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Math Help - First time with trigonometric identities

  1. #1
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    First time with trigonometric identities

    Hi, I've been going around in circles with this question, coming back to the original equation over and over because I can't honestly figure out what I'm supposed to be doing.
    The question is:
    Find all the values of x in the interval [0,2\pi] such that 2sin^2x=1

    I realize that I am meant to be looking for values between 0 and 2\pi but I've no idea how to manipulate the trig identities in order to do that.

    Any help would be appreciated.
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  2. #2
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    Hi

    There are several ways to solve this equation
    You can see that it is equivalent to \sin x = \pm \frac{\sqrt{2}}{2}
    Draw a circle and you will find the solution between 0 and 2\pi : \frac{\pi}{4} ; \frac{3\pi}{4} ; \frac{5\pi}{4} ; \frac{7\pi}{4}

    Another way is to use the trig identity \cos(2x) = 1- 2 \sin^2x
    The equation is equivalent to \cos(2x)=0 that you surely know how to solve
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  3. #3
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    Quote Originally Posted by mattty View Post
    Hi, I've been going around in circles with this question, coming back to the original equation over and over because I can't honestly figure out what I'm supposed to be doing.
    The question is:
    Find all the values of x in the interval [0,2\pi] such that 2sin^2x=1

    I realize that I am meant to be looking for values between 0 and 2\pi but I've no idea how to manipulate the trig identities in order to do that.

    Any help would be appreciated.
    2\sin^2{x} = 1<br />

    \sin^2{x} = \frac{1}{2}

    \sin{x} = \pm \frac{1}{\sqrt{2}}

    x = \frac{\pi}{4}

    x = \frac{3\pi}{4}

    x = \frac{5\pi}{4}

    x = \frac{7\pi}{4}
    Last edited by skeeter; March 7th 2009 at 05:18 AM. Reason: too late
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Another way is to use the trig identity \cos(2x) = 1- 2 \sin^2x
    The equation is equivalent to \cos(2x)=0 that you surely know how to solve
    I would love to say that your absolutely right and i know exactly how to solve it from that point, but the sad fact is I have no clue at all. I know i have to use the identity method, but this was just covered in passing and pretty much assumed that we knew how to go about it. I never learnt about it though. So any basic tips or links would probably be most helpful at this point.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    \sin^2{x} = \frac{1}{2}

    \sin{x} = \pm \frac{1}{\sqrt{2}}
    Thanks skeeter, that's put me on the verge, if I could understand how the step I quoted is done i think it would help a lot, is that simply a square rooting of both sides? If not can you please explain it?
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  6. #6
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    Quote Originally Posted by mattty View Post
    Thanks skeeter, that's put me on the verge, if I could understand how the step I quoted is done i think it would help a lot, is that simply a square rooting of both sides? If not can you please explain it?
    Do you know that: for all real numbers x, \sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}.
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  7. #7
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    another way to look at it ...

    2\sin^2{x} = 1

    2\sin^2{x} - 1 = 0<br />

    factor ...

    (\sqrt{2} \sin{x} - 1)(\sqrt{2} \sin{x} + 1) = 0

    set each factor equal to 0 ...

    \sin{x} = \frac{1}{\sqrt{2}} ... \sin{x} = -\frac{1}{\sqrt{2}}
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  8. #8
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    Quote Originally Posted by james_bond View Post
    Do you know that: for all real numbers x, \sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}.
    Indeed I do, I think what I'm having trouble with is getting used to the notation of sin^2x and what it means. I guess i thought if i took the square root of both sides then the answer would be sin\sqrt{x} =\frac{ 1}{\sqrt{2}} rather than sinx =\frac{ 1}{\sqrt{2}}

    Thanks for the help with the clarification.

    Quote Originally Posted by skeeter View Post
    another way to look at it ...
    thanks for spelling it out, i understand now, muchly appreciated.
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  9. #9
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    Quote Originally Posted by mattty View Post
    I would love to say that your absolutely right and i know exactly how to solve it from that point, but the sad fact is I have no clue at all. I know i have to use the identity method, but this was just covered in passing and pretty much assumed that we knew how to go about it. I never learnt about it though. So any basic tips or links would probably be most helpful at this point.
    The solutions of
    \cos (x) = \cos (a) are
    x = a +2k \pi and x = -a +2k \pi

    Here you have
    \cos (2x) = 0 = \cos \left(\frac{\pi}{2}\right)

    Then
    2x = \frac{\pi}{2} + 2k \pi or 2x = -\frac{\pi}{2} + 2k \pi

    x = \frac{\pi}{4} + k \pi or x = -\frac{\pi}{4} + k \pi

    Finally the solutions between 0 and 2\pi are
    \frac{\pi}{4} ; \frac{5\pi}{4} and \frac{3\pi}{4} ; \frac{7\pi}{4}
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  10. #10
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    Quote Originally Posted by mattty View Post
    Indeed I do, I think what I'm having trouble with is getting used to the notation of sin^2x and what it means. I guess i thought if i took the square root of both sides then the answer would be sin\sqrt{x} =\frac{ 1}{\sqrt{2}} rather than sinx =\frac{ 1}{\sqrt{2}}
    \sin^2x=\left(\sin x\right)^2 just like \ln^2 x=\left(\ln x\right)^2 or f^2(x)=\left(f(x)\right)^2 (and of course the other trigonometric functions ).
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