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Math Help - proving trig identities

  1. #1
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    proving trig identities

    a) 1/cos^2Θ + 1/sin^2Θ = (tanΘ + 1/tanΘ)^2

    b) cos^4Θ - sin^4Θ = cos^2Θ - sin^2Θ


    thank youu.
    Last edited by checkmarks; April 14th 2007 at 06:17 PM.
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  2. #2
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    Quote Originally Posted by checkmarks View Post
    b) cos^4Θ - sin^4Θ = cos^2Θ - sin^2Θ
    .
    Difference of two square

    (cos^2 x+sin^2 x)(cos^2 x - sin^2 x)

    Pythagoren Identity

    cos^2 x - sin^2 x
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  3. #3
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    Hello, checkmarks!

    a) .1/cos²θ + 1/sin²θ .= .(tanθ + 1/tanθ)²

    This is a messy one . . . Start with the right side.

    . . . . . . . . .sinθ . . cosθ . . . . . . sin²θ + cos²θ . . . . . . . . . 1
    We have: .(------ + ------)² . = . (-----------------)² . = . --------------
    . . . . . . . . .cosθ . .sinθ . . . . . . . . sinθ·cosθ . . . . . . . sin²θ·cos²θ

    . . . . .sin²θ + cos²θ . . . . . .sin²θ . . . . . . .cos²θ . . . . . . . .1 . . . . . 1
    . . = . ---------------- . = . -------------- + --------------- . = . ------- + -------
    . . . . . .sin²θ·cos²θ . . . . sin²θ·cos²θ . . sin²θ·cos²θ . . . . .cos²θ . . sin²θ

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  4. #4
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    thanks!!

    i also need help with these 3:

    a) tan^2Θ - sin^2Θ = tan^2Θ(sin^2Θ)

    b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)
    Last edited by checkmarks; April 15th 2007 at 08:51 AM.
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  5. #5
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    Quote Originally Posted by checkmarks View Post
    a) tan^2Θ - sin^2Θ = tan^2Θ(sin^2Θ)
    Consider LHS
    tan^2Θ - sin^2Θ = sin^2 Θ/cos^2 Θ – sin^2 Θ
    …………………= (sin^2 Θ – cos^2 Θsin^2 Θ)/cos^2 Θ
    …………………= [sin^2 Θ(1 – cos^2 Θ)]/cos^2 Θ
    …………………= (sin^2 Θ * sin^2 Θ)/cos^2 Θ
    …………………= sin^2 Θ * sin^2 Θ/cos^2 Θ
    …………………= sin^2 Θ * tan^2 Θ
    …………………= RHS

    Thus we see, LHS = RHS

    b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)
    Use the fact that x^3 + y^3 = (x + y)(x^2 – xy + y^2)

    Consider LHS
    cos^3Θ + sin^3Θ = (cos Θ + sin Θ)(cos^2 Θ - sin Θcos Θ + sin^2 Θ) …sin^2 Θ + cos^2 Θ = 1
    ………………….= (cos Θ + sin Θ)(1 - sin Θcos Θ)
    ………………….= RHS

    Thus we see, LHS = RHS
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    OR

    If you didn’t remember the rule x^3 + y^3 = (x + y)(x^2 – xy + y^2), it would be a little harder

    b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)
    Consider RHS
    (cosΘ + sinΘ)(1 - sinΘcosΘ) = cosΘ – sinΘcos^2Θ + sinΘ – sin^2ΘcosΘ
    ……………………………...= cosΘ(1 – sinΘcosΘ) + sinΘ(1 – sinΘcosΘ)
    ……………………………...= cosΘ(sin^2Θ + cos^2Θ – sinΘcosΘ) + sinΘ(sin^2Θ + cos^2Θ – sinΘcosΘ)
    ……………………………...= cosΘsin^2Θ + cos^3Θ – sinΘcos^2Θ + sin^3Θ + sinΘcos^2Θ – sin^2ΘcosΘ
    ……………………………...= cos^3Θ + sin^3Θ
    ...........................= LHS
    Thus we see, LHS = RHS
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  7. #7
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    Tips for approaching these kinds of problems:

    1) Always start on the most complicated side
    2) Change everything to sine and cosine. These are the trig functions all students are most familiar with, it will be easy to see patterns with them
    3) Note that you may have to work on both sides. Working on one side and trying to get the other side does not always work. If you find you can't get the other side, stop and work on the other side to get where you left off on the first side.
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  8. #8
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    Hello again, checkmarks!

    a) .cscθ(cscθ + cotθ) .= .1/(1 - cosθ)

    . . . . . . . . . 1. . . . .1 . . . cosθ . . . . .1 + cosθ . . . . 1 + cosθ
    We have: . ----- · (------ + ------) . = . ---------- . = . ------------
    . . . . . . . . sinθ . . sinθ . . sinθ . . . . . . sin²θ . . . . . 1 - cos²θ


    . . . . . . . . . . . . . . . . . . .1 + cosθ . . . . . . . . . .1
    Factor and reduce: . ------------------------ . = . ----------
    . . . . . . . . . . . . . . .(1 - cosθ)(1 + cosθ) . . . .1 - cosθ



    b) .tan²θ - sin²θ .= .tan²θ·sin²θ

    The right side is: .tan²θ(1 - cos²θ) . = . tan²θ - tan²θ·cos²θ

    . . . . . . . . . . . .sin²θ
    . . . = . tan²θ - --------·cos²θ . = . tan²θ - sin²θ
    . . . . . . . . . . . .cos²θ



    c) .cos³θ + sin³θ .= .(cosθ + sinθ)·(1 - sinθ·cosθ)

    The left side is the sum of two cubes . . . factor it!

    . . . cos³θ + sin³θ . = . (cosθ + sinθ)·(cos²θ - sinθ·cosθ + sin²θ)


    But sin²θ + cos²θ .= .1

    . . Hence, we have: .(cosθ + sinθ)·(1 - sinθ·cosθ)



    Hmmm, too slow . . . again!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hmmm, too slow . . . again!
    Sorry Soroban. As usual, I did not realize you were working on the problem, or I would have left it to you--you can give a lot more help than I can.
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  10. #10
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    thanks so much, both of you
    ive been given around 40-50 of these problems for the weekend, these are just the ones im a little stuck on
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  11. #11
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    Quote Originally Posted by checkmarks View Post
    thanks so much, both of you
    ive been given around 40-50 of these problems for the weekend, these are just the ones im a little stuck on
    how can you be a little stuck, either you're stuck or you're not
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  12. #12
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    Quote Originally Posted by Jhevon View Post
    how can you be a little stuck, either you're stuck or you're not
    haha hey, give me a little slack
    i usually start off correctly, i just dont know how to continue..hence..."a little stuck"
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  13. #13
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    Quote Originally Posted by checkmarks View Post
    haha hey, give me a little slack
    i usually start off correctly, i just dont know how to continue..hence..."a little stuck"
    ok, i'll accept that
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