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Proof of trig identity (difficult)

  • Thread starter Amad27
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  • #1
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Homework Statement


Prove that

[tan(a) + 1][cot(a+pi/4) + 1] = 2


Homework Equations



[tan(a) + 1][cot(a+pi/4) + 1] = 2


The Attempt at a Solution



This was very hard, I tried my best at expanding.

[tan(a) + 1][cot(a+pi/4) + 1] = tan(a)cot(a+pi/4) + tan(a) + cot(a+pi/4) + 1

The issue is that there is no cofuntion identity for cot(a + pi/4) so I did

cot(a + pi/4) = cot([a - pi/4] + pi/2]) Let b = a - pi/4 thus,

cot(b + pi/2) = -tan(b) = -tan(a - pi/4)

This doesnt help AT ALL.


Any help?

Thanks
 

Answers and Replies

  • #2
ehild
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  • #3
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Do you know the addition rule for tangent?
[tex]\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-tan(\alpha)\tan(\beta)}[/tex]
and the cotangent is reciprocal of tangent.

http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

ehild

Can't believe I didnt think of that, so

cot(a + b) = 1-tan(a)tan(b)/tan(a)+tan(b)

cot(a + pi/4) = [1-tan(a)]/[tan(a)+1]

[tan(a) + 1][cot(a+pi/4) + 1] = tan(a)cot(a+pi/4) + tan(a) + cot(a+pi/4) + 1

= [tan(a)][1-tan(a)]/[tan(a)+1] + tan(a) + [1-tan(a)]/[tan(a)+1] + 1

= [1-tan(a)]/[tan(a)+1][ [tan(a)] + 1] + [tan(a) + 1]

= [tan(a) +1][ [1-tan(a)]/[tan(a)+1] + 1]

One thing that could help is finding

[1-tan(a)]/[tan(a)+1]. But what is a possible way to do so?

WolframAlpha gives a very complicated formula involving sines and cosines.

One "good" formula it gives is,

[1-tan(a)]/[tan(a)+1] = 2/[tan(a) +1] - 1

But how is this result derived?

= [tan(a) +1][ 2/[tan(a) +1]] = 2

This works, but how do you derive that?

[1-tan(a)]/[tan(a)+1] = 1/[tan(a) +1] - tan(a)/[tan(a) +1]

1/[tan(a) +1] = 1/[sin/cos + 1] = 1/[ [sin(a) + cos(a)]/cos(a)] = cos(a)/[sin(a) + cos(a)]

tan(a)/[tan(a) +1] = [itex]\frac{sin(a)/cos(a)}{[sin(a) + cos(a)]/cos(a)]}[/itex]

= [itex]\frac{sin(a)}{cos(a)+sin(a)}[/itex]

1/[tan(a) +1] - tan(a)/[tan(a) +1] = [itex]\frac{cos(a)-sin(a)}{sin(a)+cos(a)}[/itex]

Thats as far as I can go, any advice?
 
  • #4
jbunniii
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I just worked out a solution. This problem is quite straightforward if you express ##\tan## and ##\cot## in terms of ##\sin## and ##\cos##. You will want to use the appropriate trig identity to simplify the resulting ##\cos(a + \pi/4)## and ##\sin(a + \pi/4)##.
 
  • #5
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I just worked out a solution. This problem is quite straightforward if you express ##\tan## and ##\cot## in terms of ##\sin## and ##\cos##. You will want to use the appropriate trig identity to simplify the resulting ##\cos(a + \pi/4)## and ##\sin(a + \pi/4)##.

sin(x + pi/4) = sin(x)sqrt(2)/2 + cos(x)sqrt(2)/2
cos(x + pi/4) = cos(x)sqrt(2)/2 - sin(x)sqrt(2)/2

We originally had,

[tan(a) + 1][cot(a+pi/4) + 1] = tan(a)cot(a+pi/4) + tan(a) + cot(a+pi/4) + 1

cot(a + pi/4) = sqrt(2)/2[cos(x) - sin(x)]/sqrt(2)/2[sin(x) + cos(x)] = [cos(x) - sin(x)/[sin(x) +cos(x)]

What now?
 
  • #6
verty
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How about this method?

Let b = a + pi/4.

##(tan(a) + 1)(cot(b) + 1) = 2##

Multiply by tan(b):

##(tan(a) + 1)(1 + tan(b)) = 2 \; tan(b)##
 
  • #7
jbunniii
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sin(x + pi/4) = sin(x)sqrt(2)/2 + cos(x)sqrt(2)/2
cos(x + pi/4) = cos(x)sqrt(2)/2 - sin(x)sqrt(2)/2

We originally had,

[tan(a) + 1][cot(a+pi/4) + 1] = tan(a)cot(a+pi/4) + tan(a) + cot(a+pi/4) + 1

cot(a + pi/4) = sqrt(2)/2[cos(x) - sin(x)]/sqrt(2)/2[sin(x) + cos(x)] = [cos(x) - sin(x)/[sin(x) +cos(x)]

What now?
OK, so:
$$\begin{align}
\left[\tan(a) + 1\right]\left[\cot(a+\pi/4) + 1\right] &= \left[\frac{\sin(a)}{\cos(a)} + 1\right]\left[\frac{\cos(a) - \sin(a)}{\cos(a) + \sin(a)} + 1\right] \\
&= \frac{1}{\cos(a)}\left[\sin(a) + \cos(a)\right]\left[\frac{\cos(a) - \sin(a)}{\cos(a) + \sin(a)} + 1\right] \\
&= ??? \\
\end{align}$$
 
  • #8
146
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OK, so:
$$\begin{align}
\left[\tan(a) + 1\right]\left[\cot(a+\pi/4) + 1\right] &= \left[\frac{\sin(a)}{\cos(a)} + 1\right]\left[\frac{\cos(a) - \sin(a)}{\cos(a) + \sin(a)} + 1\right] \\
&= \frac{1}{\cos(a)}\left[\sin(a) + \cos(a)\right]\left[\frac{\cos(a) - \sin(a)}{\cos(a) + \sin(a)} + 1\right] \\
&= ??? \\
\end{align}$$
Thanks, I got it

The [sin(a) + cos(a)] cancel out, leaving you with

[1/cos(a)] * [2cos(a)/1] = 2

THANKS
 
  • #9
jbunniii
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Thanks, I got it

The [sin(a) + cos(a)] cancel out, leaving you with

[1/cos(a)] * [2cos(a)/1] = 2

THANKS
Right. I'm sure the other suggested methods working directly with ##\tan## and ##\cot## will also work, but I only memorize the trig identities involving ##\sin## and ##\cos##, so converting to them is usually my first step. :tongue:
 
  • #10
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Right. I'm sure the other suggested methods working directly with ##\tan## and ##\cot## will also work, but I only memorize the trig identities involving ##\sin## and ##\cos##, so converting to them is usually my first step. :tongue:
I think it is a better STRATEGY to convert into sines and cosines. Since they are the easiest to work around with aren't they?

This expression was part of the integral

INT[ln(x + 1)/x^2 + 1] 0 --> 1

I used trig substitution so I needed this, PM or something if you are interested in seeing the solution or want to CHECK the solution.

THe only reason I posted this question in the Pre-Calculus section is because it is pre-calculus work, otherwise this was FOR calculus =)

Thanks a lot, your tipis were VERY helpful.
 
  • #11
jbunniii
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I think it is a better STRATEGY to convert into sines and cosines. Since they are the easiest to work around with aren't they?
Sines and cosines are probably the functions most people work with most often, so they have the advantage of familiarity, which can make it easier to spot what to do next when simplifying a complicated expression. This is just a general observation, I'm sure there are exceptions when working directly with tangents or secants or whatever can seem more natural. After you learn about complex exponentials, sometimes they can be even easier to work with than sines and cosines.
This expression was part of the integral

INT[ln(x + 1)/x^2 + 1] 0 --> 1

I used trig substitution so I needed this, PM or something if you are interested in seeing the solution or want to CHECK the solution.
I'll be happy to check your solution to the calculus problem if you want to post it. I think it would be fine to continue using this thread even though it's in the precalculus section, since it's all in the context of the same problem. But it would also be fine to create a new thread in the calculus section if you prefer.
 
  • #12
ehild
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Can't believe I didnt think of that, so

cot(a + b) = 1-tan(a)tan(b)/tan(a)+tan(b)

cot(a + pi/4) = [1-tan(a)]/[tan(a)+1]

[tan(a) + 1][cot(a+pi/4) + 1] = tan(a)cot(a+pi/4) + tan(a) + cot(a+pi/4) + 1
Do not expand, substitute cot(a+pi/4)=(1-tan(a))/(1+tan(a)) instead.

##(\tan(a)+1)\left(\cot(a+\pi/4)+1\right) \rightarrow (\tan(a)+1)\left(\frac{1-\tan(a)}
{1+\tan(a)}+1\right)##

Expand and simplify by tan(a)+1:

##(1-\tan(a))+(1+\tan(a))##...equal to ??? :tongue:

ehild
 
  • #13
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Do not expand, substitute cot(a+pi/4)=(1-tan(a))/(1+tan(a)) instead.

##(\tan(a)+1)\left(\cot(a+\pi/4)+1\right) \rightarrow (\tan(a)+1)\left(\frac{1-\tan(a)}
{1+\tan(a)}+1\right)##

Expand and simplify by tan(a)+1:

##(1-\tan(a))+(1+\tan(a))##...equal to ??? :tongue:

ehild
That is a very interesting approach. Okay, lets see,

cot(a + pi/4) = [1-tan(a)]/[tan(a)+1]

[tan(a) +1][2/tan(a) +1] = 2

Very nice, I must say. One of best approaches, it was very quick. Thanks =)
 
  • #14
146
1
Sines and cosines are probably the functions most people work with most often, so they have the advantage of familiarity, which can make it easier to spot what to do next when simplifying a complicated expression. This is just a general observation, I'm sure there are exceptions when working directly with tangents or secants or whatever can seem more natural. After you learn about complex exponentials, sometimes they can be even easier to work with than sines and cosines.

I'll be happy to check your solution to the calculus problem if you want to post it. I think it would be fine to continue using this thread even though it's in the precalculus section, since it's all in the context of the same problem. But it would also be fine to create a new thread in the calculus section if you prefer.
Hi, Okay I'll post it here, I just noticed an error in my method, can you see to it? Please?


Okay, so we have INT[ln(x+1)/x^2+1] 0 --> 1

Next I drew a right triangle,

| \
1| \
| a \ [where "a" is the angle]
|------------
x

cot(a) = x

Now I think the problem arrives, I drew another triangle

| \
x| \
| a \ [where "a" is the angle]
|------------
1

tan(a) = x

The question here is,

I just realized that I CANT use the same angle "a" so the whole approach gets messed up! Advice?

Thanks

**Please dont mind the weird lines drawn in the triangle, there was a formatting issue.
 
Last edited:
  • #15
ehild
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That is a very interesting approach. Okay, lets see,

cot(a + pi/4) = [1-tan(a)]/[tan(a)+1]

[tan(a) +1][2/tan(a) +1] = 2
Your equation is wrong.:grumpy:

[tan(a) +1][2/tan(a) +1] = 2+2/tan(a)+tan(a)+1. Do not forget the parentheses!!!!

ehild
 
  • #16
ehild
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I think it is a better STRATEGY to convert into sines and cosines. Since they are the easiest to work around with aren't they?

This expression was part of the integral

INT[ln(x + 1)/x^2 + 1] 0 --> 1
What is the integral exactly? Is it as you wrote without enough parentheses
[tex]\int_0^1\left(\frac{\ln(x+1)}{x^2}+1\right)dx[/tex]
or is it
[tex]\int_0^1\frac{\ln(x+1)}{x^2+1}dx[/tex]

ehild
 
  • #17
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What is the integral exactly? Is it as you wrote without enough parentheses
[tex]\int_0^1\left(\frac{\ln(x+1)}{x^2}+1\right)dx[/tex]
or is it
[tex]\int_0^1\frac{\ln(x+1)}{x^2+1}dx[/tex]

ehild
Hi there,

It is [tex]\int_0^1\frac{\ln(x+1)}{x^2+1}dx[/tex]
 
  • #18
146
1
Your equation is wrong.:grumpy:

[tan(a) +1][2/tan(a) +1] = 2+2/tan(a)+tan(a)+1. Do not forget the parentheses!!!!

ehild
WAIT! How?

cot(a + pi/4) = [1-tan(a)]/[tan(a)+1]

[tan(a) + 1][[1-tan(a)]/[tan(a)+1] + (1)]

[[1-tan(a)]/[tan(a)+1] + (1)] = [1-tan(a)]/[tan(a) +1] + [tan(a) + 1]/[tan(a) + 1]

= 2/[tan(a) + 1] and then

[tan(a) + 1][[1-tan(a)]/[tan(a)+1] + (1)] = [tan(a) + 1][1/[tan(a) +1]][2] = 2

The tangent +1 cancels.

mmm.. I cant do LaTex, can you write

[tan(a) +1][2/tan(a) +1] in LaTex?

It is very hard to see where you mean [2/tan(a)] + 1 or 2/[tan(a) +1] ??

Thanks
 
  • #19
olivermsun
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mmm.. I cant do LaTex, can you write

[tan(a) +1][2/tan(a) +1] in LaTex?
You just write
Code:
[itex][\tan(a) +1][2/\tan(a) +1][/itex]
which gives you: [itex][\tan(a) +1][2/\tan(a) +1][/itex].

Or better yet:

You just write
Code:
[itex]\left[\tan(a) +1\right] \left[\dfrac{2}{\tan(a)} +1 \right][/itex]
which typesets the "display" fraction and auto-sizes the square brackets: ##\left[\tan(a) +1\right] \left[\dfrac{2}{\tan(a)} +1 \right]##.
 
  • #20
ehild
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[tan(a) +1][2/tan(a) +1] in LaTex?

It is very hard to see where you mean [2/tan(a)] + 1 or 2/[tan(a) +1] ??

Thanks
You need to use parentheses/brackets where multiplication/division and sum are involved. Multiplication/division comes first. Addition/subtraction later.
So 2/tan(a)+1 corresponds to ##\frac {2}{\tan(a)}+1## instead of ##\frac {2}{\tan(a)+1}##

You can write Maths formulae easily with Physicsforum's TeX:
https://www.physicsforums.com/misc/howtolatex.pdf
https://www.physicsforums.com/showthread.php?t=8997

ehild
 
  • #21
jbunniii
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Okay, so we have INT[ln(x+1)/x^2+1] 0 --> 1

Next I drew a right triangle,

| \
1| \
| a \ [where "a" is the angle]
|------------
x

cot(a) = x

Now I think the problem arrives, I drew another triangle

| \
x| \
| a \ [where "a" is the angle]
|------------
1

tan(a) = x

The question here is,

I just realized that I CANT use the same angle "a" so the whole approach gets messed up! Advice?
Right, either of these is valid, but you have to choose one and stick with it. So let's say you put ##x = \tan(a)##. So ##dx = \sec^2(a) da = da/\cos^2(a)##. Your integral becomes
$$\begin{align}
\int_0^{\pi/4} \left(\frac{1}{\cos^2(a)}\right) \left(\frac{\ln(\tan(a) + 1)}{\tan^2(a) + 1}\right) da &=
\int_0^{\pi/4} \left(\frac{\cos^2(a)}{\cos^2(a)}\right) \ln(\tan(a) + 1) da\\
&= \int_0^{\pi/4} \ln(\tan(a) + 1) da \\
\end{align}$$
I'm not sure where you can take this from here. Are you sure that a trig substitution is the right approach?

What if you try integration by parts: say ##u = \ln(x+1)## and ##dv = dx/(x^2+1)##. Then ##du = dx/(x+1)## and ##v = \arctan(x)##. Your integral becomes
$$\begin{align}
\left. uv \right|_{x=0}^{x=1} - \int_{x=0}^{x=1} v du &=
\left. \ln(x+1)\arctan(x)\right|_{0}^{1} - \int_{0}^{1} \frac{\arctan(x)}{x+1} dx \\
&= \frac{\pi}{4}\ln(2) - \int_0^1 \frac{\arctan(x)}{x+1} dx \\
\end{align}$$
According to Wolfram Alpha, the answer is supposed to be ##\frac{\pi}{8}\ln(2)##, so if that's true, then somehow we must have
$$\int_0^1 \frac{\ln(x+1)}{x^2+1} dx = \int_0^1 \frac{\arctan(x)}{x+1} dx$$
If you can prove that then you're done. I don't know enough stupid integration tricks to prove it off the top of my head. :tongue:
 
Last edited:
  • #22
146
1
Right, either of these is valid, but you have to choose one and stick with it. So let's say you put ##x = \tan(a)##. So ##dx = \sec^2(a) da = da/\cos^2(a)##. Your integral becomes
$$\begin{align}
\int_0^{\pi/4} \left(\frac{1}{\cos^2(a)}\right) \left(\frac{\ln(\tan(a) + 1)}{\tan^2(a) + 1}\right) da &=
\int_0^{\pi/4} \left(\frac{\cos^2(a)}{\cos^2(a)}\right) \ln(\tan(a) + 1) da\\
&= \int_0^{\pi/4} \ln(\tan(a) + 1) da \\
\end{align}$$
I'm not sure where you can take this from here. Are you sure that a trig substitution is the right approach?

What if you try integration by parts: say ##u = \ln(x+1)## and ##dv = dx/(x^2+1)##. Then ##du = dx/(x+1)## and ##v = \arctan(x)##. Your integral becomes
$$\begin{align}
\left. uv \right|_{x=0}^{x=1} - \int_{x=0}^{x=1} v du &=
\left. \ln(x+1)\arctan(x)\right|_{0}^{1} - \int_{0}^{1} \frac{\arctan(x)}{x+1} dx \\
&= \frac{\pi}{4}\ln(2) - \int_0^1 \frac{\arctan(x)}{x+1} dx \\
\end{align}$$
According to Wolfram Alpha, the answer is supposed to be ##\frac{\pi}{8}\ln(2)##, so if that's true, then somehow we must have
$$\int_0^1 \frac{\ln(x+1)}{x^2+1} dx = \int_0^1 \frac{\arctan(x)}{x+1} dx$$
If you can prove that then you're done. I don't know enough stupid integration tricks to prove it off the top of my head. :tongue:

Actually, interestingly,this is not needed at all. Recognizing these two facts is the most important,
(A) tan(a) = x
(B) cot(a) = x

If you use the first option (A) then

I = ∫ln(tan(a) + 1) da from 0 to pi/4

If you use the second (B) then

∫ln(cot(a) + 1) da from pi/4 to pi/2 = ∫ln(cot(a + pi/4) + 1) da from 0 to pi/4 = I

2I = ∫ln(cot(a + pi/4) + 1) da + ∫ln(tan(a) + 1) da

2I = ∫ln( [cot(a + pi/4) + 1][tan(a) + 1] da ) from 0 to pi/4

We proved [cot(a + pi/4) + 1][tan(a) + 1] = 2 thus,

2I = ∫ln(2) da from 0 to pi/4

2I = ln(2)*pi/4

I = ln(2)* (pi/8) which is correct.
 
  • #23
jbunniii
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Actually, interestingly,this is not needed at all.
Very nice!
 

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