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Math Help - [SOLVED] New trigonometry problems

  1. #1
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    [SOLVED] New trigonometry problems

    First, what is the value of cos20.cos40.cos60.cos80 = ?
    Second, determine of the value tan9-tan27-tan63-tan81=
    Third, if sin(A+B).cosC=2sinA .cos (B-C) prove that tanA - \frac{sin(B-A)}{2sinAsinB}
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  2. #2
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    I revise for the third problems :
    Third, if sin(A+B).cosC=2sinA .cos (B-C) prove that tanC = tanA - \frac{sin(B-A)}{2sinAsinB}
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  3. #3
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    Quote Originally Posted by wizard654zzz View Post
    First, what is the value of cos20.cos40.cos60.cos80 = ?
    Second, determine of the value tan9-tan27-tan63-tan81=
    Third, if sin(A+B).cosC=2sinA .cos (B-C) prove that tanA - \frac{sin(B-A)}{2sinAsinB}
    ill try the first one....
    cos20.cos40.cos60.cos80=

    =cos20.\frac{sin20}{sin20}.cos40.cos60.cos80

    =\frac{cos20.sin20}{sin20}.cos40.\frac{1}{2}.cos80

    =\frac{\frac{1}{2}sin40}{sin20}.cos40.\frac{1}{2}.  cos80

    remember:
    2sinXcosX=sin2X
    so, sinXcosX=\frac{sin2X}{2}
    =\frac{1}{2}\frac{1}{2} \frac{sin40.cos40}{sin20}.cos80

    =\frac{1}{4} \frac{\frac{1}{2}sin80}{sin20}.cos80

    =\frac{1}{4}\frac{1}{2} \frac{sin80.cos80}{sin20}

    =\frac{1}{8} \frac{\frac{1}{2}sin160}{sin20}

    =\frac{1}{16} \frac{sin160}{sin20}

    =\frac{1}{16} \frac{sin(180-20)}{sin20}

    remember:
    Sin(180-x)=Sin(x)
    \frac{1}{16} \frac{sin20}{sin20}

    =\frac{1}{16}.1

    =\frac{1}{16}

    ........hope it help.........
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  4. #4
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    Quote Originally Posted by wizard654zzz View Post
    First, what is the value of cos20.cos40.cos60.cos80 = ?
    Second, determine of the value tan9-tan27-tan63-tan81=
    Third, if sin(A+B).cosC=2sinA .cos (B-C) prove that tanA - \frac{sin(B-A)}{2sinAsinB}
    HI

    For (1)

    \cos 20\cdot \cos 40\cdot \cos 60\cdot \cos 80

    =\frac{1}{2}(\cos 20\cdot \cos 40\cdot \cos 80)

    =\frac{1}{2}(\cos 20 \cdot \frac{1}{2}(\cos 120+\cos 40))

    =\frac{1}{2}(\cos 20\cdot \frac{1}{2}(-\frac{1}{2}+\frac{1}{2}\cos 40))

    =\frac{1}{2}(-\frac{1}{4}\cos 20+\frac{1}{2}\cos 40\cos 20)

    =\frac{1}{2}[-\frac{1}{4}\cos 20+\frac{1}{2}(\frac{1}{2}(\cos 60+\cos 20))]

    =\frac{1}{2}(\frac{1}{8}+\frac{1}{4}\cos 20-\frac{1}{4}\cos 20)

     <br />
=\frac{1}{16}<br />
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  5. #5
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    Quote Originally Posted by wizard654zzz View Post
    I revise for the third problems :
    Third, if sin(A+B).cosC=2sinA .cos (B-C) prove that tanC = tanA - \frac{sin(B-A)}{2sinAsinB}

    THis is the furthest i can get ..

    \cos C\sin (A+B)=2\sin A\cos (B-C)

    \cos C[\sin A\cos B+\cos A\sin B]=2\sin A[\cos B\cos C+\sin B\sin C]

    \sin A\cos B\cos C+\cos A\cos C\sin B=2\sin A\cos B\cos C+2\sin A\sin B\sin C

    Divide both sides by \cos C

    \sin A\cos B+\cos A\sin B=2\sin A\cos B+2\sin A\sin B\tan C

    \sin B\cos A-\cos B\sin A=2\sin A\sin B\tan C

    \sin (B-A)=2\sin A\sin B\tan C

     <br />
\tan C=\frac{\sin (B-A)}{2\sin A\sin B}<br />

    \tan A ??
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  6. #6
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    Thank you

    I would like to say thanks to pencil09 and mathaddict.
    mathaddict you are right tanC = \frac{sin(B-A)}{2sinAsinB} i mistyped the third problem..
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