# Thread: find boundaries of sine and cos without graphing

1. ## find boundaries of sine and cos without graphing

Hi!

Is there a way to find the boundaries without graphing?

I think 3sinx=2sin^2 may work, but i'm not sure how to solve for zero. Could some show me how?

Here is what i have:

3sinx=2sin^2
3sin - 2sin^2 - 0

2. ## Re: find boundaries of sine and cos without graphing

Factor: $\sin x (3-2\sin x)=0$

Now you have either $\sin x=0$ or $3-2\sin x=0$. The latter has no solutions over the reals. So the intersections occur at $x=0,\pi,2\pi$

Next plug in $\dfrac{\pi}{2} ,\dfrac{3\pi}{2}$ to figure out which function is bigger in each interval. This gives you

$\displaystyle A=\int_0^\pi(3\sin x-2\sin^2 x)dx+\int_\pi^{2\pi}(2\sin^2 x-3\sin x)dx$

3. ## Re: find boundaries of sine and cos without graphing

Thanks for the response, SlipEternal!!

In the Area, why did you write 3sinx-2sin^2x twice? And why are they reversed in the second part? Also why are the boundaries different in each of the cases?
Originally Posted by SlipEternal
Factor: $\sin x (3-2\sin x)=0$

Now you have either $\sin x=0$ or $3-2\sin x=0$. The latter has no solutions over the reals. So the intersections occur at $x=0,\pi,2\pi$

Next plug in $\dfrac{\pi}{2} ,\dfrac{3\pi}{2}$ to figure out which function is bigger in each interval. This gives you

$\displaystyle A=\int_0^\pi(3\sin x-2\sin^2 x)dx+\int_\pi^{2\pi}(2\sin^2 x-3\sin x)dx$

4. ## Re: find boundaries of sine and cos without graphing

Originally Posted by RBB22
Thanks for the response, SlipEternal!!

In the Area, why did you write 3sinx-2sin^2x twice? And why are they reversed in the second part? Also why are the boundaries different in each of the cases?
As slip eternal said, the two curves, 3sin(x) and $2 sin^2(x)$ cross at x= 0, $x= \pi$, and $x= 2\pi$. He then said "Next, plug in $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ to figure out which function is bigger in each interval". When $x= \frac{\pi}{2}$, which is between 0 and $\pi$, $3sin(x)= 3sin(\pi/2)= 3$ while $2 sin^2(x)= 2 sin^2(\pi/2)= 2$. Since 3sin(x) is greater than $2sin^2(x)$ at $x= \frac{\pi}{2}$ (and both functions are continuous so the only place they can change order is where they are equal) $3sin(x)$ is larger than $2sin^2(x)$ for all x between 0 and $\pi$. So the area between them is $\int_0^{\pi} 3sin(x)- 2sin^2(x) dx$.

At $x= \frac{3\pi}{2}$, $3 sin(x)= 3sin(3\pi/2)= -3$ while $2sin^2(x)= 2 sin^2(3\pi/2)= 2$ so now $2 sin^2(x)$ is larger. The area between them is $\int_{\pi}^{2\pi} 2sin^2(x)- 3 sin(x) dx$. The total area is the sum of those two.

5. ## Re: find boundaries of sine and cos without graphing

@RBB22 ,
One question: what does "without graphing" have to do with this question?

6. ## Re: find boundaries of sine and cos without graphing

RBB22 probably had used some graphing utility to graph the two functions and identified where they crossed from that. Now he was asking how to get the "boundary" if you don't have such a utility.