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Thread: trig cont.

  1. #1
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    trig cont.

    if 0<a<pi/2

    prove that tana = root all (1-cos2a)/(1+cos2a)
    ( the square root is above the entire right hand side)

    and

    if 4tan(a-b) = 3tana, prove that tanb = sin2a/(7+cos2a)

    thanks for any help, i really appreciate it ! =D
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  2. #2
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    Quote Originally Posted by flyinhigh123 View Post
    if 0<a<pi/2

    prove that tana = root all (1-cos2a)/(1+cos2a)
    ( the square root is above the entire right hand side)
    Hi

    Use the formula $\displaystyle \cos 2a = 1 - 2 \sin^2a = 2 \cos^2a - 1$
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  3. #3
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    Quote Originally Posted by flyinhigh123 View Post
    if 4tan(a-b) = 3tana, prove that tanb = sin2a/(7+cos2a)

    thanks for any help, i really appreciate it ! =D
    $\displaystyle \begin{aligned}
    4\tan(a - b) &= 3\tan a \\
    4\left(\frac{\tan a - \tan b}{1 + \tan a \tan b}\right) &= 3\tan a \\
    4(\tan a - \tan b) &= 3\tan a(1 + \tan a \tan b) \\
    4\tan a - 4\tan b &= 3\tan a + 3\tan^2 a \tan b \\
    \tan a &= 3\tan^2 a \tan b + 4 \tan b \\
    \tan a &= \tan b(3\tan^2 a + 4) \\
    \tan b &= \frac{\tan a}{3\tan^2 a + 4}
    \end{aligned}$

    Now we need to prove that
    $\displaystyle \frac{\tan a}{3\tan^2 a + 4} = \frac{\sin 2a}{7 + \cos 2a}$.

    Starting from the left side:
    $\displaystyle \frac{\tan a}{3\tan^2 a + 4}$

    $\displaystyle \begin{aligned}
    &= \frac{\tan a}{3\tan^2 a + 4} \cdot \frac{2\cos^2 a}{2\cos^2 a} \\
    &= \frac{2\sin a \cos a}{6\sin^2 a + 8\cos^2 a} \\
    &= \frac{\sin 2a}{6\sin^2 a + 6\cos^2 a + 2\cos^2 a}
    \end{aligned}$

    $\displaystyle \begin{aligned}
    &= \frac{\sin 2a}{6 + 2\cos^2 a} \\
    &= \frac{\sin 2a}{7 + 2\cos^2 a - 1} \\
    &= \frac{\sin 2a}{7 + \cos 2a}
    \end{aligned}$


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  4. #4
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    Hello, flyinhigh123!

    Prove: .$\displaystyle \tan A \:=\: \sqrt{\frac{1-\cos2A}{1+\cos2A}} $
    We need these two identities: .$\displaystyle \begin{array}{ccccccc}\sin^2\theta &=& \dfrac{1-\cos2\theta}{2} & \Rightarrow & \sin\theta &=& \sqrt{\dfrac{1-\cos2\theta}{2}} \\ \\[-3mm] \cos^2\theta &=& \dfrac{1+\cos2\theta}{2} & \Rightarrow & \cos\theta &=& \sqrt{\dfrac{1+\cos2\theta}{2}} \end{array}$


    We have: .$\displaystyle \tan A \;=\;\frac{\sin A}{\cos A} \;=\;\frac{\sqrt{\frac{1-\cos2A}{2}} }{\sqrt{\frac{1+\cos2A}{2}}} \;=\; \sqrt{\frac{\frac{1-\cos2A}{2}} {\frac{1+\cos2A}{2}}}$

    . . Therefore: .$\displaystyle \boxed{\tan A \;=\;\sqrt{\frac{1-\cos2A}{1+\cos2A}}} $




    If $\displaystyle 4\tan(A-B) \:=\: 3\tan A$, prove that: .$\displaystyle \tan B \:=\:\frac{\sin2A}{7+\cos2A}$
    We are given: .$\displaystyle 4\tan(A-B) \:=\:3\tan A $

    . . $\displaystyle 4\left(\frac{\tan A - \tan B}{1 + \tan A\tan B}\right) \;=\;3\tan A $

    . . $\displaystyle 4(\tan A - \tan B) \;=\;3\tan A(1 + \tan A\tan B)$

    . . $\displaystyle 4\tan A - 4\tan B \;=\;3\tan A + 3\tan^2\!A\tan B$

    . . $\displaystyle 3\tan^2\!A\tan B + 4\tan B \;=\;\tan A$

    . . $\displaystyle \tan B(3\tan^2\!A + 4) \;=\;\tan A$

    . . $\displaystyle \tan B \;=\;\frac{\tan A}{3\tan^2\!A + 4} \;=\; \frac{\frac{\sin A}{\cos A}} {3\frac{\sin^2\!A}{\cos^2\!A} + 4} $

    $\displaystyle \text{Multiply by }\frac{\cos^2\!A}{\cos^2\!A}\!:\quad\tan B \;=\; \frac{\sin A\cos A}{3\sin^2\!A + 4\cos^2\!A} \;=\;\frac{\overbrace{\sin A\cos A}^{\text{This is }\frac{1}{2}\sin2A}}{3(1-\cos^2\!A) + 4\cos^2\!A}
    $

    . . . . $\displaystyle \tan B \;=\;\frac{\frac{1}{2}\sin2A}{3 + \cos^2\!A} \;=\;\frac{\frac{1}{2}\sin2A}{3 + \frac{1+\cos2A}{2}}$

    Multiply by $\displaystyle \frac{2}{2}\!:\quad \tan B \;=\; \frac{\sin2A}{6 + 1 + \cos2A} \quad\Rightarrow\quad \boxed{\tan B \;=\;\frac{\sin2A}{7 + \cos2A}} $

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