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Math Help - trigonometry (1)

  1. #1
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    trigonometry (1)

    Without using mathematical tables or calculator , show that
     \theta=\frac{\pi}{10} satisfies the equation
    \sin2\theta=\cos3\theta .



    Thanks for ur help .
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  2. #2
    Moo
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    Hello,

    Consider : \cos^2 3\theta-\sin^2 2\theta
    ------------------------------------------------------
    Remember these identities :
    \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)

    \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)

    Multiply them :

    \begin{aligned}<br />
\cos(a-b)\cos(a+b) &=\cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\<br />
&=\cos^2(b)(1-\sin^2(a))-\sin^2(a)\sin^2(b) \\<br />
&=\cos^2(b)-\sin^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\<br />
&=\cos^2(b)-\sin^2(a)(\cos^2(b)+\sin^2(b)) \\<br />
&=\boxed{\cos^2(b)-\sin^2(a)}<br />
\end{aligned}<br />
    ------------------------------------------------------

    Hence \cos^2 3\theta-\sin^2 2\theta=\cos \theta \cos 5\theta

    And if \theta=\frac{\pi}{10}, then \cos 5\theta=0

    So 0=\cos^2 3\theta-\sin^2 2\theta=(\cos 3\theta-\sin 2\theta)(\cos 3\theta+\sin 2\theta) (1)

    Now notice that :
    3\theta=\frac{3\pi}{10} and 2\theta=\frac\pi 5 are both in the first quadrant (because <\frac\pi 2)
    So \cos 3\theta>0 and \sin 2\theta>0

    Hence \cos 3\theta+\sin 2\theta>0


    And it follows from (1) that \boxed{\cos 3\theta-\sin 2\theta=0} if \theta=\frac{\pi}{10}


    Looks clear to you ?
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  3. #3
    Flow Master
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    Quote Originally Posted by thereddevils View Post
    Without using mathematical tables or calculator , show that

     \theta=\frac{\pi}{10} satisfies the equation
    \sin2\theta=\cos3\theta .




    Thanks for ur help .
    Since it's a 'show question' (rather than a 'solve question') the following is perfectly acceptable:

    \cos \left( \frac{3 \pi}{10} \right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{10} \right) = \sin \left( \frac{2 \pi}{10} \right)
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Consider : \cos^2 3\theta-\sin^2 2\theta
    ------------------------------------------------------
    Remember these identities :
    \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)

    \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)

    Multiply them :

    \begin{aligned}<br />
\cos(a-b)\cos(a+b) &=\cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\<br />
&=\cos^2(b)(1-\sin^2(a))-\sin^2(a)\sin^2(b) \\<br />
&=\cos^2(b)-\sin^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\<br />
&=\cos^2(b)-\sin^2(a)(\cos^2(b)+\sin^2(b)) \\<br />
&=\boxed{\cos^2(b)-\sin^2(a)}<br />
\end{aligned}<br />
    ------------------------------------------------------

    Hence \cos^2 3\theta-\sin^2 2\theta=\cos \theta \cos 5\theta

    And if \theta=\frac{\pi}{10}, then \cos 5\theta=0

    So 0=\cos^2 3\theta-\sin^2 2\theta=(\cos 3\theta-\sin 2\theta)(\cos 3\theta+\sin 2\theta) (1)

    Now notice that :
    3\theta=\frac{3\pi}{10} and 2\theta=\frac\pi 5 are both in the first quadrant (because <\frac\pi 2)
    So \cos 3\theta>0 and \sin 2\theta>0

    Hence \cos 3\theta+\sin 2\theta>0


    And it follows from (1) that \boxed{\cos 3\theta-\sin 2\theta=0} if \theta=\frac{\pi}{10}


    Looks clear to you ?

    What a beautiful solution ! Thanks Moo . Thanks Mr F too , for that simple and easy approach
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