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Math Help - Trig Identity Homework Help Please!

  1. #1
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    Trig Identity Homework Help Please!

    I'm having trouble with some trig hwk problems, these are the specific ones. If anyone can help me i would appreciate it. Btw, this is my first post on here, heard about the forum from a friend, hopefully people can help me a lot on here! I need all the extra help i can get! Thanks!

    3) If sinA=-5/13 with A in QIII, find the exact value of:

    a) cos A/2(half-angle formula) b) sin2A(Double-Angle formula)


    7) Find all RADIAN solutions. Give your anwers as EXACT values.

    2cos²x-5sinx-4=0

    8) Find all solutions Θ for 0°≤Θ<360° NOT 0°≤Θ≤360°.

    √3sinΘ+cosΘ=1
    Last edited by norcal1.8t; November 30th 2008 at 06:52 PM.
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  2. #2
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    did i post too many problems at once?
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  3. #3
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    ok i made the list smaller because i was able to get some worked out. nobody has time to help? i need this for tommorow
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  4. #4
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    yo moderator, help a newb out! Im falling into the darkness of the sinx world!!!!!!!
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  5. #5
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    Re :

    Quote Originally Posted by norcal1.8t View Post
    I'm having trouble with some trig hwk problems, these are the specific ones. If anyone can help me i would appreciate it. Btw, this is my first post on here, heard about the forum from a friend, hopefully people can help me a lot on here! I need all the extra help i can get! Thanks!

    3) If sinA=-5/13 with A in QIII, find the exact value of:

    a) cos A/2(half-angle formula) b) sin2A(Double-Angle formula)


    7) Find all RADIAN solutions. Give your anwers as EXACT values.

    2cos²x-5sinx-4=0

    8) Find all solutions Θ for 0°≤Θ<360° NOT 0°≤Θ≤360°.

    √3sinΘ+cosΘ=1


    (1) cos2A=2cos^2 A-1
    cosA=2cos^2(A/2)-1
    cos(A/2)=SQRT((cosA+1)/2 )
    note that cos A = -12/13
    sin2A= 2sinAcosA

    (2)Remember cos^2 x + sin^2 x = 1
    2 ( 1-sin^2 X) - 5sin X -4 = 0
    i guess you can work this out
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  6. #6
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    Re :

    Quote Originally Posted by norcal1.8t View Post
    8) Find all solutions Θ for 0°≤Θ<360° NOT 0°≤Θ≤360°.

    √3sinΘ+cosΘ=1
    (√3sinΘ+cosΘ)^2=(1)^2
    3sin^2Θ+2SQRT3sinΘcosΘ+cos^2Θ=1
    3sin^2Θ+2SQRT3sinΘcosΘ=sin^2Θ
    2sin^2Θ(sinΘ+SQRT3cosΘ)=0
    nvm , then solve for Θ .
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  7. #7
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    Thumbs up

    phew, thanks a lot of, that saved me lots of trouble! Appreciate it!
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