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Math Help - solving Natural Logs for x and a Trig identity question.

  1. #1
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    solving Natural Logs for x and a Trig identity question.

    Hey.

    Code:
    2Ln2 - Ln(x-1) = ln(2x)
    That is what im having trouble sorting out. I know its something to do with the quadratic formula but its just not clicking in my head.

    also, i cant seem to finish this off:
    Code:
    Prove that (1/cot^2 θ) - (1/csc^2 θ) = (sin^4 θ)/(cos^2 θ)
    This is what i have so far:
    Code:
    -> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
    -> 1/(tan^2 θ - sin^2 θ)
    -> (1/((cos^2 θ)/(sin^2 θ))) - (1/sin^2 θ)
    -> ((sin^2 θ)/(cos^2 θ)) + ((sin^2 θ)/ 1)
    i am stuck there... unless of course the actually DO add together to make
    Code:
    (sin^4 θ)/(cos^2 θ)
    Any help would be greatly appreciated!
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by squrrilslayer View Post
    Hey.

    Code:
    2Ln2 - Ln(x-1) = ln(2x)
    That is what im having trouble sorting out. I know its something to do with the quadratic formula but its just not clicking in my head.


    b ln(a) = ln(a^b)

    Thus

    ln(4) - ln(x-1) = ln(2x)

    But
    ln(a) - ln(b) =ln(a/b)

    Thus

    ln(4/(x-1)) = ln(2x)

    Thus

    4/x-1 = 2x

    Hence
    2x^2 - 2x = 4

    x^2 - x - 2 = 0

    x^2 - 2x + x -2 = 0

    x(x-2) + (x-2) = 0

    (x+1)(x-2)=0

    x = -1 or x= 2

    But since ln(-ve) is not defined thus x = 2
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  3. #3
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    Quote Originally Posted by squrrilslayer View Post


    Code:
    Prove that (1/cot^2 θ) - (1/csc^2 θ) = (sin^4 θ)/(cos^2 θ)
    This is what i have so far:


    Code:
    -> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
    -> tan^2θ - sin^2 θ 
    -> sin^2 θ /cos^2 θ - sin^2 θ 
    -> sin^2θ( 1/cos^2 θ -1)
    -> sin^2 θ ( sec^2 θ -1) 
    -> sin^2θ (tan^2 θ) 
    -> sin^2 θ x sin^2 θ/cos^2 θ
    -> sin^4 θ/cos^2 θ
    I have corrected in code
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  4. #4
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    wow thanks that was quick. I see now on the first one how you did that
    as for the second, you got the answer so im reading through step by step analyzing what you did.

    Thanks again!
    [resolved]
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  5. #5
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    sorry re on trig idents

    i still dont understand what's been done:

    -> (1/(1/tan^2 θ)) - (1/(1/sin^2 θ))
    -> tan^2θ - sin^2 θ
    -> sin^2 θ /cos^2 θ - sin^2 θ <<here
    -> sin^2θ( 1/cos^2 θ -1) <<here
    -> sin^2 θ ( sec^2 θ -1) <<here
    -> sin^2θ (tan^2 θ) <<here
    -> sin^2 θ x sin^2 θ/cos^2 θ
    -> sin^4 θ/cos^2 θ

    like how did

    sin^2 θ /cos^2 θ - sin^2 θ

    become:

    sin^2θ( 1/cos^2 θ -1) ?

    and so on so forth?

    Thanks if someone could clear this up!
    Last edited by mr fantastic; March 18th 2009 at 04:19 AM.
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  6. #6
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    Talking

    Line (3) is (effectively) "x/y - x".
    Line (4) is (effectively) "x(1/y - 1)".
    The author factored the "x" out front.

    To go from Line (5) to Line (6), apply one of the trig identities they gave you.
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  7. #7
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    tan^2(\theta) - sin^2(\theta) = \frac{sin^2(\theta)}{cos^2(\theta)} - sin^2 (\theta)

    \frac{sin^2(\theta)}{cos^2(\theta)} - sin^2(\theta)

    = sin^2(\theta)\times \frac{1}{cos^2(\theta)} - sin^2(\theta) \times 1

    Take sin^2(\theta) common

    = sin^2(\theta) (\frac{1}{cos^2(\theta)} -1) -------(1)

    Using the rule that

    sec(x) = 1/cos(x)

    Replace 1/cos^2(\theta) by sec^2(\theta)

    = sin^2 θ ( sec^2 θ -1)......................(2)

    Using the property that

    sec^2 (x) = 1 + tan^2 (x)

    Replace sec^2(\theta) by tan^2(\theta) - 1

    which gives

    = sin^2(\theta)\times (tan^2(\theta))

      \frac{sin^2 \theta \times sin^2\theta}{cos^2 \theta}

    -----------------------------

    Tell the step you fail to understand now
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