ $cos(theta)$ must then be negative.

Remember that ${sin}^{2} (θ) + {cos}^{2} (θ) = 1$ $sin^2(theta)+cos^2(theta)=1$ . Divide both sides by ${cos}^{2} (θ)$ $cos^2(theta)$ to get $\frac{{sin}^{2} (θ)}{{cos}^{2} (θ)} + 1 = \frac{1}{{cos}^{2} (θ)}$ $sin^2(theta)/cos^2(theta)+1=1/cos^2(theta)$ . Since $tan (θ) = \frac{sin (θ)}{cos (θ)}$ $tan(theta)=sin(theta)/cos(theta)$ , this is simply ${tan}^{2} (θ) + 1 = \frac{1}{{cos}^{2} (θ)}$ $tan^2(theta)+1=1/cos^2(theta)$ . Thus, ${cos}^{2} (θ) = \frac{1}{{tan}^{2} (θ) + 1}$ $cos^2(theta)=1/(tan^2(theta)+1)$ .

Since $tan (θ) = \frac{3}{4}$ $tan(theta)=3/4$ , ${cos}^{2} (θ) = \frac{1}{{(\frac{3}{4})}^{2} + 1} = \frac{16}{25}$ $cos^2(theta)=1/((3/4)^2+1)=16/25$ . Since $cos (θ)$ $cos(theta)$ is negative, $cos (θ) = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$ $cos(theta)=-sqrt(16/25)=-4/5$ .

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If $tan (θ) = \frac{3}{4}$ $tan(theta) = 3/4$ and $sin (θ) < 0$ $sin(theta) < 0$ , how do you find $cos (θ)$ $cos(theta)$ ?

1 Answer

Explanation:

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If $tan (θ) = \frac{3}{4}$ $tan(theta) = 3/4$ and $sin (θ) < 0$ $sin(theta) < 0$ , how do you find $cos (θ)$ $cos(theta)$ ?