How do you use fundamental identities to find the values of the trigonometric values given angle in quadrant III, such that sec(x) = -3?

Dec 23, 2016

Explanation:

Use the identity $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$:

$\cos \left(x\right) = - \frac{1}{3}$

Use the identity $\sin \left(x\right) = \pm \sqrt{1 - {\cos}^{2} \left(x\right)}$

Because we are given that x is in the third quadrant, we know that the sine function is negative so we remove the $+$sign:

$\sin \left(x\right) = - \sqrt{1 - {\cos}^{2} \left(x\right)}$

Substitute ${\left(- \frac{1}{3}\right)}^{2} \text{ for } {\cos}^{2} \left(x\right)$

$\sin \left(x\right) = - \sqrt{1 - {\left(- \frac{1}{3}\right)}^{2}}$

$\sin \left(x\right) = - \sqrt{1 - \frac{1}{9}}$

$\sin \left(x\right) = - \sqrt{\frac{8}{9}}$

$\sin \left(x\right) = \frac{- 2 \sqrt{2}}{3}$

Use the identity $\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$

$\csc \left(x\right) = \frac{- 3 \sqrt{2}}{4}$

Use the identity $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

$\tan \left(x\right) = \frac{\frac{- 2 \sqrt{2}}{3}}{- \frac{1}{3}} = 2 \sqrt{2}$

Use the identity $\cot \left(x\right) = \frac{1}{\tan} \left(x\right)$:

$\cot \left(x\right) = \frac{1}{2 \sqrt{2}} = \frac{\sqrt{2}}{4}$