# Trig Identities

• Nov 23rd 2009, 11:13 AM
goliath
Trig Identities
Hey guy, I need help with another problem,

Suppose that $\displaystyle 3\pi /2 < \Theta < 2\pi$ and $\displaystyle \sec \Theta = 3$ Evaluate the following

$\displaystyle a. \sin \Theta$
$\displaystyle b. \cos (-\Theta)$
$\displaystyle c. \tan (\pi /2 - \Theta)$
$\displaystyle d. \cot ( \pi + \Theta )$
$\displaystyle e. \csc (2\pi + \Theta )$
• Nov 23rd 2009, 04:06 PM
apcalculus
Quote:

Originally Posted by goliath
Hey guy, I need help with another problem,

Suppose that $\displaystyle 3\pi /2 < \Theta < 2\pi$ and $\displaystyle \sec \Theta = 3$ Evaluate the following

$\displaystyle a. \sin \Theta$
$\displaystyle b. \cos (-\Theta)$
$\displaystyle c. \tan (\pi /2 - \Theta)$
$\displaystyle d. \cot ( \pi + \Theta )$
$\displaystyle e. \csc (2\pi + \Theta )$

The angle is in the third quadrant, where the sine if negative and the cosine is positive. The rest of the tri ratios have signs that depend on the sine and cosine.

Secant is 1/cosine, so cosine(theta) is 1/3.

Use the Trig Identity $\displaystyle \sin^2x + \cos^2 x = 1$ and definitions/properties to find the remaining measures.

Good luck!
• Nov 23rd 2009, 06:40 PM
Soroban
Hello, goliath!

Quote:

Suppose that $\displaystyle \tfrac{3\pi}{2} \,<\, \theta \,<\, 2\pi$ and $\displaystyle \sec\theta \,=\, 3$.
$\displaystyle \sec\theta \:=\:\frac{3}{1} \:=\:\frac{hup}{adj}$

We have: .$\displaystyle adj = 1,\;hyp = 3$
And Pythagorus says: .$\displaystyle opp = \pm2\sqrt{2}$

Since $\displaystyle \theta$ is in Quadrant 4, $\displaystyle opp = -2\sqrt{2}$

And we have this list: .$\displaystyle \begin{Bmatrix}\sin\theta &=&-\dfrac{2\sqrt{2}}{3} \\ \\[-3mm] \cos\theta &=& \dfrac{1}{3} \\ \\[-3mm] \tan\theta &=& -2\sqrt{2} \end{Bmatrix}$

Quote:

Evaluate the following.

$\displaystyle a.\;\sin\theta$

From our list: .$\displaystyle \sin\theta \:=\:-\frac{2\sqrt{2}}{3}$

Quote:

$\displaystyle b.\;\cos (-\theta)$
$\displaystyle \cos(-\theta) \;=\;\cos\theta \;=\;\frac{1}{3}$

Quote:

$\displaystyle c.\;\tan \left(\tfrac{\pi}{2} - \theta\right)$
$\displaystyle \tan\left(\tfrac{\pi}{2}-\theta\right) \;=\;\cot\theta \;=\;\frac{1}{\tan\theta} \;=\;-\frac{1}{2\sqrt{2}}$

Quote:

$\displaystyle d.\;\cot(\pi + \theta)$
$\displaystyle \cot(\pi + \theta) \;=\;\cot\theta \;=\;-\frac{1}{2\sqrt{2}}$

Quote:

$\displaystyle e.\;\csc(2\pi + \theta )$
$\displaystyle \csc(2\pi + \theta) \;=\;\csc\theta \;=\;\frac{1}{\sin\theta} \;=\;\frac{1}{\text{-}\frac{2\sqrt{2}}{3}} \;=\;-\frac{3}{2\sqrt{2}}$

• Nov 23rd 2009, 06:45 PM
goliath
Thank you!, kudos