# proving a trig identity

• Apr 17th 2009, 12:30 PM
skeske1234
proving a trig identity
This is my work so far for proving this trig identity. What should I do next? Is there anything I did wrong thus far?

cotx=tan(-(x-pi/2))

Right side:

tan(-x+pi/2)
[-tanx+tanpi/2]/[1-(-tanx)(tanpi/2)]
[-sinx/cosx+tanpi/2]/[1+(sinx/cosx)(tanpi/2)]

also, tanpi/2 is undefined... what does that do to the question?

• Apr 17th 2009, 12:50 PM
JoanF
Quote:

Originally Posted by skeske1234
This is my work so far for proving this trig identity. What should I do next? Is there anything I did wrong thus far?

cotx=tan(-(x-pi/2))

Right side:

tan(-x+pi/2)
[-tanx+tanpi/2]/[1-(-tanx)(tanpi/2)]
[-sinx/cosx+tanpi/2]/[1+(sinx/cosx)(tanpi/2)]

also, tanpi/2 is undefined... what does that do to the question?

Hi!

http://www.postimage.org/Pq2fWB6r.jpg

tan(pi/2) is undefined...well, I think you don't need that to solve this problem. If there is something you don't understand, just say it.
• Apr 17th 2009, 01:58 PM
Soroban
Hello, skeske1234!

You didn't do anything wrong . . .

Quote:

Prove: . $\cot x\:=\:\tan\left[\text{-}\left(x-\tfrac{\pi}{2}\right)\right]$

The right side is: . $\tan\left(\tfrac{\pi}{2} - x\right) \;=\;\frac{\sin\left(\frac{\pi}{2}-x\right)}{\cos\left(\frac{\pi}{2}-x\right)} \;= \;\frac{\sin\frac{\pi}{2}\cos x - \cos\frac{\pi}{2}\sin x} {\cos\frac{\pi}{2}\cos x + \sin\frac{\pi}{2}\sin x }$

. . $=\;\frac{1\cdot\cos x - 0\cdot\sin x}{0\cdot\cos x + 1\cdot\sin x} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x$

• Apr 17th 2009, 02:52 PM
JoanF
So, forget what my resolution. I didn't know that you can use these formulas ( sin(a-b) and cos(a-b). Here, in Portugal, you can't use these formulas to solve the problem.