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Math Help - Trigonometry identities help

  1. #1
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    Trigonometry identities help

    Section B

    Number 9) II

    How do go about answering this?

    Please see attached pdf

    What i did
    ________

    cos2x=1-2sin275

    cos 2x = cos (x+x)

    cos (x+x) = cosA*CosB-SinA*SinB

    cos( 60 + 90)= cos60*cos90 - sin60*cos90

    answer i got was wrong

    section B part III number 9
    ____________________

    cos 2x = cos2x - sin2x


    cos( x + x) = cosA*CosB-SinA*SinB

    cos(90 + 45)= cos90*cos45-sin90*sin45




    I attempted this one similarly.. wrong answer... help!
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    Last edited by Benja303; November 3rd 2012 at 04:17 PM.
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  2. #2
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    Re: Trigonometry identities help

    9. II: Note that 1 - 2 \sin^2 {75^{\circ}} = \cos(150^{\circ}).
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  3. #3
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    Re: Trigonometry identities help

    I know, sorry i made a mistake in labelling and typing , now fixed,
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  4. #4
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    Re: Trigonometry identities help

    Oh okay, I was a little confused with the 45/60 thing.

    Also, \sin(60^{\circ}) = \frac{\sqrt{3}}{2}, not \frac{2}{\sqrt{3}}. The sine/cosine of an angle can only be from -1 to 1 inclusive. Since -\frac{2}{\sqrt{3}} is outside the set [-1,1], this answer cannot be correct.
    Last edited by richard1234; November 3rd 2012 at 04:11 PM.
    Thanks from Benja303
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  5. #5
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    Re: Trigonometry identities help

    got it! thanks !
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  6. #6
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    Re: Trigonometry identities help

    what about 9) III? my answer is -1/square root of 2
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