1. ## Proving Cofunction Identities

I have no idea where to start.

tan(pi/2-x)=cotx

sec(pi/2-x)=cscx

csc(pi/2-x)=secx

For the first one, I did the tan(a-b)=tan(a)+tan(b)/1+tan(a)tan(b), but I got -tan(x)/1.

So confused!

2. Originally Posted by Thatoneguy12345
I have no idea where to start.

tan(pi/2-x)=cotx

sec(pi/2-x)=cscx

csc(pi/2-x)=secx

For the first one, I did the tan(a-b)=tan(a)+tan(b)/1+tan(a)tan(b), but I got -tan(x)/1.

So confused!

Tangent isn't defined at $\frac{\pi}{2}$. The identity doesn't seem possible in my opinion. Secant also is not defined at $\frac{\pi}{2}$.

Edit: I was wrong, the identities DO seem to be true. I was mistaken because when I went to prove it with the tangent of a difference, the $tan(\frac{\pi}{2})$ is undefined. What you need to do, is write the tangent as $\frac{sin(\frac{\pi}{2}-x)}{cos(\frac{\pi}{2}-x)}$. So when you expand, all of the functions are defined on $\frac{\pi}{2}$

3. $tan(\frac{\pi}{2}-x)=\frac{sin(\frac{\pi}{2}-x)}{cos(\frac{\pi}{2}-x)}=\frac{sin(\frac{\pi}{2})cos(x)-cos(\frac{\pi}{2})sin(x)}{cos(\frac{\pi}{2})cos(x)-sin(\frac{\pi}{2})sin(x)}=cot(x)$

4. Ahh thanks a lot for helping!

So would I do the same thing with csc and sec?

Like have it csc(pi/2 - x) = pi/2 - x / sin (pi/2-x) and solve for it that way/

5. For csc and the sec part

I did csc(pi/2-x)= 1/sin(pi/2-x) then I changed it to

1/ sin(pi/2)cos(x)-cos(pi/2)sin(x)= 1/cosx = sec(x)

I did it almost the same, but I didn't put pi/2-x on the numerator and denominator like how you did for tan. Did I do it wrong?

6. Originally Posted by Thatoneguy12345
Ahh thanks a lot for helping!

So would I do the same thing with csc and sec?

Like have it csc(pi/2 - x) = pi/2 - x / sin (pi/2-x) and solve for it that way/
Yes, just write them in terms of either sine, or cosine. Sine and cosine are defined for all real numbers, so you can't go wrong. When you tried to expand the tangent of a difference, you had terms with $tan(\frac{\pi}{2})$. The problem is that this number doesn't exists, this is why the graph of tangent has an asymptote at $\frac{\pi}{2}$.

To do prove the cofunction theorem for cosecant, just expand:

$csc(\frac{\pi}{2}-x)=\frac{1}{sin(\frac{\pi}{2}-x)}$

If you expand this, you should get $sec(x)$.

7. Originally Posted by Thatoneguy12345
For csc and the sec part

I did csc(pi/2-x)= 1/sin(pi/2-x) then I changed it to

1/ sin(pi/2)cos(x)-cos(pi/2)sin(x)= 1/cosx = sec(x)

I did it almost the same, but I didn't put pi/2-x on the numerator and denominator like how you did for tan. Did I do it wrong?
That's it!

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# trigonometric cofunction prove in simplify

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