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Math Help - Find the value between 0 and 2pi

  1. #1
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    Find the value between 0 and 2pi

    Find the values of A in the range between 0 and 2pi for which
    sinA+sin3A=cosA+cos3A

    I have been trying for two days and still can't get it yet. Any help is greatly appreciated.
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  2. #2
    Member great_math's Avatar
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    Quote Originally Posted by arze View Post
    Find the values of A in the range between 0 and 2pi for which
    sinA+sin3A=cosA+cos3A

    I have been trying for two days and still can't get it yet. Any help is greatly appreciated.
    \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)

    \cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)

    \sin A+\sin 3A=2\sin2A\cos A

    \cos A+\cos 3A=2\cos 2A\cos A

    \Rightarrow 2\sin2A\cos A=2\cos 2A\cos A

    [edited from yeongil's post]
    \begin{aligned}<br />
2\sin2A\cos A &= 2\cos 2A\cos A \\<br />
\sin2A\cos A &= \cos 2A\cos A \\<br />
\sin2A\cos A - \cos 2A\cos A &= 0 \\<br />
\cos A(\sin2A - \cos 2A) &= 0<br />
\end{aligned}

    \begin{aligned}<br />
\cos A &= 0 \\<br />
A &= \frac{\pi}{2},\;\frac{3\pi}{2}<br />
\end{aligned}

    \Rightarrow \sin 2A=\cos 2A

    \Rightarrow \sin 2A=\sin\left(\frac\pi{2}-2A\right)

    \Rightarrow 2A=n\pi+(-1)^n\left(\frac\pi{2}-2A\right)

    0<A<2\pi

    Possible values \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{8},\frac{  7\pi}{2},\frac{9\pi}{8},\frac{11\pi}{2},\frac{13\p  i}{8}
    Last edited by great_math; July 9th 2009 at 12:36 AM. Reason: Made corrections
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  3. #3
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    Quote Originally Posted by great_math View Post
    \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)

    \cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)

    \sin A+\sin 3A=2\sin2A\cos A

    \cos A+\cos 3A=2\cos 2A\cos A

    \Rightarrow 2\sin2A\cos A=2\cos 2A\cos A

    \Rightarrow {\color{red}\sin 2A=\cos 2A}
    It's not a good idea to divide both sides by cos A like that. You lose roots that way. Instead, do this:

    \begin{aligned}<br />
2\sin2A\cos A &= 2\cos 2A\cos A \\<br />
\sin2A\cos A &= \cos 2A\cos A \\<br />
\sin2A\cos A - \cos 2A\cos A &= 0 \\<br />
\cos A(\sin2A - \cos 2A) &= 0<br />
\end{aligned}

    \begin{aligned}<br />
\cos A &= 0 \\<br />
A &= \frac{\pi}{2},\;\frac{3\pi}{2}<br />
\end{aligned}

    \begin{aligned}<br />
 \sin2A - \cos 2A &= 0 \\<br />
\sin 2A &= \cos 2A<br />
\end{aligned}
    Then proceed as in the previous post.


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  4. #4
    Member great_math's Avatar
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    Quote Originally Posted by yeongil View Post
    It's not a good idea to divide both sides by cos A like that. You lose roots that way. Instead, do this:

    \begin{aligned}<br />
2\sin2A\cos A &= 2\cos 2A\cos A \\<br />
\sin2A\cos A &= \cos 2A\cos A \\<br />
\sin2A\cos A - \cos 2A\cos A &= 0 \\<br />
\cos A(\sin2A - \cos 2A) &= 0<br />
\end{aligned}

    \begin{aligned}<br />
\cos A &= 0 \\<br />
A &= \frac{\pi}{2},\;\frac{3\pi}{2}<br />
\end{aligned}

    \begin{aligned}<br />
 \sin2A - \cos 2A &= 0 \\<br />
\sin 2A &= \cos 2A<br />
\end{aligned}
    Then proceed as in the previous post.


    01
    Oops! I keep making that mistake again and again ....
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  5. #5
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    Thank you so much!
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