How Can Cos(x) Be Expressed Using Tan(x)?

[06Sport]

Homework Statement

seems simple, but i am stumped. Says write cos(x) in terms of tan(x).

Homework Equations

would this be a reciprocal equation? or a Pythagorean? I'm lost

The Attempt at a Solution

i don't even know where to begin.

 

[mathwonk]

sin/cos = tan, so cos= sin/tan. har har.

can you use derivatives?

 

[Moridin]

Write down the two formula for tan x and cos x for a right angle triangle. Are there any similar terms in those equations?

Edit: Beaten to it.

 

[mathwonk]

DO YOU KNOW WHaT TAN' IS? or 1 + tan^2?

 

[06Sport]

the angle is unknown. I think that's why its confusing me.

sin/cos = tan, so cos= sin/tan - these are what i have. But would that be the answer? tan= sin/cos ? or cos=sin/tan?

 

[mathwonk]

i was joking. read my second post.

 

[06Sport]

now I am even more confused.

would it be cos=sin/tan?

 

[arildno]

How can you express sine in terms of cosine?

 

[06Sport]

How can you express sine in terms of cosine?

i don't know

 

[arildno]

Well, what RELATION exists between the sine and cosine of an angle?

 

[@/@]

$\sin x= \sqrt{1-cos^2x}$

 

[06Sport]

hmm, sin/cos=tan, cos/sin=cot, sin^2 + cos^2=1

i need cos(theta) in terms of tan(theta) though. Unless that's what we are working up to :)

 

[arildno]

So, look at the last identity you posted.

What do you get by dividing ôn both sides with cos^{2} ?

 

[06Sport]

sin^2 = 1/cos^2?

 

[@/@]

fundamental relationship of trigonometry
$$sin^2 x + cos^2 x = 1$$

look
http://en.wikipedia.org/wiki/List_of_trigonometric_identities

 

[arildno]

sin^2 = 1/cos^2?

Don't you know how to divide an equation with a number?

 

[blackcat]

you can express sine in terms of cosine as cos (x-90) where x is in degrees or in radians cos (x-pi/2).

 

[drpizza]

Am I the first person who thinks it can't be done? Maybe I'm overlooking something, but I'm seeing a sign problem. (+/- when you solve)

 

[drpizza]

$\sin x= \sqrt{1-cos^2x}$

Only works for 1st and 2nd quadrant angles, that is, angles between 0 and 180 degrees. (or between 0 and 2Pi). Plus, it works for 0 degrees and 180 degrees. If you're in the 3rd or 4th quadrant, then you'd have to use a negative square root.

 

[@/@]

yes,
how to use +- in latex?

 

[cristo]

how to use +- in latex?

In latex the command for +/- is \pm