Question

Please help!!!!
How do you verify cot(x-pi/2) = -tan(x)

Answer 1

\(\cot \theta = \cfrac{\cos \theta}{\sin \theta}\)

Therefore,

\(\cot \left( x- \pi /2 \right) = \cfrac{\cos (x - \pi/2)}{\sin (x- \pi /2)}\)

Answer 2

Now, what is the formula for : \(\cos (x- \pi/2) \) and \(\sin (x-\pi/2)\) ?

Answer 3

Uhm... Shoot I can't remember what the formula for it is! Give me a second

Answer 4

Sin(π/2-x) = cosx cos(π/2-x) = sinx?

Answer 5

Yep. that's right, but notice that we have : \(\cos (x - \pi/2) \) and \(\sin (x -\pi /2)\) instead of \(\cos(\pi/2 - x)\) and \(\sin (\pi/2 -x)\) .

Answer 6

So, let us play with what we have.

\(\cos(x - \pi/2)\) can be written as : \(\cos (-(\pi/2 - x)) \) , right?

Answer 7

I think so

Answer 8

Okay. So, remember this that :

\(\cos (-\theta) = \cos (\theta)\)

while

\(\sin (-\theta) = - \sin (\theta)\)

Answer 9

So, that would make it cos(x)/-sin(x) right?

Answer 10

Therefore, I can write : \(\cos(-(x - \pi/2) ) \) as \(\cos (\pi/2 -x)\)

and

\(\sin (-(x-\pi/2)) \) as \( - \sin (\pi/2 - x) \)

Answer 11

What we have now is :

\(\cfrac{\cos (\pi/2 - x)}{- \sin (\pi/2 -x) }\)

Answer 12

Use this now : Sin(π/2-x) = cosx cos(π/2-x) = sinx

what do u get?

Answer 13

sin(x)/-cos(x) ?

Answer 14

Excellent ! So, you get : \(\cfrac{\sin (x) }{-\cos (x)}\)

What is \(\cfrac{\sin (x)}{\cos (x)}\) ?

Answer 15

tan(x)

Answer 16

So, that would mean that it would be the -tan(x), right?

Answer 17

good, so there was a minus in denominator, it becomes tan(x)

GOOD WORK!

Answer 18

So, all together it would be
cos(x-π/2)/sin(x-π/2)
cos(π/2-x)/-sin(π/2-x)
sin(x)/-cos(x) = -tan(x)
-tan(x)=-tan(x)
Would this be the work for it then?

Answer 19

Yeah, right!

Answer 20

Thank you so much for your help!!!

Answer 21

You're welcome. Also, Welcome to OpenStudy!

Answer 22

Thank you!! You are incredibly helpful!

Answer 23

:-) Good to hear that I was of some help.