How do you find the values of the six trigonometric functions given $cot θ = - 3$ $cottheta=-3$ and $cos θ > 0$ $costheta>0$ ?

Question

How do you find the values of the six trigonometric functions given $cot θ = - 3$ $cottheta=-3$ and $cos θ > 0$ $costheta>0$ ?

1 Answer

Impact of this question

5568 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-27T06:07:36

Find trig function values

Explanation:

Use tri identity:
${sin}^{2} t = \frac{1}{1 + {cot}^{2} t} = \frac{1}{1 + 9} = \frac{1}{10}$ $sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 9) = 1/10$ -->
$sin t = \pm \frac{1}{\sqrt{10}} = \pm \frac{\sqrt{10}}{10}$ $sin t = +- 1/sqrt10 = +- sqrt10/10$
Since cot = - 3 < 0 and cos t > 0, there for, sin t < 0.
$sin t = - \frac{\sqrt{10}}{10}$ $sin t = - sqrt10/10$
${cos}^{2} t = 1 - {sin}^{2} t = 1 - \frac{1}{10} = \frac{9}{10}$ $cos^2 t = 1 - sin^2 t = 1 - 1/10 = 9/10$ -->
$cos t = \frac{3}{\sqrt{10}} = \frac{3 \sqrt{10}}{10}$ $cos t = 3/sqrt10 = (3sqrt10)/10$
$tan t = \frac{1}{cot t} = \frac{1}{- 3} = - \frac{1}{3}$ $tan t = 1/cot t = 1/-3 = - 1/3$
$sec t = \frac{1}{cos} = \frac{10}{3 \sqrt{10}} = \frac{\sqrt{10}}{3}$ $sec t = 1/(cos) = 10/(3sqrt10) = sqrt10/3$
$csc t = - \frac{10}{\sqrt{10}} = - \sqrt{10}$ $csc t = - 10/sqrt10 = - sqrt10$

Science

Math

Humanities

... and beyond

How do you find the values of the six trigonometric functions given $cot θ = - 3$ $cottheta=-3$ and $cos θ > 0$ $costheta>0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the values of the six trigonometric functions given cotθ=−3cottheta=-3 and cosθ>0costheta>0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the values of the six trigonometric functions given $cot θ = - 3$ $cottheta=-3$ and $cos θ > 0$ $costheta>0$ ?