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Math Help - Series in Trigonometry

  1. #1
    Junior Member mrmohamed's Avatar
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    Lightbulb Series in Trigonometry

    Hi all mathlinkers
    here i will post trigonometry problems and i wish that it will be good for al

    (1)


    with my best wishes
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  2. #2
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    Hello, mrmohamed!

    1) Prove that: . \cos(15^o-x)\sec15^o - \sin(15^o-x)\csc15^o \:=\:4\sin x

    We have: . \frac{\cos(15-x)}{\cos15} - \frac{\sin(15-x)}{\sin15}

    . . . . . =\;\; \frac{\cos15\cos x + \sin15\sin x}{\cos15} - \frac{\sin15\cos x - \cos15\sin x}{\sin 15}

    . . . . . =\;\; \frac{\sin15(\cos15\cos x + \sin15\sin x) - \cos15(\sin15\cos x - \cos15\sin x)}{\sin15\cos15}

    . . . . . =\;\; \frac{{\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \sin^2\!15\sin x - {\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \cos^2\!15\sin x}{\sin15\cos15}

    . . . . . =\;\; \frac{\overbrace{(\sin^2\!15 + \cos^2\!15)}^{\text{This is 1}}\sin x}{\sin15\cos15} \;\;=\;\;\frac{\sin x}{\sin15\cos15}

    . . . . . =\;\;\frac{\sin x}{\frac{1}{2}\sin30} \;\;=\;\;\frac{\sin x}{\frac{1}{2}\cdot\frac{1}{2}} \;\;=\;\;4\sin x

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  3. #3
    Junior Member mrmohamed's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, mrmohamed!


    We have: . \frac{\cos(15-x)}{\cos15} - \frac{\sin(15-x)}{\sin15}

    . . . . . =\;\; \frac{\cos15\cos x + \sin15\sin x}{\cos15} - \frac{\sin15\cos x - \cos15\sin x}{\sin 15}

    . . . . . =\;\; \frac{\sin15(\cos15\cos x + \sin15\sin x) - \cos15(\sin15\cos x - \cos15\sin x)}{\sin15\cos15}

    . . . . . =\;\; \frac{{\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \sin^2\!15\sin x - {\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \cos^2\!15\sin x}{\sin15\cos15}

    . . . . . =\;\; \frac{\overbrace{(\sin^2\!15 + \cos^2\!15)}^{\text{This is 1}}\sin x}{\sin15\cos15} \;\;=\;\;\frac{\sin x}{\sin15\cos15}

    . . . . . =\;\;\frac{\sin x}{\frac{1}{2}\sin30} \;\;=\;\;\frac{\sin x}{\frac{1}{2}\cdot\frac{1}{2}} \;\;=\;\;4\sin x
    thank u mr soroban
    very good solution
    still there another methode
    i will wait
    mrmohamed
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  4. #4
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    Quote Originally Posted by mrmohamed View Post
    thank u mr soroban
    very good solution
    still there another methode
    i will wait
    mrmohamed
    If you already know the solution and want to post a problem just to challenge others, we have a particular forum for that. I'm moving this thread there.
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  5. #5
    Junior Member mrmohamed's Avatar
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    Quote Originally Posted by Jameson View Post
    If you already know the solution and want to post a problem just to challenge others, we have a particular forum for that. I'm moving this thread there.
    hi sir
    at first im sorry to put the topic in the wrong place
    second
    i dislike to challenge others but im looking for more ideas to learn more
    thanks sir and u cane move the topic as u liks
    mrmohamed
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  6. #6
    Junior Member mrmohamed's Avatar
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    Lightbulb

    Hi all
    my methode

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  7. #7
    Junior Member mrmohamed's Avatar
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    Hi all

    ( 2 )



    with my best wishes
    mrmohamed


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  8. #8
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    Note that


     \cos(\frac{5\pi}{8}) = \cos( \pi - \frac{3\pi}{8} )  = -\cos(\frac{3\pi}{8})

    Similarly ,


     \cos(\frac{7\pi}{8} = -\cos(\frac{\pi}{8})

    so we have to consider

     ( 1 + u)(1+v)(1-v)(1-u)

     u = \cos(\frac{\pi}{8})  , v = \cos(\frac{3\pi}{8})

     = (1 - u^2)(1 - v^2 ) = \sin^2(\frac{\pi}{8})  \sin^2(\frac{3\pi}{8})

     = \frac{1}{4} ( 1 - \cos(\frac{\pi}{4}))(1 - \cos(\frac{3\pi}{4})) = \frac{1}{8}


    But i guess you have a better and interesting method to solve it
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  9. #9
    Junior Member mrmohamed's Avatar
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    Quote Originally Posted by simplependulum View Post
    Note that


     \cos(\frac{5\pi}{8}) = \cos( \pi - \frac{3\pi}{8} ) = -\cos(\frac{3\pi}{8})

    Similarly ,


     \cos(\frac{7\pi}{8} = -\cos(\frac{\pi}{8})

    so we have to consider

     ( 1 + u)(1+v)(1-v)(1-u)

     u = \cos(\frac{\pi}{8}) , v = \cos(\frac{3\pi}{8})

     = (1 - u^2)(1 - v^2 ) = \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8})

     = \frac{1}{4} ( 1 - \cos(\frac{\pi}{4}))(1 - \cos(\frac{3\pi}{4})) = \frac{1}{8}



    But i guess you have a better and interesting method to solve it

    Hi all
    I wish the idea will be new
    unusual methode


    with my best wishes
    mrmohamed
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  10. #10
    Junior Member mrmohamed's Avatar
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    Hi all

    (3)


    with my best wishes
    mrmohamed
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  11. #11
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     \cos(\frac{A}{2} ) \cos(\frac{B-C}{2} )

     = \cos(\frac{ \pi - (B+C)}{2} ) \cos(\frac{B-C}{2} )


     = \sin(\frac{B+C}{2})\cos(\frac{B-C}{2} )

     =  \frac{1}{2} [ \sin(B) + \sin(C) ]

    similar to the others

    finally we have

     \frac{1}{2} [   \sin(B) + \sin(C) +  \sin(A) + \sin(C) +  \sin(B) + \sin(A )


     =  \sin(A) + \sin(B) + \sin(C)
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