Question

Verify the identity.

cotx minus pi divided by two. = -tan x

Answer 1

cot x = cos x / sin x

tan x = sin x / cos x

Answer 2

\(\large\color{black}{\cot\left(\begin{matrix} x-\frac{\LARGE\pi}{\LARGE2} \\ \end{matrix}\right)= -\tan x }\)

Answer 3

split those up and use the sum/difference formulas

Answer 4

Yea, that's the one Z.

Answer 5

I'm totally lost.

Answer 6

Do you know the sum and difference formulas?

Answer 7

I was thinking of perhaps,
\(\large\color{black}{\sin\left(\begin{matrix} x-\frac{\LARGE\pi}{\LARGE2} \\ \end{matrix}\right)= -\cos x }\)
and
\(\large\color{black}{\cos\left(\begin{matrix} x-\frac{\LARGE\pi}{\LARGE2} \\ \end{matrix}\right)= \sin x }\)
and proving each, but you can only use one side at a time, right?

Answer 8

Not really, minaj.

Answer 9

ah that works too

Answer 10

I am just re-writing tan and cot in sines and cosines, but I might not be allowed to do this, will see what the user tells me.

Answer 11

Could you do one side so I could have an example. I'm still kinda lost

Answer 12

I just want to know, the rules I posted should they also be proven?

Answer 13

http://www.clarku.edu/~djoyce/trig/sumformulas.jpg

Answer 14

\(\large\color{black}{\cot\left(\begin{matrix} x-\frac{\LARGE\pi}{\LARGE2} \\ \end{matrix}\right)= -\tan x }\)
if I can use
\(\large\color{black}{1)~~\cos\left(\begin{matrix} x-\frac{\LARGE\pi}{\LARGE2} \\ \end{matrix}\right)= -\sin x }\)
\(\large\color{black}{2)~~\sin\left(\begin{matrix} x-\frac{\LARGE\pi}{\LARGE2} \\ \end{matrix}\right)= \cos x }\) ??

Answer 15

can I use rules 1 and 2, or I would have to verify them too?
But you see how it comes out right away, (using just the left side) into the -cot(x), yes?

Answer 16

cot(x) = cos(x)/sin(x)

Answer 17

Oh, I see.

Answer 18

x in this case is x - pi/2

Answer 19

so, if you are able to use those rules, 1 and 2. but if you have to show why these rules work, then we can do that too, if you like.

Answer 20

cot(x-pi/2) = -tan(x)
cos(x-pi/2)/sin(x-pi/2) = - tan(x)

as @SolomonZelman said

cos(x-pi/2) = -sin(x)
sin(x-pi/2) = cos(x)

so plugging that in gives us

-sin(x)/cos(x) = -tan(x)

and since tan(x) = sin(x)/cos(x)

-tan(x) = -tan(x)

Answer 21

All I am asking, is that can you use rules 1 and 2 I posted as a given.

Answer 22

If not, then you would have to verify the rules, buy expanding each of the rules' left sides.

Answer 23

and if you can't use the rules that @SolomonZelman posted, then you'd have to use the sum/difference formulas

Answer 24

yes, which is not hard, knowing that:
\(\large\color{black}{ \cos(\frac{\LARGE \pi }{\LARGE 2})=0 }\), and\(\huge\color{white}{ \rm \left| \right| }\)
\(\large\color{black}{ \sin(\frac{\LARGE \pi }{\LARGE 2})=1 }\)\(\huge\color{white}{ \rm \left| \right| }\)