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Math  /  Trigonometry

QuestionGiven that sin⁡θ=−13\sin \theta=-\frac{1}{3} and that cos⁡θ<0\cos \theta<0, then determine the exact value of tan⁡θ\tan \theta Select one: a. −8-\sqrt{8} b. 18\frac{1}{\sqrt{8}} c. 8\sqrt{8} d. −18-\frac{1}{\sqrt{8}}

Studdy Solution

STEP 1

What is this asking? If the sine of an angle is −1/3-1/3 and its cosine is negative, what's the tangent of that angle? Watch out! Remember sin⁡\sin is negative in the third and fourth quadrants, while cos⁡\cos is negative in the second and third quadrants.
Both are negative only in the third quadrant!

STEP 2

1. Find Cosine
2. Calculate Tangent

STEP 3

We know that sin⁡θ=−1/3\sin \theta = -1/3 and cos⁡θ<0\cos \theta < 0.
Let's **visualize** this on the unit circle!
Since both sine (the *y*-coordinate) and cosine (the *x*-coordinate) are negative, our angle θ\theta must be in the **third quadrant**!

STEP 4

Let's use the **Pythagorean identity**: sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1 We're doing this because it connects sine and cosine, and we know one and want to find the other!

STEP 5

**Substitute** the known value of sin⁡θ\sin \theta: (−1/3)2+cos⁡2θ=1 (-1/3)^2 + \cos^2 \theta = 1

STEP 6

**Simplify** and **solve** for cos⁡θ\cos \theta: 1/9+cos⁡2θ=1 1/9 + \cos^2 \theta = 1 cos⁡2θ=1−1/9 \cos^2 \theta = 1 - 1/9 cos⁡2θ=8/9 \cos^2 \theta = 8/9 cos⁡θ=±8/9=±83=±223 \cos \theta = \pm \sqrt{8/9} = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3} Since we know cos⁡θ<0\cos \theta < 0 (from the problem statement!), we take the negative solution: cos⁡θ=−223 \cos \theta = -\frac{2\sqrt{2}}{3}

STEP 7

Now, let's **recall** the definition of tangent: tan⁡θ=sin⁡θcos⁡θ \tan \theta = \frac{\sin \theta}{\cos \theta} We're using this because we want to find tan⁡θ\tan \theta, and we know both sin⁡θ\sin \theta and cos⁡θ\cos \theta!

STEP 8

**Substitute** the values we found: tan⁡θ=−1/3−22/3 \tan \theta = \frac{-1/3}{-2\sqrt{2}/3}

STEP 9

**Simplify** the fraction by multiplying the numerator and denominator by −3-3 (to divide to one): tan⁡θ=122 \tan \theta = \frac{1}{2\sqrt{2}}

STEP 10

**Rationalize** the denominator by multiplying the numerator and denominator by 2\sqrt{2}: tan⁡θ=122⋅22=22⋅2=24=18 \tan \theta = \frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2 \cdot 2} = \frac{\sqrt{2}}{4} = \frac{1}{\sqrt{8}} We do this to avoid having a square root in the denominator, which is often preferred for cleaner presentation.

STEP 11

The exact value of tan⁡θ\tan \theta is 18\frac{1}{\sqrt{8}}, which corresponds to answer choice *b*.

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