1. ## Sinusodial functions

Find the regions where sinx < x

2. Originally Posted by math123456
Find the regions where sinx < x

Maybe this will help. $\displaystyle sinx\leq{x}\Rightarrow{sinx}-x\leq0$

I took this image from http://webgraphing.com/

3. ## sinusodial functions

i'm still confused, i would you solve the function to find those regions.

4. Originally Posted by math123456
Find the regions where sinx < x
Are exact values of the endpoints of the intervals required?

5. Originally Posted by math123456
i'm still confused, i would you solve the function to find those regions.
sinx is always between -1 and 1. It follows then, that when |x|>1, the graph will either become more and more negative, or more and more positive. Know that, all you must do is analyze the behavior of the graph for -1<x<1 and determine the behavior of the function for those values.

6. Let $\displaystyle f(x)=\sin x-x$

$\displaystyle f'(x)=\cos x-1=-2\sin^2\frac{x}{2}\leq 0$

Thus $\displaystyle f(x)$ is a decreasing funtion(i.e. if $\displaystyle x_{1}<x_{2}$ ,then $\displaystyle f(x_{1})>f(x_{2}))$

Thus for $\displaystyle x\geq 0$

$\displaystyle f(x)\leq f(0)$

$\displaystyle f(x)\leq 0$

$\displaystyle \sin x-x\leq 0$

$\displaystyle \sin x\leq x$ for $\displaystyle x\in [0,\infty)$

7. Originally Posted by pankaj
Let $\displaystyle f(x)=\sin x-x$

$\displaystyle f'(x)=\cos x-1=-2\sin^2\frac{x}{2}\leq 0$

Thus $\displaystyle f(x)$ is a decreasing funtion(i.e. if $\displaystyle x_{1}<x_{2}$ ,then $\displaystyle f(x_{1})>f(x_{2}))$

Thus for $\displaystyle x\geq 0$

$\displaystyle f(x)\leq f(0)$

$\displaystyle f(x)\leq 0$

$\displaystyle \sin x-x\leq 0$

$\displaystyle \sin x\leq x$ for $\displaystyle x\in [0,\infty)$
This was my initial mode of attack as well, but this is the pre-calculus subforum, so I disregarded the use of the first derivative test.

8. Originally Posted by VonNemo19
This was my initial mode of attack as well, but this is the pre-calculus subforum, so I disregarded the use of the first derivative test.
O.K. then.

Now consider acircle of radius 1 unit.Let a chord AB of this circle subtend an angle x at the centre O of the circle.

Clearly,Ar($\displaystyle \triangle AOB$)<Area(\sector AOB).

Ar($\displaystyle \triangle AOB$)=$\displaystyle \frac{1}{2}\sin x$

Area of sector AOB=$\displaystyle \frac{1}{2}.1^2.x=\frac{1}{2}.x$

$\displaystyle \frac{1}{2}\sin x\leq \frac{1}{2}x$

$\displaystyle \sin x\leq x$

which is true for all non-negative $\displaystyle x$.If $\displaystyle x<0$,then $\displaystyle \sin x>x$

9. Hello : this graphic -solution