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Proof of trig identity (difficult)

  1. Aug 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that

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


    2. Relevant equations

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


    3. 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
     
  2. jcsd
  3. Aug 6, 2014 #2

    ehild

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  4. Aug 6, 2014 #3

    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?
     
  5. Aug 6, 2014 #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)##.
     
  6. Aug 6, 2014 #5

    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?
     
  7. Aug 6, 2014 #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)##
     
  8. Aug 6, 2014 #7

    jbunniii

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    OK, so:
    $$\begin{align}
    \left[\tan(a) + 1\right]\left[\cot(a+\pi/4) + 1\right] &= \left[\frac{\sin(a)}{\cos(a)} + 1\right]\left[\frac{\cos(a) - \sin(a)}{\cos(a) + \sin(a)} + 1\right] \\
    &= \frac{1}{\cos(a)}\left[\sin(a) + \cos(a)\right]\left[\frac{\cos(a) - \sin(a)}{\cos(a) + \sin(a)} + 1\right] \\
    &= ??? \\
    \end{align}$$
     
  9. Aug 6, 2014 #8
    Thanks, I got it

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

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

    THANKS
     
  10. Aug 6, 2014 #9

    jbunniii

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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:
     
  11. Aug 6, 2014 #10
    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.
     
  12. Aug 6, 2014 #11

    jbunniii

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    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.
     
  13. Aug 7, 2014 #12

    ehild

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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
     
  14. Aug 7, 2014 #13
    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 =)
     
  15. Aug 7, 2014 #14
    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: Aug 7, 2014
  16. Aug 7, 2014 #15

    ehild

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    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
     
  17. Aug 7, 2014 #16

    ehild

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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
     
  18. Aug 7, 2014 #17
    Hi there,

    It is [tex]\int_0^1\frac{\ln(x+1)}{x^2+1}dx[/tex]
     
  19. Aug 7, 2014 #18
    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
     
  20. Aug 7, 2014 #19

    olivermsun

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    You just write
    Code (Text):
    [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 (Text):
    [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]##.
     
  21. Aug 7, 2014 #20

    ehild

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    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
     
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