# Six trigonometric equations

• Jul 21st 2010, 01:06 PM
FailingTrig
Six trigonometric equations
Given: cot theta = sqrt2/4 , pi < theta < 3pi/2
find the exact value of the other five trig functions of theta, leave answers in radical form.

could someone show me how to do this, when i did it i got:
sin theta= -2sqrt2/3
cos theta= -1/3
tan theta= 2/sqrt2
cot theta= sqrt2/4
sec theta= -3
csc theta= -3sqrt2/4

is this correct?

a friend in class said that it wasn't and that i had to start by doing cot theta= cos/sin cos = sqrt2 sin = 4....
• Jul 21st 2010, 02:47 PM
skeeter
image link doesn't work. some you got correct.

$\displaystyle \cot{t} = \frac{\sqrt{2}}{4}$

noted that $\displaystyle t$ is in quad III ...

$\displaystyle \tan{t} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

using the identity $\displaystyle \csc^2{t} = 1 + \cot^2{t}$ ...

$\displaystyle \csc{t} = -\frac{3}{2\sqrt{2}}$

$\displaystyle \sin{t} = -\frac{2\sqrt{2}}{3}$

$\displaystyle \cos{t} = -\frac{1}{3}$

$\displaystyle \sec{t} = -3$
• Jul 21st 2010, 02:52 PM
eumyang
$\displaystyle cot \theta = \frac{x}{y} = \frac{\sqrt{2}}{4}$, and we're given that θ is in quadrant III, so $\displaystyle x = -\sqrt{2}$ and $\displaystyle y = -4$

Use $\displaystyle x^2 + y^2 = r^2$ to find r:
$\displaystyle (-\sqrt{2})^2 + (-4)^2 = r^2$
$\displaystyle 2 + 16 = r^2$
$\displaystyle 18 = r^2$
$\displaystyle r = 3\sqrt{2}$

So...
$\displaystyle sin \theta = \frac{y}{r} = \frac{-4}{3\sqrt{2}} = ...$
$\displaystyle cos \theta = \frac{x}{r} = ...$
...
• Jul 21st 2010, 04:03 PM
Soroban
Hello, FailingTrig!

Here's another approach . . .

Quote:

Given: .$\displaystyle \cot\theta \,=\, \frac{\sqrt{2}}{4},\;\; \pi < \theta < \frac{3\pi}{2}$

Find the exact value of the other five trig functions of $\displaystyle \theta$.

We are given: .$\displaystyle \cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}$

That is, $\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj = \sqrt{2},\;opp = 4$

Pythagorus says: .$\displaystyle hyp^2 \:=\:(\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}$

Since $\displaystyle \theta$ is in Quadrant 3, only $\displaystyle \tan\theta$ and $\displaystyle \cot\theta$ are positive.

Therefore:

. . $\displaystyle \begin{array}{cccccc} \sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm] \cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm] \tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm] \cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm] \sec\theta &=& -3 \\ \\[-3mm] \csc\theta &=& -\dfrac{3\sqrt{2}}{4}\end{array}$

• Jul 21st 2010, 08:17 PM
FailingTrig
Quote:

Originally Posted by Soroban
Hello, FailingTrig!

Here's another approach . . .

We are given: .$\displaystyle \cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}$

That is, $\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj = \sqrt{2},\;opp = 4$

Pythagorus says: .$\displaystyle hyp^2 \:=\:(\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}$

Since $\displaystyle \theta$ is in Quadrant 3, only $\displaystyle \tan\theta$ and $\displaystyle \cot\theta$ are positive.

Therefore:

. . $\displaystyle \begin{array}{cccccc} \sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm] \cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm] \tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm] \cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm] \sec\theta &=& -3 \\ \\[-3mm] \csc\theta &=& -\dfrac{3\sqrt{2}}{4}\end{array}$

Did it right then! Awesome, thanks for showing me everyone =)