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QuestionFind tan⁡θ\tan \theta if sin⁡θ=−12\sin \theta = -\frac{1}{2} and θ\theta is in quadrant IV.

Studdy Solution

STEP 1

Assumptions1. The given value of sin⁡θ\sin \theta is −1-\frac{1}{} . θ\theta is in quadrant IV3. We need to find the exact value of tan⁡θ\tan \theta
4. We need to rationalize denominators when applicable

STEP 2

In any right triangle, the tangent of an angle is the ratio of the side opposite the angle to the side adjacent to the angle. This can be represented astan⁡θ=oppositeadjacent\tan \theta = \frac{opposite}{adjacent}

STEP 3

We know that sin⁡θ=−12\sin \theta = -\frac{1}{2} and sin⁡θ\sin \theta is the ratio of the side opposite the angle to the hypotenuse. This can be represented assin⁡θ=oppositehypotenuse\sin \theta = \frac{opposite}{hypotenuse}

STEP 4

From the given value of sin⁡θ\sin \theta, we can say that the side opposite to the angle is -1 and the hypotenuse is2.

STEP 5

We also know that in a right triangle, the Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. This can be represented ashypotenuse2=opposite2+adjacent2hypotenuse^2 = opposite^2 + adjacent^2

STEP 6

We can use the Pythagorean theorem to find the length of the side adjacent to the angle. Plug in the values for the hypotenuse and the side opposite to the angle.
22=(−1)2+adjacent22^2 = (-1)^2 + adjacent^2

STEP 7

olve the equation to find the length of the side adjacent to the angle.
4=1+adjacent24 =1 + adjacent^2adjacent2=4−1adjacent^2 =4 -1adjacent2=3adjacent^2 =3

STEP 8

Since θ\theta is in quadrant IV, the side adjacent to the angle should be positive. So, we take the positive square root of3.
adjacent=3adjacent = \sqrt{3}

STEP 9

Now that we have the lengths of the sides opposite and adjacent to the angle, we can find the value of tan⁡θ\tan \theta. Plug in the values for the sides opposite and adjacent to the angle.
tan⁡θ=−3\tan \theta = \frac{-}{\sqrt{3}}

STEP 10

Rationalize the denominator by multiplying the numerator and the denominator by 3\sqrt{3}.
tan⁡θ=−×33×3\tan \theta = \frac{- \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

STEP 11

implify the expression to find the exact value of tan⁡θ\tan \theta.
tan⁡θ=−33\tan \theta = \frac{-\sqrt{3}}{3}The exact value of tan⁡θ\tan \theta is −33-\frac{\sqrt{3}}{3}.

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