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Math Help - Help please! (double angle formulas)

  1. #1
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    Help please! (double angle formulas)

    Im so confused! im sure these r the correct answers!! but it keeps saying no! is it really worng or is is the programs error!? (some times hte program does make errors -_-) thxs for the help

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  2. #2
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    Hello S2Krazy
    Quote Originally Posted by S2Krazy View Post
    Im so confused! im sure these r the correct answers!! but it keeps saying no! is it really worng or is is the programs error!? (some times hte program does make errors -_-) thxs for the help
    What you need to know:

    • Double-angle formulae. These are the ones you need here:

    \tan2A = \frac{2\tan A}{1-\tan^2A} and \cos2A= 2\cos^2A-1
    • When the various trig functions are positive. Here's the diagram:

    \begin{array}{c|c}S & A\\ \hline T & C \end{array}
    (Remember ACTS starting in QI and going clockwise.)

    • \cot A = \frac{1}{\tan A} and \sec A = \frac{1}{\cos A}


    So for #3, A is in QIII, so only tangent is positive: sine and cosine are both negative. So if \sin A = -\frac{5}{13}, \cos A = -\frac{12}{13} (Pythagoras) and \tan A = \frac{5}{12}.

    Using \tan2A = \frac{2\tan A}{1-\tan^2A}:
    \tan 2A = \frac{2\times \frac{5}{12}}{1-(\frac{5}{12})^2} = ... = \frac{120}{119}
    For # 4, x is in QIV, so only cosine is positive; sine and tangent are negative. So \cos x = \frac{2}{\sqrt5}\Rightarrow\sin x = - \frac{1}{\sqrt5} (Pythagoras) \Rightarrow \tan x = -\frac12
    \Rightarrow \tan2x=...=-\frac43 (You can check this out.)

    \Rightarrow \cot2x = -\frac34
    For #5, all functions are positive in QI. So \cos\theta = \frac{12}{13} (Pythagoras)
    \Rightarrow \cos2\theta = 2\cos^2\theta - 1 = ... = \frac{119}{169} (Check my working)

    \Rightarrow \sec2\theta = \frac{169}{119}
    Grandad
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  3. #3
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    Hello, S2Krazy!

    Sorry, you made errors on all of them . . .


    Let \sin A = -\frac{5}{13} .with A in Q3. . Find \tan(2A)
    We need the identity: . \tan(2A) \:=\:\frac{2\tan A}{1 - \tan^2\!A}

    We have: . \sin A \:=\:-\frac{5}{13} \;=\;\frac{opp}{hyp}

    So:. opp = -5,\;hyp = 13
    . . Pythagorus says: . adj = \pm 12
    Since A is in Q3, adj = -12

    . . Hence: . \tan A \:=\:\frac{opp}{adj} \:=\:\frac{-5}{-12} \:=\:\frac{5}{12}

    Therefore: . \tan2A \;=\;\frac{2(\frac{5}{12})}{1 - (\frac{5}{12})^2} \;=\;\frac{120}{119}



    Let \cos x = \frac{2}{\sqrt{5}} .with x in Q4. . Find \cot(2x)

    We have: . \cos x \:=\:\frac{2}{\sqrt{5}} \:=\:\frac{adj}{hyp}

    So: . adj = 2,\;\;hyp = \sqrt{5}
    . . Pythagorus says: . opp \,=\,\pm1
    Since x is in Q4: . opp = -1

    . . Hence: . \tan x \:=\:\frac{opp}{adj} \:=\:-\frac{1}{2}

    Then: . \tan(2x) \:=\:\frac{2(-\frac{1}{2})}{1 - (-\frac{1}{2})^2} \;=\;-\frac{4}{3}

    Therefore: . \cot2x \:=\:-\frac{3}{4}



    Let \tan\theta = \frac{5}{12} .with \theta in Q1. . Find \sec2\theta

    We have: . \tan\theta \:=\:\frac{5}{12}\:=\:\frac{opp}{adj}
    So: . opp = 5,\;adj = 12
    . . Pythagorus says: . hyp = 13

    Hence: . \cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{12}{13}

    Then: . \cos2\theta \;=\;2\cos^2\theta - 1 \;=\;2\left(\frac{12}{13}\right)^2 - 1 \;=\;\frac{119}{169}

    Therefore: . \sec2\theta \;=\;\frac{169}{119}



    Edit: Too fast for me, Grandad . . . Nice job, too!
    .
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