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Math Help - Trig identity question

  1. #1
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    Trig identity question

    Prove that

    4 cos (θ + 60) cos (θ + 30) = √3 - 2sinθ

    I have said

    4 cos (θ + 60) cos (θ + 30)

    = 4(cosθ cos60 − sinθ sin 60) (cosθ cos30 − sinθ sin 30)

    = 4(cosθ 1/2 - sinθ √3/2) (cosθ √3/2 − sinθ 1/2 )

    not sure where to go now

    thanks
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by gracey View Post
    Prove that

    4 cos (θ + 60) cos (θ + 30) = √3 - 2sinθ

    I have said

    4 cos (θ + 60) cos (θ + 30)

    = 4(cosθ cos60 − sinθ sin 60) (cosθ cos30 − sinθ sin 30)

    = 4(cosθ 1/2 - sinθ √3/2) (cosθ √3/2 − sinθ 1/2 )

    not sure where to go now

    thanks
    Hi gracey,

    Are you sure the right side of the equation is not \sqrt{3}-2 \sin {\color{red}2}\theta?
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  3. #3
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    Hello, Gracey!

    masters is right . . . you left out a "2" . . .


    Prove that: . 4\cos(\theta + 60)\cos(\theta+30) \:=\:\sqrt{3} - 2\sin{\color{red}2}\theta


    I have said:

    4\cos(\theta+60)\cos(\theta+30) \;=\; 4(\cos\theta\cos60 - \sin\theta\sin60) (\cos\theta\cos30 - \sin\theta\sin30)

    . . . . . = \;4\left(\cos\theta\cdot\tfrac{1}{2} - \sin\theta\cdot\tfrac{\sqrt{3}}{2}\right)\left(\co  s\theta\cdot\tfrac{\sqrt{3}}{2} - \sin\theta\cdot \tfrac{1}{2}\right)

    not sure where to go now . . . Really?

    There is only one thing you can do ... multiply it out!


    You have: . 4\left(\frac{\cos\theta - \sqrt{3}\sin\theta}{2}\right)\left(\frac{\sqrt{3}\  cos\theta - \sin\theta}{2}\right)

    . . . . . . =\;(\cos\theta - \sqrt{3}\sin\theta)(\sqrt{3}\cos\theta - \sin\theta)

    . . . . . . = \;\sqrt{3}\cos^2\!\theta - \sin\theta\cos\theta - 3\sin\theta\cos\theta + \sqrt{3}\sin^2\!\theta


    . . . . . . = \;\sqrt{3}\underbrace{(\cos^2\!\theta + \sin^2\!\theta)}_{\text{This is 1}} - 4\sin\theta\cos\theta

    . . . . . . = \;\sqrt{3} - 2\underbrace{(2\sin\theta\cos\theta)}_{\text{This is }\sin2\theta}

    . . . . . . =\;\sqrt{3}-2\sin2\theta

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