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Thread: [SOLVED] New trigonometry problems

  1. #1
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    [SOLVED] New trigonometry problems

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle =\frac{1}{16}.1 $

    $\displaystyle =\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 $\displaystyle cos20.cos40.cos60.cos80$ = ?
    Second, determine of the value $\displaystyle tan9-tan27-tan63-tan81=$
    Third, if $\displaystyle sin(A+B).cosC=2sinA .cos (B-C)$ prove that $\displaystyle tanA - \frac{sin(B-A)}{2sinAsinB}$
    HI

    For (1)

    $\displaystyle \cos 20\cdot \cos 40\cdot \cos 60\cdot \cos 80$

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

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

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

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

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

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

    $\displaystyle
    =\frac{1}{16}
    $
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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 $\displaystyle sin(A+B).cosC=2sinA .cos (B-C)$ prove that $\displaystyle tanC = tanA - \frac{sin(B-A)}{2sinAsinB}$

    THis is the furthest i can get ..

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

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

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

    Divide both sides by $\displaystyle \cos C$

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

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

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

    $\displaystyle
    \tan C=\frac{\sin (B-A)}{2\sin A\sin B}
    $

    $\displaystyle \tan A$ ??
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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 $\displaystyle tanC = \frac{sin(B-A)}{2sinAsinB}$ i mistyped the third problem..
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