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Math Help - Trig help.

  1. #1
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    Trig help.

    I need help on these questions please.

    1. given that sin^2 x + cos^2 x = 1 show that 1+cot^2 x = cosec^2 x

    2. Solve for values x in the interval -pi<x<pi for the following in 2 decimal places:

    a. 2cosec(x/2) - 5 =0
    b. 2cot^2 x - cot x -7 = 0

    Thanks
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  2. #2
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    For the first question I got:

    sin^2 x + cos^2 x = 1

    so (sin^2 x)/(sin^2 x) + (cos^2 x)/(cos^2 x) = 1/(sin^2 x)

    so 1 + (cos x/ sin x)^2 = (1/ sin x)^2

    so 1 + cot^2 x = cosec ^2 x.

    Is that right?

    But im really stuck on the other two questions
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  3. #3
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    1) Divide by \sin^{2}(x) and see what happens.

    2a) sin(x) = 1/csc(x)

    2b) Solve this first: 2Y^2 - Y - 7 = 0. I see four solutions in your future.
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  4. #4
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    for 2a I got:

    2cosec(x/2) - 5 = 0

    so 2(1/sin(x/2)) - 5 = 0

    so 2(2/sin x) - 5 = 0

    so 4/sin x -5 = 0

    so 4 = 5sin x

    so sin x = 4/5

    so x = inverse sin(4/5)

    So the only solution I got for x is 53.13 to 2 decimal places.

    Is this right?
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  5. #5
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    sorry tk hunny but 2b does not factorise.
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  6. #6
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    I wasn't suggesting factoring, unless that happens to work.

    Solve it, nonetheless.
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