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Math Help - 3 pretty easy questions

  1. #1
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    3 pretty easy questions

    1.)



    2.)


    3.)
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  2. #2
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    Quote Originally Posted by stiitches View Post
    [3 images]
    This looks a lot like you asking us to do your homework for you. What have you tried? Where are you stuck?
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  3. #3
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    For the third one I did

    1/(5/4) since cos=1/sec
    so its cos(y)=4/5
    then did cos(2y)=8/5
    but it was wrong

    for the other two, not sure how to start them and what the Qs are exactly asking.
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  4. #4
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    Quote Originally Posted by stiitches View Post
    For the third one I did

    1/(5/4) since cos=1/sec
    so its cos(y)=4/5
    then did cos(2y)=8/5
    but it was wrong

    for the other two, not sure how to start them and what the Qs are exactly asking.
    Thanks.

    Q3, it's right that cos(y)=4/5, but the next step is not right because it's not generally true that cos(2*theta) = 2*cos(theta).

    Use identity \cos(2y) = \cos^2(y)-\sin^2(y) = 2\cos^2(y)-1 = 2\cdot(4/5)^2-1 =\ ?

    Q1 You need to know interval notation to see what form the answer is supposed to be in. And as for getting the answer you can do:

    x^3+3x<4x^2 \ \ \ \iff

    x^3-4x^2+3x < 0\ \ \ \iff

    x(x^2-4x+3) < 0 \ \ \ \iff

    \;?

    Factor the quadratic and see where the inequality holds.

    Q2 asks you to find some trig function values based on a single given value. You need some trig identities. One such is that \sin^2(\theta)+\cos^2(\theta)=1. Can you continue?
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  5. #5
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    Quote Originally Posted by stiitches View Post
    1.)
    x^3- 4x^2+ 3x= x(x- 3)(x- 4)
    and a product of numbers is positive if and only if an even number of its factors are negative.



    2.)
    csc is defined as "hypotenuse over opposite side" so you can imagine a right triangle with one vertex at (0, 0) on a coordinate system, right angle at (3, 0) and hypotenus of length 5. Where is the third vertex?


    3.)
    cos(2y)= cos^2(y)- sin^2(y). You are told what sin(y) is and, of course, cos(y)= 1/sec(y).
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