## anonymous one year ago Verify the identity. cotx minus pi divided by two. = -tan x

1. misty1212

HI!!

2. misty1212

if you can use the cofunction identity this is real easy

3. misty1212

since cotangnent is even, you know $\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)$

4. misty1212

damn i meant "odd"

5. misty1212

in any case , as cotangent is odd, you have $\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)$ and the cofunction identity tells gives $\cos(\frac{\pi}{2}-x)=\tan(x)$ and you are done

6. anonymous

yea, its tan x right?

7. misty1212

damn i am making a lot of typos

8. misty1212

let me try again in one line $\cot(x-\frac{\pi}{2})=-\cot(\frac{\pi}{2}-x)=-\tan(x)$

9. anonymous

yea, -tan x right?

10. misty1212

the first equal sign because cotangent is odd the second equal sign in the cofunction identity

11. misty1212

is this clear? it is really only two steps, but maybe they are not obvious

12. anonymous

the identity is -tan x or am I lost.

13. anonymous

Is it cos(pi/2-x) ?

14. misty1212

i thought that might be the case lets go slow and take them one at a time the cofunction identities say $\cos(\frac{\pi}{2}-x)=\sin(x)\\ \csc(\frac{\pi}{2}-x)=\sec(x)\\ \cot(\frac{\pi}{2}-x)=\tan(x)$

15. anonymous

cot(pi/2-x)

16. misty1212

since $\cot(\frac{\pi}{2}-x)=\tan(x)$ that is the same as $-\cot(\frac{\pi}{2}-x)=-\tan(x)$ just change the sign of both sides

17. misty1212

is that part clear?

18. anonymous

Oh ok, yes it is. Thnx!! :)

19. misty1212

ok and how about the first part $\cot(x-\frac{\pi}{2})=-\cot(\frac{\pi}{2}-x)$?

20. anonymous

yea, that's the part that kinda got me.

21. misty1212

ok lets do that slow too

22. anonymous

Ok :)

23. misty1212

cotangent in an "odd" function, which means $\cot(-x)=-\cot(x)$ in general "odd" means $f(-x)=-f(x)$

24. misty1212

sine and tangent are also "odd" so $\sin(-x)=-\sin(x)$ etc

25. misty1212

now $-(x-\frac{\pi}{2})=\frac{\pi}{2}-x$ right ?

26. anonymous

Yes...

27. misty1212

that means, since cotangent is odd, that $\cot(x-\frac{\pi}{2})=\cot(\color{red}-(\frac{\pi}{2}-x))=\color{red}-\cos(\frac{\pi}{2}-x)$

28. misty1212

notice the minus sign comes out of the function

29. anonymous

oh yea, this makes sense....

30. misty1212

it is a bit abstract, but not too hard just realizing that $$a-b=-(b-a)$$ and therefore $$\cot(a-b)=-\cos(b-a)$$

31. anonymous

oh, wow it's so much clearer now. Thank you @misty1212 for your help.

32. misty1212

$\huge \color\magenta\heartsuit$

33. anonymous

<3

34. anonymous

Lol

35. misty1212

thanks