I have to do this packet for an AP class I'm taking, and my teacher didn't give me anything to work with. I'm desperate.

sin(theta)= (square root)3/2 and tan(theta) > 0 ; Find cos(theta).

- Aug 27th 2012, 10:37 AMMaddyVFind the requested function value meeting all of the given conditions??
I have to do this packet for an AP class I'm taking, and my teacher didn't give me anything to work with. I'm desperate.

sin(theta)= (square root)3/2 and tan(theta) > 0 ; Find cos(theta). - Aug 27th 2012, 11:06 AMMaxJasperRe: Find the requested function value meeting all of the given conditions??
Darw a circle with radius=1 and center at (0,0) then on sin axis go up as much as squrt(3)/2, draw a horizontal line to intersect circle at two points...now draw tan axis and check which of the two points on circle qualifies for tan>0

- Aug 27th 2012, 12:34 PMSorobanRe: Find the requested function value meeting all of the given conditions??
Hello, MaddyV!

Quote:

$\displaystyle \sin\theta\,=\,\frac{\sqrt{3}}{2}\,\text{ and }\,\tan\theta > 0.\;\;\text{ Find }\,\cos\theta.$

$\displaystyle \sin\theta$ is positive: .$\displaystyle \theta$ is in Quadrant 1 or Quadrant 2.

$\displaystyle \tan\theta$ is positive: .$\displaystyle \theta$ is in Quardrant 1 or Quadrant 3.

. . Hence, $\displaystyle \theta$ is in Quadrant 1.

$\displaystyle \sin\theta \:=\:\frac{\sqrt{3}}{2} \:=\:\frac{opp}{hyp}$

$\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = \sqrt{3},\;hyp = 2$

Pythagorus tell us that: .$\displaystyle adj = 1$

Therefore: .$\displaystyle \cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{1}{2}$

- Aug 27th 2012, 01:11 PMMaddyVRe: Find the requested function value meeting all of the given conditions??
Okay, so how about:

cos(theta) = - [(square root)2]/2 and tan(theta) < 0 ; find cot(theta) ? - Aug 27th 2012, 05:27 PMskeeterRe: Find the requested function value meeting all of the given conditions??
did you not learn this in your pre-AP class?

you are expected to already know the values for sine and cosine on the unit circle and the definitions/identities for basic all six trig ratios ...

http://media.wiley.com/Lux/83/323183.image0.jpg - Aug 27th 2012, 07:28 PMMaddyVRe: Find the requested function value meeting all of the given conditions??
It was actually recommended that I skip Pre-Calculus and go straight to AP Calculus AB. I was given a summer packet with these problems specifically so I could learn how to do these things.

- Aug 27th 2012, 09:09 PMProve ItRe: Find the requested function value meeting all of the given conditions??
Why was this recommended? The only way I would recommend something like this is if you have shown that you already know how to do the topics in Pre-Calculus...

- Aug 28th 2012, 02:56 AMskeeterRe: Find the requested function value meeting all of the given conditions??