Board 8 > how to solve this trig/calc problem?

(+) Yeah! (+)

Find all the angles between 0 and 2pi satisfying the given condition

tan(theta) = -1

I think I have a good grasp on how to do sin/cos.. but how do I do tangent? I have the unit circle in front of me

--
---

If you have the unit circle in front of you, then consider that tan(theta) = y/x, so you need y and x equal but with opposite signs. This occurs midway through quadrants 2 and 4, so the angles you need are 135 degrees and 315 degrees, or 3/4*pi and 7/4*pi, since you're using radians.

--
The RPG Duelling League: www.rpgdl.com
An unparalleled source for RPG information and discussion

and like how would you find the value of tan of (2pi/3) and tan of (3pi/4)

for tan and cot i only have up till pi/2, what's the way of finding the complement or whatever it's called?

--
---

if you know sin and cos well tan = sin/cos

HeroicLinusReed posted...
and like how would you find the value of tan of (2pi/3) and tan of (3pi/4)

for tan and cot i only have up till pi/2, what's the way of finding the complement or whatever it's called?
---

like on the table i have all the values of the 6 trig functions from 0 to pi/2, but how do i find ones when it goes beyond pi/2, mainly for tan?

--
---

and when doing the signs, does the ASTC acronym going from quadrants 1 to 4 work for the reciprocals? For example, if a value positive where sin would be positive would it be positive for csc as well?

--
---

if your looking at a circle, all the angles are relative to the x axis.

so pi/6 is the same amount as 5pi/6 or 7pi/6 and 11pi/6

all are positive in quadrant 1
sin is positive in 2
tan in 3
and cos in 4.

it's symmetric along the x axis, except with the signs changed

so tan(30) = - tan(150) for example

--
Growlithe-Tackling Mantine - Go To Meowth - Gardevoir + Mamoswine
Woo Woo Woo! You Know It!

ninja'd

--
Growlithe-Tackling Mantine - Go To Meowth - Gardevoir + Mamoswine
Woo Woo Woo! You Know It!

Colegreen_c12 posted...
if your looking at a circle, all the angles are relative to the x axis.

so pi/6 is the same amount as 5pi/6 or 7pi/6 and 11pi/6

all are positive in quadrant 1
sin is positive in 2
tan in 3
and cos in 4.

so is csc + in 2, cot + in 3, and sec + in 4?

--
---

From: HeroicLinusReed | #010
so is csc + in 2, cot + in 3, and sec + in 4?

yes

how would you solve this problem?

tan^-1 (tan 3pi/4)

"compute without using a calculator"

--
---

uhh, it just cancels out to 3pi/4

--
_foolmo_
[NO BARKLEY NO PEACE]

ok last question, this one is about logs..

how do you deal with logs with different bases?

for example,

log (base 8) 2 + log (base 4) 2 = ?

--
---

for log (base x) y, you can do:

log (base 10) y / log (base 10) x

or ln y/ ln x

--
Growlithe-Tackling Mantine - Go To Meowth - Gardevoir + Mamoswine
Woo Woo Woo! You Know It!

so how would that apply to the question i asked?

--
---

8^(log_8(2))=2. In other words, log_8(2) is the power you have to take 8 to in order to get 2.
...if you don't know how to do this stuff calc is probably not going to be fun.

--
http://img255.imageshack.us/img255/2636/ivotedphoenixyi0.png

it probably wont be, cause i still dont understand that very well

--
---

log (base x) y = 1/(log (base y) x)

hope that helps

--
Black Turtle shoots the lights out

From: HeroicLinusReed | #014
log (base 8) 2 + log (base 4) 2 = ?

For this, you just need to figure out the exact value of each. If you can't use a calculator, you should hope that the values are clean (like they are in this case)

log (base 8) 2 = x
8^x = 2
2^(3x) = 2^(1)
3x = 1
x = 1/3
log (base 8) 2 = 1/3

similarly you can find log (base 4) 2 = 1/2

--
_foolmo_
'To be foolmo'd is to be better opinion'd.' - Blairville

"For this, you just need to figure out the exact value of each."


oooooooooooh right ok ok now i got it.. i thought you had to get similar bases or whatever for some reason.. ok this makes sense

--
---