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Thread: Trigonometry exam

  1. #1
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    Trigonometry exam

    So there is this problem
    If sinα*cosα = 1/2,
    and α ranges from (π ; 3π/2),
    how much is sinα + cosα?
    Thanks a lot!

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  2. #2
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    Re: Trigonometry exam

    $\sin(a)\cos(a) = \dfrac 1 2$

    apply magic identity $\sin(2x)=2\sin(x)\cos(x)$

    $\dfrac 1 2 \sin(2a) = \dfrac 1 2$

    $\sin(2a) = 1$

    $2a = \left\{\dfrac \pi 2 ,\dfrac{5\pi}{2}\right\}$

    $a = \left\{\dfrac \pi 4, \dfrac{5\pi}{4}\right\}$
    Thanks from carrie9987
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  3. #3
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    Re: Trigonometry exam

    Quote Originally Posted by romsek View Post
    $\sin(a)\cos(a) = \dfrac 1 2$

    apply magic identity $\sin(2x)=2\sin(x)\cos(x)$

    $\dfrac 1 2 \sin(2a) = \dfrac 1 2$

    $\sin(2a) = 1$

    $2a = \left\{\dfrac \pi 2 ,\dfrac{5\pi}{2}\right\}$

    $a = \left\{\dfrac \pi 4, \dfrac{5\pi}{4}\right\}$
    Thanks!

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  4. #4
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    Re: Trigonometry exam

    Quote Originally Posted by carrie9987 View Post
    Thanks!

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    I do hope that you weren't taking the exam at the time... :/
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  5. #5
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    Re: Trigonometry exam

    What I would do: $\displaystyle (sin(a)+ cos(a))^2= sin^2(a)+ 2sin(a)cos(a)+ cos^2(a)= 1+ 2sin(a)cos(a)$. Since we are given that $\displaystyle sin(a)cos9(a)= \frac{1}{2}$, $\displaystyle (sin(a)+ cos(a))^2= 1+ 1= 2$ so $\displaystyle sin(a)+ cos(a)$ is either $\displaystyle \sqrt{2}$, or $\displaystyle -\sqrt{2}$.

    Between $\displaystyle \pi$ and $\displaystyle 3\pi/2$, both sin(x) and cos(x) are negative so
    $\displaystyle sin(x)+ cos(x)= -\sqrt{2}$.
    Thanks from carrie9987
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    Re: Trigonometry exam

    Quote Originally Posted by HallsofIvy View Post
    What I would do: $\displaystyle (sin(a)+ cos(a))^2= sin^2(a)+ 2sin(a)cos(a)+ cos^2(a)= 1+ 2sin(a)cos(a)$. Since we are given that $\displaystyle sin(a)cos9(a)= \frac{1}{2}$, $\displaystyle (sin(a)+ cos(a))^2= 1+ 1= 2$ so $\displaystyle sin(a)+ cos(a)$ is either $\displaystyle \sqrt{2}$, or $\displaystyle -\sqrt{2}$.

    Between $\displaystyle \pi$ and $\displaystyle 3\pi/2$, both sin(x) and cos(x) are negative so
    $\displaystyle sin(x)+ cos(x)= -\sqrt{2}$.
    Thank you!
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  7. #7
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    Re: Trigonometry exam

    Quote Originally Posted by romsek View Post
    I do hope that you weren't taking the exam at the time... :/
    Haha, i wasn't, it's in two days
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