# Thread: Find the regions where sin(x) is greater than or equal to X

1. ## Find the regions where sin(x) is greater than or equal to X

I know the graphs intersect three times, one at (0,0) and the other 2 points are reflected about the y-axis. Is there a solution for this exercise? I can't think of any of these points where a value of X is less than its sin value. And also, is there any way I can solve for the other 2 points algebraically? Because I've been approximating and it seems my approximates aren't that good...

Thanks in advance guys, I appreciate it.

2. Have you tried graphing the functions? The answer
Spoiler:
$\displaystyle x < 0$
should then be obvious.

3. Originally Posted by Pupil
I know the graphs intersect three times, one at (0,0) and the other 2 points are reflected about the y-axis. Is there a solution for this exercise? I can't think of any of these points where a value of X is less than its sin value. And also, is there any way I can solve for the other 2 points algebraically? Because I've been approximating and it seems my approximates aren't that good...

Thanks in advance guys, I appreciate it.
$\frac{d}{dx}Sinx=Cosx$

The maximum value of the derivative of $Sinx$ is 1.

The derivative of $f(x)=x$ is always 1.

Hence $Sinx=x$ only when $x=0$

Therefore $Sinx>x$ if $x<0$

4. modem error....

5. Alright guys, thanks for the help. I understand the inequality now.

Just one more question: Is there any way I can get the exact values by algebraic methods? If I try to do it graphically, my approximations are not as nearly as accurate. I know the graphs intersect 3 times and one is (0,0), but how would I determine the other points without approximating? Because frankly, my graphing software sucks. ><

6. Originally Posted by Pupil
Alright guys, thanks for the help. I understand the inequality now.

Just one more question: Is there any way I can get the exact values by algebraic methods? If I try to do it graphically, my approximations are not as nearly as accurate. I know the graphs intersect 3 times and one is (0,0), but how would I determine the other points without approximating? Because frankly, my graphing software sucks. ><
If the graph of the functions intersect to the right of the y-axis,
then Sinx would have had to have a gradient >1 at some stage but it doesn't!

Why are you convinced that there is more than one point of intersection?

7. Originally Posted by Archie Meade
If the graph of the functions intersect to the right of the y-axis,
then Sinx would have had to have a gradient >1 at some stage but it doesn't!

Why are you convinced that there is more than one point of intersection?
Ah, it was just a mistake with my software then. But I have other questions where there is more than 1 intersection, and my approximation with the software is not that good.

So how would I be able to solve for the exact points of intersection? Ex: finding the 4 points where $cos(x) = 1/2^x$ Thanks for all the help, by the way.

8. Originally Posted by Pupil
Ah, it was just a mistake with my software then. But I have other questions where there is more than 1 intersection, and my approximation with the software is not that good.

So how would I be able to solve for the exact points of intersection? Ex: finding the 4 points where $cos(x) = 1/2^x$ Thanks for all the help, by the way.
The first point of intersection is quite straightforward, x=0.

You could use Newton's method to find the other two from $0$ to $2{\pi}$, by solving

$cosx-\frac{1}{2^x}=0$

However, that will give "very close approximations" !

9. Originally Posted by Archie Meade
The first point of intersection is quite straightforward, x=0.

You could use Newton's method to find the other two from $0$ to $2{\pi}$, by solving

$cosx-\frac{1}{2^x}=0$

However, that will give "very close approximations" !
Okay, it seems similar to solving trig equations. However, since we can't use the unit circle, so how would I go about approximating the points? Oh and its $0.5^x$

10. Originally Posted by Pupil
Okay, it seems similar to solving trig equations. However, since we can't use the unit circle, so how would I go about approximating the points? Oh and its $0.5^x$

The Newton-Raphson formula is

$\displaystyle\ x_{n+1}=x_n-\frac{f\left(x_n\right)}{f'\left(x_n\right)}$

$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$

$x_2=x_1-\frac{f\left(x_1\right)}{f'\left(x_1\right)}$

$x_3=x_2-\frac{f\left(x_2\right)}{f'\left(x_2\right)}$

It's an iterative process. $x_0$ is a first guess at a solution.

Then $x_1$ is the calculated first approximation. $x_2$ will be much closer, as will $x_3$

$\frac{1}{2^x}=2^{-x}=0.5^x$

$f(x)=cosx-0.5^x$

$f'(x)=-sinx+\left(0.5^x\right)ln(0.5)$

Take an initial guess at the next root of $cosx-0.5^x=0,$ say $x=\frac{\pi}{3}$.

Then

$\displaystyle\ x_1=\frac{\pi}{3}-\frac{cos\left(\frac{\pi}{3}\right)-0.5^{\frac{\pi}{3}}}{-sin\left(\frac{\pi}{3}\right)+\left(0.5^{\frac{\pi }{3}}\right)ln(0.5)}$

Use the result as $x_1$ in the 2nd iteration, which should be enough.

Use $\frac{3{\pi}}{2}$ as an initial guess for the 3rd root of $cosx-0.5^x=0$

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