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QuestionGiven sin⁡θ=−79\sin \theta=-\frac{7}{9} in Quadrant III, find sin⁡θ\sin \theta, csc⁡θ\csc \theta, cos⁡θ\cos \theta, sec⁡θ\sec \theta, tan⁡θ\tan \theta, and cot⁡θ\cot \theta.

Studdy Solution

STEP 1

Assumptions1. The sine of θ\theta is given as −79-\frac{7}{9} . θ\theta is in Quadrant III3. We are to find the values of the six trigonometric functions for θ\theta

STEP 2

In Quadrant III, both sine and cosine are negative, and tangent is positive. Since we know the value of sine, we can use the Pythagorean identity to find the value of cosine.
The Pythagorean identity is given assin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta =1

STEP 3

Rearrange the Pythagorean identity to solve for cos⁡θ\cos\thetacos⁡2θ=1−sin⁡2θ\cos^2\theta =1 - \sin^2\theta

STEP 4

Substitute the given value of sin⁡θ\sin\theta into the equationcos⁡2θ=1−(−79)2\cos^2\theta =1 - \left(-\frac{7}{9}\right)^2

STEP 5

implify the equationcos⁡2θ=1−4981\cos^2\theta =1 - \frac{49}{81}

STEP 6

Calculate the value of cos⁡2θ\cos^2\thetacos⁡2θ=3281\cos^2\theta = \frac{32}{81}

STEP 7

Since θ\theta is in Quadrant III where cosine is negative, we take the negative square root to find cos⁡θ\cos\thetacos⁡θ=−3281\cos\theta = -\sqrt{\frac{32}{81}}

STEP 8

implify the square root to find the value of cos⁡θ\cos\thetacos⁡θ=−42\cos\theta = -\frac{4\sqrt{2}}{}

STEP 9

Now that we have the values of sine and cosine, we can find the values of the other trigonometric functions.The reciprocal of sine is cosecant, socsc⁡θ=sin⁡θ\csc\theta = \frac{}{\sin\theta}

STEP 10

Substitute the given value of sin⁡θ\sin\theta into the equationcsc⁡θ=−79\csc\theta = \frac{}{-\frac{7}{9}}

STEP 11

Calculate the value of csc⁡θ\csc\thetacsc⁡θ=−97\csc\theta = -\frac{9}{7}

STEP 12

The reciprocal of cosine is secant, sosec⁡θ=cos⁡θ\sec\theta = \frac{}{\cos\theta}

STEP 13

Substitute the value of cos⁡θ\cos\theta into the equationsec⁡θ=−29\sec\theta = \frac{}{-\frac{\sqrt{2}}{9}}

STEP 14

Calculate the value of sec⁡θ\sec\thetasec⁡θ=−942\sec\theta = -\frac{9}{4\sqrt{2}}

STEP 15

The ratio of sine to cosine is tangent, sotan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}

STEP 16

Substitute the values of sin⁡θ\sin\theta and cos⁡θ\cos\theta into the equationtan⁡θ=−9−429\tan\theta = \frac{-\frac{}{9}}{-\frac{4\sqrt{2}}{9}}

STEP 17

Calculate the value of tan⁡θ\tan\thetatan⁡θ=742\tan\theta = \frac{7}{4\sqrt{2}}

STEP 18

The reciprocal of tangent is cotangent, socot⁡θ=tan⁡θ\cot\theta = \frac{}{\tan\theta}

STEP 19

Substitute the value of tan⁡θ\tan\theta into the equationcot⁡θ=174\cot\theta = \frac{1}{\frac{7}{4\sqrt{}}}

STEP 20

Calculate the value of cot⁡θ\cot\thetacot⁡θ=47\cot\theta = \frac{4\sqrt{}}{7}So, the values of the trigonometric functions aresin⁡θ=−79\sin \theta=-\frac{7}{9} csc⁡θ=−97\csc \theta=-\frac{9}{7} cos⁡θ=−49\cos \theta=-\frac{4\sqrt{}}{9} sec⁡θ=−94\sec \theta=-\frac{9}{4\sqrt{}} tan⁡θ=74\tan \theta=\frac{7}{4\sqrt{}} cot⁡θ=47\cot \theta=\frac{4\sqrt{}}{7}

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