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

1. ## 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!

2. 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

3. Sorry,
I have no idea what you said

4. 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.

5. Originally Posted by jayshizwiz
Sorry,
I have no idea what you said
Sorry, I have no idea what your syllabus covers.

Fernando Revilla

6. Originally Posted by jayshizwiz
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.

7. 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:

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

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

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

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

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.

8. 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.