# CosX in terms of Tanx?

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

2. Homework Equations
would this be a reciprocal equation? or a Pythagorean? I'm lost

3. The Attempt at a Solution

i dont even know where to begin.

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mathwonk
Homework Helper
sin/cos = tan, so cos= sin/tan. har har.

can you use derivatives?

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

the angle is unknown. I think thats 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
Homework Helper
i was joking. read my second post. now im even more confused.

would it be cos=sin/tan?

arildno
Homework Helper
Gold Member
Dearly Missed
How can you express sine in terms of cosine?

How can you express sine in terms of cosine?
i dont know arildno
Homework Helper
Gold Member
Dearly Missed
Well, what RELATION exists between the sine and cosine of an angle?

@/@
$\sin x= \sqrt{1-cos^2x}$ Last edited:
hmm, sin/cos=tan, cos/sin=cot, sin^2 + cos^2=1

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

arildno
Homework Helper
Gold Member
Dearly Missed
So, look at the last identity you posted.

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

sin^2 = 1/cos^2?

arildno
Homework Helper
Gold Member
Dearly Missed
sin^2 = 1/cos^2?
Don't you know how to divide an equation with a number?

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

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)

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

Last edited:
@/@
yes,
how to use +- in latex?

cristo
Staff Emeritus