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Thread: Prove sinx+cosx=x has a single solution

  1. #1
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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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    MHF Contributor FernandoRevilla's Avatar
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    Consider:

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

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


    Fernando Revilla
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    Sorry,
    I have no idea what you said
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    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.

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

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

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

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

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

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

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

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

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

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

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

    then the line $\displaystyle y=x$ cuts the curve only once.
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    Alternatively,

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

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

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

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

    The maximum value of $\displaystyle f(x),$ occurs when

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

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

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

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

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