Given that cos t=(3/4) and that P(t) is a point in the fourth quadrant, what is sin (t)?

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Given that cos t=(3/4) and that P(t) is a point in the fourth quadrant, what is sin (t)?

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Answer 1 · 2018-04-05T21:09:31

$sin (θ) = - \frac{\sqrt{7}}{4}$ $sin(theta) = -sqrt7/4$

Explanation:

Start with the identity:

$sin (θ) = \pm \sqrt{1 - {cos}^{2} (θ)}$ $sin(theta) = +-sqrt(1-cos^2(theta))$

We are told that $θ$ $theta$ is in the fourth quadrant and we know that the sine function is negative in the fourth quadrant, therefore, we shall choose the negative case of the identity:

$sin (θ) = - \sqrt{1 - {cos}^{2} (θ)}$ $sin(theta) = -sqrt(1-cos^2(theta))$

Substitute ${cos}^{2} (θ) = {(\frac{3}{4})}^{2}$ $cos^2(theta) = (3/4)^2$ :

$sin (θ) = - \sqrt{1 - {(\frac{3}{4})}^{2}}$ $sin(theta) = -sqrt(1-(3/4)^2)$

$sin (θ) = - \sqrt{\frac{16}{16} - \frac{9}{16}}$ $sin(theta) = -sqrt(16/16-9/16)$

$sin (θ) = - \sqrt{\frac{7}{16}}$ $sin(theta) = -sqrt(7/16)$

$sin (θ) = - \frac{\sqrt{7}}{4}$ $sin(theta) = -sqrt7/4$

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Given that cos t=(3/4) and that P(t) is a point in the fourth quadrant, what is sin (t)?

1 Answer

Explanation:

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