Question

Verify the identity.

cotx minus pi divided by two. = -tan x

Answer 1

You have to be more specific here. Is it\[cotx-\frac{ \pi }{ 2 }=-tanx\] or is it \[\frac{ cotx-\pi }{ 2 }=-tanx\]? please be specific.

Answer 2

\[\cot \left( x-\pi/2 \right)= -tanx\] @IMStuck

Answer 3

Okay lets start with the left side because it is more complicated...

We know that cotx = cosx/sinx

so use that identity on cot(x-pi/2)

Answer 4

So since cotangent is cosx/sinx?
x would be cos? and -pi/2 would be sinx?
@timely

Answer 5

nope, the x would be (x-pi/2)

and we know that cotx=cosx/sinx

so how could you make cot(x-pi/2) be in the form of cosx/sinx?

Answer 6

I'm really not sure! This lesson kind of threw me off at some points. Like this one for instance... Since Cotangent is cos/sin would it be like... (x-pi/2)/(1+x-pi/2)? @timely

Answer 7

We have a function cot(x-pi/2)

the cotangent it applying to EVERYTHING inside its brackets, the entire x-pi/2

so if cotx=cosx/tanx

and we know x = x-pi/2

we can sub in (x-pi/2) for x

Answer 8

for example, if we have cot(pi)

we can write that as

cot(pi)=cos(pi)/sin(pi)

they are both equal, just different ways of expressing it!

Answer 9

Okay so what you're saying is that cotx=cosx/sinx? and this essentially creates the same answer or just one? but how is it equal to the -tanx? I'm confused! lol @timely

Answer 10

we still have more steps to go...

our end result should say -tanx=-tanx and then we know that we have verified the identity.

But the first step is to make the cot(x-pi/2) look like cosx/sinx and then we can do more steps to make it look like -tanx.

So have you figured out how to make it like cosx/sinx yet? the x is the (x-pi/2)

Answer 11

Okay so if Cot(x)=Cos(x)/sin(x)
Then its like saying Cot(x-pi/2)=cos(x-pi/2)/sin(x-pi/2)
I don't understand though... Is this what you are meaning? @timely

Answer 12

exactly!

we KNOW that cot(x)=cos(x)/sin(x)

so whatever we write for x (in our case it is x-pi/2) we know that it will be equal

if you were to put cos(x-pi/2)/sin(x-pi/2) into your calculator, you would see that it equals cot(x-pi/2)


so now we have cos(x-pi/2)/sin(x-pi/2)

Answer 13

Do you have a formula sheet for math? Something with all the different trig identities on it?

Answer 14

I have some of them written down in my notes... I was not given a formula sheet so I've been trying to write them as I go... Which trig identity type are we going to need to use? @timely

Answer 15

\[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\]

and

\[\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)\]

Answer 16

Okay yes I have these. but how do I know which is a and b? @timely

Answer 17

a=x
b=-pi/2?
@timely

Answer 18

okay the site is back up now :)

if we are using cos(a-b) then

a=x
b=pi/2

you do not need to include the negative sign

Answer 19

@timely okay i wasnt the only one with that problem haha! but... okay so if we plug it back into the trig formulas I would leave out the negative?

Answer 20

if you were using cos(a+b) then yes you could make b=-pi/2 but we would never need to do that because it works out perfect with our cos(a-b)=cosacosb+sinasinb

Answer 21

yes use the proper trig identities I gave you on the top and bottom of the fraction we have now

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

Answer 22

okay so when they are erm "filled in" or substituted rather they come out to be cos(x)cos(pi/2)+sin(x)sin(pi/2)

and

sin(x)cos(pi/2) - cos(x)sin(pi/2) ?
@timely

Answer 23

yes, exactly!

so now you might notice that we can simplify it a bit... specifically the terms with pi/2 in them

Answer 24

do the cos(pi/2) and sin(pi/2) equal to one because cotangent is cos/sin?

Answer 25

umm no not exactly... remember that we have a large fraction with multiple terms

\[\frac{ \cos(x)\cos(\pi/2)+\sin(x)\sin(\pi/2) }{ \sin(x)\cos(\pi/2)-\cos(x)\sin(\pi/2) }\]

starting with the top, what does cos(pi/2) equal?

Answer 26

I'm kind of unsure as to what you are asking? @timely

Answer 27

cos(pi/2) = ???



so when x is pi/2 what is the y?