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Homework Help: Trig identities

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove this identity

    m) Sin^2 x / sin^2 x + cos x^2 = tan^2 x / 1 + tan^ x
    2. Relevant equations

    http://i52.tinypic.com/105tdtk.jpg Letter (m) on the top

    3. The attempt at a solution

    I dont know to much about trig identities, he barely taught anything. But apparently 1 is = to cos/cos and plenty of other things.

    I tried to solve the right side, the first thing i did was turn the 1 into sin^ x + cos ^ x.

    Then i turned the denominator into 1-sin^x - 1 + cos^x by subbing the values of cos^x , since cos^x = 1 - sin^2 .

    I know that tan^x = sin^x / cos^x , but the things that I am subbing in keep cancelling eachother out, does anyone know how to solve this?
  2. jcsd
  3. Jul 25, 2011 #2


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    Letter m is

    [tex]\frac{1+tan^2x}{1-tan^2x} = \frac{1}{cos^2x-sin^2x}[/tex]

    If you take the left side and divide both the numerator and denominator by cos2x what will you get?
  4. Jul 25, 2011 #3
    Divide the tan bracket by cos^2x?

    you would get the original tan bracket, with a division of cos^2x ....
  5. Jul 25, 2011 #4


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    Sorry, wrong operation.

    rewrite tanx as sin/cosx and then multiply the numerator and denominator by cos2x. It should easily work out.
    Last edited: Jul 25, 2011
  6. Jul 25, 2011 #5
    Its not tan x though, Its 1 + tan^2x , cant write tan^2x as sin/cos
  7. Jul 25, 2011 #6


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    tanx = sinx/cosx

    squaring both sides

    tan2x = (sinx/cosx)2 = ?
  8. Jul 25, 2011 #7
    tru, i dont get what to do when theres a 1. it seem as tho sometimes people change the 1 into sinx^2 + cosx^2, sometimes they keep the one, honostly I dont even get how to do it. Am i always looking at both side of the equation? or just focusing on one? If i pick leftside do i ignore right side to the extent of just trying to solve for it? wth do i do with the ones?
  9. Jul 25, 2011 #8


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    How about: taking the left side and multiplying [STRIKE]divide[/STRIKE] both the numerator and denominator by cos2x ?
  10. Jul 25, 2011 #9
    does 1 - sin = cos?
  11. Jul 25, 2011 #10


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    No. But 1-sin2x=cos2x
  12. Jul 26, 2011 #11
    f) 2sin^2x - 1 = sin^2 x - cos^2 x

    I dont know how to solve this.

    I tried taking the right side and simplyiing it to 1-cos^2x -1 - sin^2x ... then didnt know where to go with it,

    And I have no idea where to even begin with the left side.
  13. Jul 26, 2011 #12
    Can anyone tell me if i did this question right?

    It goes :: 1 / cosx - cosx = sinx*tanx

    I took the right side and simplified it to 1-cosx.

    Then i took the left side and multiplied the lone cos value with cos to give me

    1/cos - cos^2x / cos

    1 / cos - cos^2x / cos

    1-cos^2x / cos

    = 1 - cos.

    ls = rs

    Is this correct?
  14. Jul 26, 2011 #13


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    If you take the left side, what can you replace '1' by? (something with sine and cosine in it)

    That is one way to do it, but you should generally use one side and prove the other.
  15. Jul 26, 2011 #14
    Are you ever supposed to recipricol subtraction into addition? Or is it jus for division into multiplication?
  16. Jul 26, 2011 #15


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    I am not sure what you mean by that.

    When you multiply the numerator and denominator by the same quantity you are essentually multiplying by 1.
  17. Jul 26, 2011 #16
    Is this one right?

    I solved it two ways, but im uncertain if one of these ways is right or not.

    f) Tan^2x - sin^2x = sin^2x*tan^2x


    ________ - 1-cos^2x = 1-cos^2x + sin^2x / cos^2x

    ^ factor out the negetive from the cos^2x
    Sin^2x - 1

    rs- cancel out the two cos^2x, and factor out the negitive giving -1 sin^2x in the end.

    Ls = sin^2x -1 , rs = -1cos^2x

    Ls = RS?

    The other way is the way the teacher did it, just solving one side. etc.
    Last edited: Jul 26, 2011
  18. Jul 26, 2011 #17


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    You could take this and then bring them both down the same denominator of cosx.
  19. Jul 26, 2011 #18
    Like... when you have division, you take the term on the right and flip it,, do you do that for subtraction into addition as wel?

    I edited my last post for a different question, please take a look @ it.
  20. Jul 26, 2011 #19


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    It seems like you are working with both sides at the same time. It is better to just take one side and use that side to prove the other. How you wrote it is confusing.

    ________ - 1-cos^2x = 1-cos^2x + sin^2x / cos^2x


    [tex]\frac{sin^2x}{cos^2x} - (1-cos^2x) = \frac{sin^2x}{cos^2x} -sin^2x[/tex]

    best to not change sin2x as yet. If you factor out the sin2x, what are you left with?

    Can you see an identity that will help to get tan2x?
  21. Jul 26, 2011 #20
    I already have that one solved, was just wondering if my way is valid to. And yes i know, my teaccher doesnt care though.

    How about this one?

    1/sinx^2x + 1/cos^2x = 1/sin^2x*cos^2x

    I tried turning 1/sin^2x into cosecent^2x + secent^x = 1/sin^2x*cos^2x

    Also just converted the two values at the bottom into 1-cos^2x and 1-sin^2x , however i dont understand how they are turned into multiplication on the right side?
  22. Jul 26, 2011 #21


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    Use one side to prove the other is quicker. Reducing both sides to the same expression is equally valid when I learned it.

    That's over complicating it. Look at it like this, two terms are being reduced to one, so you'd need to bring them to a common denominator.

    [tex]\frac{1}{sin^2x}+\frac{1}{cos^2x} = \frac{?+?}{sin^2xcos^2x}[/tex]

    (I assume you are able to do this)
  23. Jul 26, 2011 #22
    yes, i took the cos fraction and multiplied the 1/sin^2x to give me

    cos^2x + 1
    ______________ ____
    sinx^2x*cos^2x cos^2x

    the opposite side cos^2xs cancel, and you get 1/sin^2x * cos^2x.

    Sorry, I am completely new to these concepts, and I have a test on them tommorow. Not much breathing room for me, kinda panicing.

    Are you able to answer a question of sin/cos graphs if I asked?
  24. Jul 26, 2011 #23
    If i had a cos function like y=4cos1/3(x+2pi) -4

    With the final points of

    -360 , 0

    -90, -4
    180, -8
    450 , -4
    720, 0

    How would i continue this for 2 cycles?

    The period is 1080, how do i get the points for the rest of the cycle? Would i have to perform all of these translations on 2 cycles of the normal x/y graph of the cos / sin function?

    ie) normally its..

    0 1
    90 0
    180 -1
    270, 0
    360, 1

    If i continued that to 450, would the next value be 0 ? then -1 ? then 0 ?
  25. Jul 26, 2011 #24


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    okay well from

    [tex]\frac{1}{sin^2x}+\frac{1}{cos^2x} = \frac{?+?}{sin^2xcos^2x}[/tex]

    I will help you partially: with 1/sin2x

    If you divide the common denominator (sin2xcos2x) by sin2x you will be left with cos2x and cos2x *1 (the numerator of 1/sin2x) is cos2x


    [tex]\frac{1}{sin^2x}+\frac{1}{cos^2x} = \frac{cos^2x+?}{sin^2xcos^2x}[/tex]

    Do the same for 1/cos2x and you will get the other '?'. Then just apply a simple trig identity.

    Every 1080 degrees then will go through one cycle. (-360 to 720 is 1080).

    So for the next cycle the 'y' values will repeat for the corresponding x values. (so the change from -360 to -90 will be added to the next value and so on)
  26. Jul 27, 2011 #25
    As you know that sin^2 x + cos^2 x = 1 hence the Left Hand Side is Sin^2 x as the denominator is 1.

    Now Prove Right Hand Side equal to sin^2 x too.

    The right side is tan^2 x / 1 + tan^2 x convert everything into sin and cos as shown below;

    (sin^2 x/cos^2 x)/(1 + sin^2 x/cos^2 x)

    = (sin^2 x/cos^2 x)/[(sin^2 x +cos^2 x)/cos^2 x]

    = (sin^2 x/cos^2 x) * cos^2 x/(sin^2 x +cos^2 x)

    = Sin^2 x which is equal to Left Hand Side

    Finally, if you can't prove Direct, simplify both sides into sin and cos and make them same.
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