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Thread: Proving Trigonometric Identities II

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

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

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

    I start by saying that $\displaystyle R cos \alpha = 4 $ and $\displaystyle 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 $\displaystyle \alpha$ is acute and that $\displaystyle tan \alpha = \frac {3}{4}$, prove that:

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

    I start by saying that $\displaystyle R cos \alpha = 4 $ and $\displaystyle 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 $\displaystyle \theta$
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  4. #4
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    Quote Originally Posted by Quacky View Post
    Given that $\displaystyle \alpha$ is acute and that $\displaystyle tan \alpha = \frac {3}{4}$, prove that:

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

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

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

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

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

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

    Then...

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

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

    $\displaystyle =\frac{9}{5}Cos\theta+\frac{12}{5}Sin\theta+\frac{ 16}{5}Cos\theta-\frac{12}{5}Sin\theta=\frac{25}{5}Cos\theta=5Cos\t heta$
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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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