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Math Help - Solving equations

  1. #1
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    Solving equations

    Solve the following equation giving values for 0-360deg
    cosA=cos2A+cos4A

    cos2A=cos(30-A)

    i probably missed something on the identities part.
    thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    cos A=cos 2A+cos 4A

    cos A=2cos 3A cos A

    cos A=0 or cos 3A=\frac{1}{2}

    A=90^0 or 3A=60^0

    A=90^0 or A=20^0
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  3. #3
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    Quote Originally Posted by arze View Post
    Solve the following equation giving values for 0-360deg
    cosA=cos2A+cos4A
    Hi

    You can use the trig identity \cos p + \cos q = 2 \cos \left(\frac{p+q}{2}\right) \: \cos \left(\frac{p-q}{2}\right)
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  4. #4
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    ah! that's what i missed in the first one. thanks!
    how about the second one?
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by arze View Post
    ah! that's what i missed in the first one. thanks!
    how about the second one?
    cos 2A = cos (30-A)

    2A=30^o-A

    3A=30^o

    A=10^o

    OR

    2A=A-30

    A=-30^0
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  6. #6
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    Quote Originally Posted by alexmahone View Post
    cos 2A = cos (30-A)

    2A=30^o-A

    3A=30^o

    A=10^o

    OR

    2A=A-30

    A=-30^0
    i'm supposed to give the answers 10^o, 130^o, 250^o, and 330<br />
^o
    i can understand 10^o and 330^o but what about the other two?
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  7. #7
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    Quote Originally Posted by arze View Post
    i'm supposed to give the answers 10^o, 130^o, 250^o, and 330<br />
^o
    i can understand 10^o and 330^o but what about the other two?
    Looking back at alexmahone's work:
    Quote Originally Posted by alexmahone View Post
    cos 2A = cos (30-A)

    2A=30^o-A

    3A=30^o
    If A is supposed to be between 0 and 360, then 3A should be between 0 and 1080 (360 times 3). So from here you also have to mention that

    3A=390^{\circ} and

    3A=750^{\circ}

    because these angles are coterminal to 30 and they are between 0 and 1080. Now divide both sides of both equations by 3 and you will get

    A=130^{\circ} and

    A=250^{\circ}.


    01
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  8. #8
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    how could i have missed that?
    thank you very much!
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