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QuestionFind cos⁡θ\cos \theta if sin⁡θ=1213\sin \theta=\frac{12}{13} and θ\theta is in quadrant II.

Studdy Solution

STEP 1

Assumptions1. The value of sin⁡θ\sin \theta is given as 1213\frac{12}{13}. . The angle θ\theta is in quadrant II.
3. We need to find the exact value of cos⁡θ\cos \theta.

STEP 2

In any right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. Similarly, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. This relationship is expressed by the Pythagorean identitysin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta =1

STEP 3

We can rearrange the Pythagorean identity to solve for cos⁡θ\cos \thetacos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1 - \sin^2 \theta}

STEP 4

Now, plug in the given value for sin⁡θ\sin \theta to calculate cos⁡θ\cos \thetacos⁡θ=1−(1213)2\cos \theta = \sqrt{1 - \left(\frac{12}{13}\right)^2}

STEP 5

Square the fraction 1213\frac{12}{13}(1213)2=144169\left(\frac{12}{13}\right)^2 = \frac{144}{169}

STEP 6

Substitute the squared fraction back into the equationcos⁡θ=1−144169\cos \theta = \sqrt{1 - \frac{144}{169}}

STEP 7

Subtract the fraction from1cos⁡θ=169169−144169\cos \theta = \sqrt{\frac{169}{169} - \frac{144}{169}}

STEP 8

implify the fraction under the square rootcos⁡θ=25169\cos \theta = \sqrt{\frac{25}{169}}

STEP 9

Take the square root of the fractioncos⁡θ=513\cos \theta = \frac{5}{13}

STEP 10

However, we know that θ\theta is in quadrant II, and in this quadrant, cosine is negative. Therefore, we must take the negative rootcos⁡θ=−513\cos \theta = -\frac{5}{13}So, the exact value of cos⁡θ\cos \theta is −513-\frac{5}{13}.

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