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Thread: Range of trig function

  1. #1
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    Range of trig function

    Find the range of $\displaystyle f(x)=\sin ^2 x (\sin (2x) +1)$.

    Ideas?
    Last edited by atreyyu; Jan 2nd 2011 at 02:21 PM. Reason: correcting ambiguity
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    What do you know about sine?
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  3. #3
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    Well, enough to know that even though the range of sine is <-1;1>, the range of f will not be $\displaystyle 1^2\cdot(1+1)$...
    What else would be any help?
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    Quote Originally Posted by atreyyu View Post
    Find the range of $\displaystyle f(x)=\sin ^2 x (\sin 2x +1)$.

    Ideas?
    Also, what do you mean here?

    $\displaystyle \sin{(2x+1)} \ \mbox{or} \ \sin{(2x)}+1$
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    $\displaystyle \sin{2x}=2\sin{x}\cos{x}$

    $\displaystyle 2\sin^2{x}\sin{x}\cos{x}+\sin^2{x}$

    $\displaystyle \sin{x}\cos{x}\neq\pm 1$

    $\displaystyle 0\leq\sin^2{x}\leq 1$

    If $\displaystyle \sin^2{x}=1$, then $\displaystyle 2\sin^2{x}\sin{x}\cos{x}=0$.

    What is the max value $\displaystyle 2\sin^2{x}\sin{x}\cos{x}$ this can be?

    Once you determine the x value that obtains the max, x_0, of $\displaystyle 2\sin^2{x}\sin{x}\cos{x}$, then $\displaystyle f(x_0)=2\sin^2{x_0}\sin{x_0}\cos{x_0}+\sin^2{x_0}= y_0$

    The question then remains is $\displaystyle y_0>1\mbox{?}$
    Last edited by dwsmith; Jan 2nd 2011 at 05:48 PM. Reason: Deleted a line
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    Quote Originally Posted by atreyyu View Post
    Find the range of $\displaystyle f(x)=\sin ^2 x (\sin (2x) +1)$.

    Ideas?
    Are you expected to use calculus to find turning points?
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  7. #7
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    Quote Originally Posted by atreyyu View Post
    Find the range of $\displaystyle f(x)=\sin ^2 x (\sin (2x) +1)$.

    Ideas?
    It's clear that f(x) has a minimum value of zero; at x=0 for instance.

    $\displaystyle \displaystyle \sin(2x) = 2(\sin x)(\cos x)$ and $\displaystyle \displaystyle 1= \sin^2 x+\cos^2 x$

    Then, $\displaystyle \displaystyle \sin(2x)+1 = \sin^2+2(\sin x)(\cos x)+\cos^2 x$

     $\displaystyle \displaystyle \text{ } = (\sin x+ \cos x)^2$
    At this point we have that:

    $\displaystyle \displaystyle f(x)=(\sin ^2 x) (\sin (2x) +1)=(\sin ^2 x)(\sin x+ \cos x)^2\,.$

    A not so well known trig identity gives:

    $\displaystyle \sin x+ \cos x=\sqrt{2}\ \sin\left(x+{{\pi}\over{4}}\right)$

    We now have: $\displaystyle \displaystyle f(x)=(\sin ^2 x)\left(\sqrt{2}\ \sin\left(x+{{\pi}\over{4}}\right)\right)^2=2(\sin ^2 x) \sin^2\left(x+{{\pi}\over{4}}\right) \,.$

    Now use a product to sum identity on: $\displaystyle \displaystyle (\sin x) \sin\left(x+{{\pi}\over{4}}\right)$
    $\displaystyle \displaystyle = {{1}\over{2}}\left[\cos\left({{\pi}\over{4}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]$

    $\displaystyle \displaystyle = {{1}\over{2}}\left[\left({{1}\over{\sqrt{2}}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]$

    Finally, $\displaystyle \displaystyle f(x)=2\cdot {{1}\over{4}}\left[\left({{1}\over{\sqrt{2}}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]^2 $

    $\displaystyle \displaystyle ={{1}\over{2}}\left[\left({{1}\over{\sqrt{2}}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]^2$
    I suggest finding the minimum and maximum values of the expression inside the brackets. Square those then divide by 2.
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