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Math Help - Equations involving trig identities/formulae

  1. #1
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    Equations involving trig identities/formulae

    I just need someone to guide me on how to get started on these. Generally I change the identities I spot into its equivalent, before proceeding to get lost.

    Anyway, hopefully I can format these questions correctly:

    a) 2sinx -cotx -cscx = 0

    b) sin2x + cosx = 0

    c) sinx - cosx = 0
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  2. #2
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    Quote Originally Posted by Cheagles View Post
    I just need someone to guide me on how to get started on these. Generally I change the identities I spot into its equivalent, before proceeding to get lost.

    Anyway, hopefully I can format these questions correctly:

    a) 2sinx -cotx -cscx = 0

    b) sin2x + cosx = 0

    c) sinx - cosx = 0
    For number 1, try multiplying all by sinx, and then using sin^2x+cos^2x=1. For number two, use the double angle formula for sin2x, do you know it? Oh, and for number 3 I'd suggest multiplying by (sinx+cosx) and see what you can do from there. Remember, you're just looking for ways to reduce to a single identity.
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  3. #3
    Member arpitagarwal82's Avatar
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    1)

    2sinx -cotx -cscx = 0

    2sinx - cosx/sinx - 1/sinx =

    (2\sin^2 x - \cos x - 1)/sinx = 0

    1 - 2\cos^2 x - \cos x -1 = 0

    2 \cos^2x + \cos x = 0

    Now its simple

    2) sin2x + cosx = 0
    2\cos^2 x - 1 + \cos x = 0
    Its simple quadratic equation in cos x


    3) sinx - cosx = 0

    \sin x = cos x
    \tan x = 1
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  4. #4
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    Thank you, JeWiSh and arpitagarwal82. I just got done with the homework packet and ended up with only two mistakes, that I'm still trying to wrap my head around:

    1) tanx + secx = 1

    and

    2) sin2x + cosx = 0 (I realize I already asked this, but I got it wrong. After applying the quadratic formula, I ended up with 1/2 and -1. I looked at the unit circle and got 60, 180, and 300 degrees. Where did I screw up?)

    Thanks again.
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