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Math Help - Applications of compound angle formula, cmposite functions.

  1. #1
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    Exclamation Applications of compound angle formula, cmposite functions.

    Hello everyone. I need some help with these math problems. I wasn't sure if I should put this in the Pre-Calc or Calc section (its a Calc course, but the beginning of the book).

    1) Let a, b be two angles. If b=a-π, how much is sin(2a+b)?


    This is what I did, I'm not sure if I'm on the right track:

    sin(2a+b) ---> sin(2a+(a-π)) ---> sin2a cos(a-π) + sin (a-π) cos2a

    As I said, I'm not sure if I'm on the right track. If I am on the right track, how would I simplify that, because my answer does not match any of the multiple choice.

    2) Using the fact that 75=2 x 60-45, compute cos(5π/12) and write the answer in its most simplified form.

    I'm not even sure where to begin here. When I make 2π = 360, I end up with cos(900/12) or cos(75) which equals 0.9217512697.

    I'm totally lost here to be honest.

    3) If f(x)=e^(x+1) and g(x)=x^(2)+1, what is the range of the composite function f(g(x))?

    I think I might know what I'm doing here.

    f(g(x)) --> f(x^(2)+1) = e^(x+1) = e^(x^(2)+1)+1

    Judging by what I got there, I want to say that the range is the set of all numbers. I graphed it and got a parabola. However, please help me out if I did the calculations wrong.

    Thank you very much in advance to whoever helps me out on this!
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  2. #2
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    Quote Originally Posted by FullBox View Post
    1) Let a, b be two angles. If b=a-π, how much is sin(2a+b)?
    \sin(x-\pi) = -\sin(x) \Rightarrow \sin(2a+b) = \sin(2a+a-\pi) = -\sin(3a)
    2) Using the fact that 75=2 x 60-45, compute cos(5π/12) and write the answer in its most simplified form.
    \cos[(2\times{60})^{\circ}-45^{\circ}] = \sin(60^{\circ}+60^{\circ})\sin(45^{\circ})+\cos(6  0^{\circ}+60^{\circ})\cos(45^{\circ}).

    Use \sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y) for \sin(60^{\circ}+60^{\circ})

    and \cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y) for \cos(60^{\circ}+60^{\circ})

    then just plug in the standard values.

    3) If f(x)=e^(x+1) and g(x)=x^(2)+1, what is the range of the composite function f(g(x))? I think I might know what I'm doing here. f(g(x)) --> f(x^(2)+1) = e^(x+1) = e^(x^(2)+1)+1
    Right. The domain is obviously the set of all real numbers. That's what you got. But the range is not the
    set of all real numbers. We have [Math]f(g(x)) = e^{x^2+2}[/tex]. If x = 0, f(g(x)) = e^2. If x is positive, then
    e^{x^2+2} > e^2. If x is negative, then again e^{x^2+2} > e^2. I think the range of the function is clear then.
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  3. #3
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    Thank you for the help! I just wanted to make sure about several things, though.

    1) In regards to the first question, I totally understand it now. Even when using the other longer method, I obtained (sin3a(cosπ)) - (sin π (cos3a)) which equals the answer you gave me; -sin3a. However, even with the answer in two different forms, none of the answers in the multiple choice match either formats. Is there any other way I could express the answer?

    2) I'm still totally lost on this one. After following the instructions I ended up with:

    sin (60+60) became ---> sin60cos60 + cos60sin60
    cos (60+60) became ---> cos60cos60 - sin60sin60

    After plugging them into the original equation, I got the following:

    (((sin60cos60)+(cos60sin60))sin45) + (((cos60cos60)-(sin60sin60))cos45)

    I have a feeling I just made it worse.

    3) I don't have any questions in regard to the third problem. The range is e^2 to infinity, or [e^2, infinity) written in interval form. Thank you, I totally didn't realize that I was listing the domain at first.

    Sorry about the added inquiries, and thank you for all your help.
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  4. #4
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    Quote Originally Posted by FullBox View Post
    However, even with the answer in two different forms, none of the answers in the multiple choice match either formats. Is there any other way I could express the answer?
    Can post the choices that you are given?
    2) After following the instructions I ended up with:

    sin (60+60) became ---> sin60cos60 + cos60sin60
    cos (60+60) became ---> cos60cos60 - sin60sin60
    Right! So, what's \cos(60^{\circ})? And what is \sin(60^{\circ})? I think what you are missing is that you are assumed
    to know that \cos(60^{\circ}) = \frac{1}{2}, \sin(60^{\circ}) = \frac{\sqrt{3}}{2}, \cos(45^{\circ}) = \frac{1}{\sqrt{2}}, [/Math] and [Math] \sin(45^{\circ}) = \frac{1}{\sqrt{2}}. Using these we have:

    \sin(60^{\circ}+60^{\circ}) = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\r  ight)+\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}  }{2}\right) = 2\left(\frac{\sqrt{3}}{4}\right) = \frac{\sqrt{3}}{2}

    \cos(60^{\circ}+60^{\circ}) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3  }}{2}\right) = \frac{1}{4}-\frac{3}{4} = -\frac{1}{2}

    Earlier on we had:

    \cos[(2\times{60})^{\circ}-45^{\circ}] = \sin(60^{\circ}+60^{\circ})\sin(45^{\circ})+\cos(6  0^{\circ}+60^{\circ})\cos(45^{\circ})

    It now becomes:

    \cos[(2\times{60})^{\circ}-45^{\circ}] = \frac{\sqrt{3}}{2}\sin(45^{\circ})-\frac{1}{2}\cos(45^{\circ})

    \cos(45^{\circ}) = \sin(45^{\circ}) = \frac{1}{\sqrt{2}}, so we have:

    \cos[(2\times{60})^{\circ}-45^{\circ}] = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqr  t{2}}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\r  ight) = \frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}.

    \displaystyle \therefore \cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{3}-1}{2\sqrt{2}}.
    Sorry about the added inquiries, and thank you for all your help.
    No problem - and you are welcome to ask anything else that you don't understand about this.
    Last edited by TheCoffeeMachine; September 5th 2010 at 06:09 PM.
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  5. #5
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    In regards to the first question, here are the choices given to me:

    a) -sina
    b) sin^(3)a - 3 sina cos^(2)a
    c) sina
    d) -sin^(3)a

    And for the second question; WOW I completely forgot about that circle that shows the values for the degrees, its on the very first page of my book! Dang, I overlook too many things when doing problems! Thank you so much for that detailed breakdown!
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  6. #6
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    -\sin{3a} = -\sin(2a+a) = -\sin{2a}\cos{a}-\cos{2a}\sin{a}.

    Using the identities \sin{2a} = 2\cos{a}\sin{a} and \cos{2a} = \cos^2{a}-\sin^2{a} will give you (b).
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  7. #7
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    Thank you very much! I'm sorry, but I have one last question. For the problem you did in the post before your previous one, if I wanted to rationalize the denominator (remove square roots from the bottom) is this how I would do it?

    Hopefully this will be easier to read; I just realized MS Word had an equation inserter and so I did the math there and screen-capped it.



    Thanks again for your helpfulness man! I know I'll be coming back here a lot when I need math related help!
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  8. #8
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    Quote Originally Posted by FullBox View Post
    Terrific! Perfect, in fact.

    Thanks again for your helpfulness man! I know I'll be coming back here a lot when I need math related help!
    You are welcome. You can learn how to use the latex in this forum here.
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