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Math Help - Prove sinx+cosx=x has a single solution

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    Prove sinx+cosx=x has a single solution

    Prove sinx+cosx=x has a single solution in [0,pi/2]


    This question seems very strange to me. How can I prove there is a single solution for every number between 0 and pi/2?

    I'm not sure what formula to use.
    I'm learning infinitesimal math.

    Can someone help me out?
    Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Consider:

    f:[0,\pi/2]\rightarrow{\mathbb{R}}\;\quad f(x)=\sin x+\cos x-x

    Use Bolzanos's theorem (at least there is a root in the interval) and prove that f is strictly decreasing (only one root).


    Fernando Revilla
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    Sorry,
    I have no idea what you said
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Another way of looking at this is graphically.

    First convert cos x + sin x into a single trig ratio.

    \sin x + \cos x = R\sin (x + \alpha)

    You'll find that R = \sqrt2 and \alpha = \dfrac{\pi}{4}

    \sin x + \cos x = \sqrt2 \sin \left(x + \dfrac{\pi}{4}\right)

    Then, you graph this with the line y = x for the domain [0,pi/2]

    You'll get only 1 intercept.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by jayshizwiz View Post
    Sorry,
    I have no idea what you said
    Sorry, I have no idea what your syllabus covers.


    Fernando Revilla
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  6. #6
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    Quote Originally Posted by jayshizwiz View Post
    Prove sinx+cosx=x has a single solution in [0,pi/2]
    Let f(x)=\sin(x)+\cos(x)-x then f(0)>0~\&~f\left(\frac{\pi}{2}\right)<0.
    So by the intermediate value theorem f has a zero between 0~\&~\frac{\pi}{2}.
    Moreover, f^' is negative there, so f is decreasing.
    That means there is a single zero.
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  7. #7
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    Quote Originally Posted by jayshizwiz View Post
    Prove sinx+cosx=x has a single solution in [0,pi/2]


    This question seems very strange to me. How can I prove there is a single solution for every number between 0 and pi/2?

    I'm not sure what formula to use.
    I'm learning infinitesimal math.

    Can someone help me out?
    Thanks!
    Consider:

    Sinx+Cosx=x;\;\;\;0<x<\frac{\pi}{2}

    Cosx=Sin\left(\frac{\pi}{2}-x\right)

    SinA+SinB=2Sin\frac{A+B}{2}Cos\frac{A-B}{2}\Rightarrow\ Sinx+Sin\left(\frac{\pi}{2}-x\right)=2Sin\frac{\pi}{4}Cos\left(x-\frac{\pi}{4}\right)

    \Rightarrow\frac{2}{\sqrt{2}}Cos\left(x-\frac{\pi}{4}\right)=\frac{2}{\sqrt{2}}Cos\left(\f  rac{\pi}{4}-x\right)=x

    This Cosine function is 1 at x=0 and rises to a maximum at x=\frac{\pi}{4}

    at which point the line y=x is rising to meet the Cosine function
    but still underneath it.

    Since the Cosine function decreases to zero at x=\frac{3{\pi}}{4}

    then the line y=x cuts the curve only once.
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  8. #8
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    Alternatively,

    f(x)=sinx+cosx\Rightarrow\ f'(x)=cosx-sinx\Rightarrow\ f''(x)=-(sinx+cosx)

    In the given domain, sinx>0,\;\;cosx>0

    The 2nd derivative of sinx+cosx is negative, so it is concave downward for the given domain.

    f(0)=1,\;\;\;f\left(\frac{\pi}{2}\right)=1

    The maximum value of f(x), occurs when

    f'(x)=cosx-sinx=0\Rightarrow\ cosx=sinx\Rightarrow\ tanx=1\Rightarrow\ x=\frac{\pi}{4}

    \Rightarrow\ f\left(\frac{\pi}{4}\right)=\sqrt{2}

    g(x)=x\Rightarrow\ g(0)=0,\;\;g\left(\frac{\pi}{2}\right)=\frac{\pi}{  2}>1

    The straight line g(x)=x line cuts the sinusoidal curve f(x)=sinx+cosx

    somewhere in the given domain
    and since f(x) is concave downward in the domain,
    which is relevant after the maximum value of f(x),
    then only one intersection occurs.
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