# Values of Angle in Standard Position

• Jun 2nd 2011, 04:43 AM
starshine84
Values of Angle in Standard Position
I'm having a little bit of difficulty with this question. I think I may have solved it, but I'm not sure about the negative symbols and whether or not they should be included in my calculations.

Determine the values of sin θ, cos θ, tan θ, csc θ, sec θ, and cot θ at
(-3, -4) on the terminal arm of an angle θ in standard position.

I know that if the angle is in standard position the vertex is at (0,0) and that the initial arm is on the positive x-axis. By plotting (-3,-4) on the Cartesian plane I know that the terminal arm is in quadrant III.

Therefore, side a= -4 units and side b= -3 units. I then used the Pythagorean theorem to solve for the hypotenuse and found it to be 5 units.

Solving for each of the trigonometric functions:

sinθ = opp/hyp
sinθ = 4/5
sinθ = 0.8
θ = sin^-1 0.8
θ = 53.1301

** OR, should the 4 be a -4?? **

cosθ = 3/5
cosθ = 0.6
θ = cos^-1 o.6
θ = 53.1301

** OR, should the 3 be a -3?? **

tanθ = 4/3
tanθ = 1.333333333
θ = tan^-1 1.333333333
θ = 53.1301

** OR, should the 4 and 3 be a -4 and -3?? **

If I make each of those values negative like they are in the right triangle then the angle measures all come out differently which has me confused.

Then,

cscθ = hyp/opp
cscθ = 5/4
cscθ = 1.25

** Can I solve for θ here??**

secθ = 5/3
secθ = 1.666666667

** Can I solve for θ here? **

cotθ = 3/4
cotθ = 0.75

** Can I solve for θ here? **

Therefore, angle θ = 180 + 53.1301 = 233.1301

Am I on the right track here or totally off-base?? Thanks!!
• Jun 2nd 2011, 06:01 AM
Plato
All of this depends upon your lecturer and/or textbook.
In my class it would go this way.
Note that $\displaystyle \theta\in III$ so $\displaystyle \theta=\arctan \left( {\frac{4}{3}} \right) - \pi$.

Now just find each of the six functional values.
• Jun 2nd 2011, 02:06 PM
skeeter
from the definitions ...

$\displaystyle \cos{\theta} = \frac{x}{r}$

$\displaystyle \sin{\theta} = \frac{y}{r}$

$\displaystyle \tan{\theta} = \frac{y}{x}$

$\displaystyle \cot{\theta} = \frac{x}{y}$

$\displaystyle \sec{\theta} = \frac{r}{x}$

$\displaystyle \csc{\theta} = \frac{r}{y}$

$\displaystyle r = \sqrt{x^2+y^2}$