Prove the identity

That cotangent looks like the oddest duck. Did you convert to sine and cosine?
 
cutterman said:
I want to prove the identity:


cot(x/2)/cos(x) - 1/sin(2x) = (2*cos(x)+1)/sin(2x)


I get soon stuck even if I try different methods.


Please include steps.
Click to expand...

Please post some of your work, so that we might determine where to begin guiding you.

Cheers:cool:
 
So the way is to start expand the member with the cot(x/2) function?

Anyway, here it is some I tried:

cot(x/2)/cos(x)-1/sin(2x)=(2*cos(x)+1)/sin(2x)

(cos(x/2)*sin(2x)-cos(x))/(sin(x/2)*cos(x)*sin(2x)) = (2*cos(x)+1)/sin(2x)

(cos(x/2)*2*sin(x)-1)/(sin(x/2)*sin(2x))= (2*cos(x)+1)/sin(2x)

(2*cos(x/2)*2*sin(x/2)*cos(x/2)-1)/(sin(x/2)*sin(2x))=(2*cos(x)+1)/sin(2x)

(4*(cos(x/2))^2*sin(x/2)-(cos(x/2))^2-(sin(x/2))^2/(sin(x/2)*sin(2x))=(2*cos(x)+1)/sin(2x)

As you can see, I'm not that good at proving identities... it seems that I expand too much and can't simplify.
 
These exercises are trial-and-error; one gets better with practice. Generally, with these exercises, if things become very convoluted, you should back up or try a different approach.

Here is an identify that is somewhat obscure, but your exercise seems to be written for it.

tan(x/2)=sin(x)1+cos(x)=1−cos(x)sin(x)\displaystyle tan(x/2) = \frac{sin(x)}{1 + cos(x)} = \frac{1 - cos(x)}{sin(x)}
 
Thanks a lot for the exercise, mmm4444bot, but it seems I get stuck another time also with this. There's something that I miss with x/2... bisection method doesn't help this time, don't know why.

Where should I start? Expanding tan(x/2) ?

EDIT:

why (1-cos(x))/sin (x) = csc(x)-cot(x) = tg(x/2) I don't understand the last step...
 
Last edited:
Found it:

if
sin2(x/2) = (1-cos(x))/2

then
2*sin2(x/2) = 1-cos(x)

so we have
tan(x/2) = (1-cos(x)) / sin (x)

tan(x/2)=2*sin2(x/2) / sin (x)

tan(x/2) = 2*sin2(x/2) / 2*sin(x/2)*cos(x/2)

tan(x/2) = tan(x/2)

Now trying the other from the exercise:

tan(x/2)=sin(x)/(1+cos(x))

should be easier knowing how to solve that.

EDIT: and here it is:

tan(x/2)=sin(x)/(1+cos(x))
tan(x/2)=sin(2)/(2*cos2(x/2))
tan(x/2)=(2*sin(x/2)*cos(x/2))/(2*cos2(x/2))
tan(x/2) = tan(x/2)

now trying my own exercise, many thanks for giving me help with yours! You know, it's funny that I asked for help with one and you gave me another :D But it's practice practice practice! Thanks!


Sorry for the multiple editing / messages - will avoid next time.
P.S.: please tell me how I can make this more readable without parenthesis and with fractions...
 
Last edited:
Answer

Finally here it is:

cot(x/2)/cos(x) - 1/sin(2x) = (2*cos(x)+1)/sin(2x)

(1+cos(x))/(sin(x)*cos(x)) - 1/(2*sin(x)*cos(x)) = (2*cos(x)+1)/sin(2x)

(2+2*cos(x)-1)/(2*sin(x)*cos(x)) = (2*cos(x)+1)/sin(2x)

(2*cos(x)+1)/sin(2x) = (2*cos(x)+1)/sin(2x)
 
cutterman said:
I want to prove the identity:


cot(x/2)/cos(x) - 1/sin(2x) = (2*cos(x)+1)/sin(2x)

edit: here you can get the answer

I get soon stuck even if I try different methods.


Please include steps.
Click to expand...

Slightly different way - here the clue I use is that the denominator of both the LHS and RHS is sin(2x)

cot(x2)cos(x) − 1sin(2x)\displaystyle \frac{cot(\frac{x}{2})}{cos(x)} \ - \ \frac{1}{sin(2x)}

= 2∗sin(x)∗cot(x2) −1sin(2x)\displaystyle = \ \frac{2*sin(x)*cot(\frac{x}{2}) \ - 1}{sin(2x)}

= 2∗sin(x)∗1+cos(x)sin(x) −1sin(2x)\displaystyle = \ \frac{2*sin(x)*\frac{1+cos(x)}{sin(x)} \ - 1}{sin(2x)}

and so on .....
 
cutterman said:
Thanks a lot for the [identity], mmm4444bot

Where should I start? Expanding tan(x/2) ?
Click to expand...

No, that's not what I had in mind.

You should already know that tan(x) is 1/cot(x).

This means that cot(x) is 1/tan(x).

In other words, cotangent is the reciprocal of tangent, and vice versa.

Hence, cot(x/2) is 1/tan(x/2).

I gave you two identities for tan(x/2).

Try replacing cot(x/2) in your exercise with the reciprocal of one of those identities for tan(x/2).
 
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