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Math Help - Help with solving these Trig Identities... Part 2

  1. #1
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    Help with solving these Trig Identities... Part 2

    How do I solve:

    Using Double angle identities

    1. Find tan2x when secx = sq.root(65); sin x > 0

    Using Pythagorean identities

    1. [(1+ cot^3 x) / ( 1 + cot x)] + cot x = csc^2 x


    Thank you very much for your time! =)
    Last edited by mr fantastic; April 5th 2009 at 04:10 AM. Reason: Questions moved from original thread
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  2. #2
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    Hello, coolacu!

    Using Double angle identities

    1.\;\;\text{Find }\tan2x\text{ when }\sec x = \sqrt{65},\;\sin x > 0
    We have: . \tan2x \:=\:\frac{\sin2x}{\cos2x} \;=\;\frac{2\sin x\cos x}{\cos^2\!x - \sin^2\!x} .[1]


    We are given: . \sec x \:=\:\frac{\sqrt{65}}{1} \:=\:\frac{hyp}{adj}

    From Pythagorus, we have: . opp \:=\:8

    Hence: . \sin x \:=\:\frac{8}{\sqrt{65}},\;\cos x \:=\:\frac{1}{\sqrt{65}}

    Substitute into [1]: . \tan2x \;=\;\frac{2\left(\frac{8}{\sqrt{65}}\right)\left(  \frac{1}{\sqrt{65}}\right)} {\left(\frac{1}{\sqrt{65}}\right)^2 - \left(\frac{8}{\sqrt{65}}\right)^2} =\;\frac{\frac{16}{65}}{\frac{1}{65} - \frac{64}{65}} \;=\;\frac{\frac{16}{65}}{\text{-}\frac{63}{65}} \;=\;-\frac{16}{63}



    Using Pythagorean identities

    1.\;\text{Prove: }\:\frac{1+ \cot^3\!x}{1 + \cot x} + \cot x \;= \;\csc^2\!x
    The numerator is the sum of two cubes . . . factor it.

    \frac{(1+\cot x)(1 - \cot x + \cot^2\!x)}{1 + \cot x} + \cot x \;=\; (1-\cot x + \cot^2\!x) + \cot x \;=\; 1 + \cot^2\!x \;=\;\csc^2\!x

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