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Math Help - Proving Trigonometric Identities II

  1. #1
    Super Member Quacky's Avatar
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    Proving Trigonometric Identities II

    Given that \alpha is acute and that tan \alpha = \frac {3}{4}, prove that:

     3 sin (\theta + \alpha) + 4 cos (\theta + \alpha) = 5 cos \theta

    I start by saying that  R cos \alpha = 4 and  R sin \alpha = 3 but then I get stuck.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Quacky View Post
    Given that \alpha is acute and that tan \alpha = \frac {3}{4}, prove that:

     3 sin (\theta + \alpha) + 4 cos (\theta + \alpha) = 5 cos \theta

    I start by saying that  R cos \alpha = 4 and  R sin \alpha = 3 but then I get stuck.
    Employ the addition formulas.
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  3. #3
    Super Member Quacky's Avatar
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    Do you mean as in Sin(A+B)=SinACosB+CosASinB? I've tried but I can't work out how to get an answer in only \theta
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  4. #4
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    Quote Originally Posted by Quacky View Post
    Given that \alpha is acute and that tan \alpha = \frac {3}{4}, prove that:

     3 sin (\theta + \alpha) + 4 cos (\theta + \alpha) = 5 cos \theta

    I start by saying that  R cos \alpha = 4 and  R sin \alpha = 3 but then I get stuck.
    Hi Quacky,

    Since \alpha is acute, draw a right-angled triangle.

    Tan\alpha=\frac{3}{4}=\frac{opp}{adj}

    Hence, the hypotenuse is \sqrt{3^2+4^2}=\sqrt{25}=5

    Hence, Sin\alpha=\frac{3}{5},\ Cos\alpha=\frac{4}{5}

    Then...

    3Sin(\theta+\alpha)+4Cos(\theta+\alpha)=3(Sin\alph  a\ Cos\theta+Cos\alpha\ Sin\theta)+4(Cos\theta\ Cos\alpha-Sin\theta\ Sin\alpha)

    =3(\frac{3}{5}Cos\theta+\frac{4}{5}Sin\theta)+4(\f  rac{4}{5}Cos\theta-\frac{3}{5}Sin\theta)

    =\frac{9}{5}Cos\theta+\frac{12}{5}Sin\theta+\frac{  16}{5}Cos\theta-\frac{12}{5}Sin\theta=\frac{25}{5}Cos\theta=5Cos\t  heta
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  5. #5
    Super Member Quacky's Avatar
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    Ah. I would never have got that. Thanks again.
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