Prove sinx+cosx=x has a single solution

**jayshizwiz** · January 15th 2011, 10:02 AM

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!

**FernandoRevilla** · January 15th 2011, 10:16 AM

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

**jayshizwiz** · January 15th 2011, 10:22 AM

Sorry,
I have no idea what you said

**Unknown008** · January 15th 2011, 10:30 AM

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.

**FernandoRevilla** · January 15th 2011, 10:31 AM

Originally Posted by jayshizwiz

Sorry,
I have no idea what you said

Sorry, I have no idea what your syllabus covers.

Fernando Revilla

**Plato** · January 15th 2011, 10:51 AM

Originally Posted by jayshizwiz

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.

**Archie Meade** · January 15th 2011, 11:15 AM

Originally Posted by jayshizwiz

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.

**Archie Meade** · January 15th 2011, 05:25 PM

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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