Is there a formula to calculate the grey area shown in the picture?

>>7845615

Your writing is incomprehensible but I'm assuming you are asking you to do this without integrating?

>>7845619
Yeah I need to calculate the grey area but I don't even know if it has a name, on google I couldn't find anything.
That's an attempt I made but I don't know if that's right.

I'll repost the text using the eqn tag, maybe it will be easier to read:

I was thinking about subtracting from the area of the bigger triangle [eqn]A_T = \frac{r (r\tan x)}{2} = r^2 \frac{\tan x}{2}[/eqn] the area of the circular sector [eqn]A_c = r^2 \frac{\pi x}{2\pi}[/eqn] and the area of the smaller triangle outside the circle [eqn]A_t = \frac{a b }{2} = \frac{(r - r\cos x) (r\tan x - r\sin x)}{2} = r^2\frac{(1 - \cos x) (\tan x - \sin x)}{2}[/eqn].
So we get [eqn]A = A_T - A_c - A_t = r^2 \frac{\tan x}{2} - r^2 \frac{\pi x}{2\pi} - r^2\frac{(1 - \cos x) (\tan x - \sin x)}{2}[/eqn].
This simplifies to [eqn]A = \frac{r^2}{2} (\tan x - x - (1 - \cos x) (\tan x - \sin x))[/eqn].
Using some trigonometry we get [eqn]A = \frac{r^2}{2} (\tan x - x - (\tan x - \sin x - \sin x + \sin x \cos x)) = \frac{r^2}{2} (- x + 2 \sin x - \sin x \cos x) = \frac{r^2}{2} (2 \sin x - \frac{1}{2} \sin 2x - x )[/eqn].
So the area of the grey surface is: [eqn]A = \frac{r^2}{2} (2 \sin x - \frac{1}{2} \sin 2x - x)[/eqn]. Did I get it right? Can I simplify it even further?

File: eqn.png (48KB, 878x634px) Image search: [Google]

48KB, 878x634px

>>7845629

Anyone?

seems right OP, and I don't think you could meaningfully simplify this

>>7845615
Calculate the area of the larger triangle, (cosx + tanx)/2, we will call this quantity A.

Next we need to calculate the area of the slice of the circle. To do this calculate the area of the circle and multiply it by the ratio of angle x/360. We will call this need quantity B.

Now calculate the area of the small triangle, given by ab/2. We call this quantity C.

Finally, perform the operation A - B - C = Area of grey region.

>>7846544
Thanks, that's how I did it, I wanted someone to check my calculations.
The area of the bigger triangle is [math]A_T = \frac{r r \tan x}{2}[/math], the base is equal to the radius, not to cos x.

>>7846588
Why?

I got [math]A = \frac{1}{2}t\tan{x} - \left(\frac{x}{2\pi} + \frac{1}{2}ab\right)[/math]

>>7846613
that should be I got [math]A = \frac{1}{2}r\tan{x} - \left(\frac{x}{2\pi}\pi r^2 + \frac{1}{2}ab\right)[/math]

>>7846618
simplified to [math]A = \frac{1}{2}\left(\tan{x} - \frac{x}r^2 + ab\right)[/math]

>>7846619
[math]A = \frac{1}{2}\left(r\tan{x} - \frac{xr^2}{\pi}+ ab\right)[/math]

The first term? is [math]\frac{r (r \tan x)}{2}[/math], the second term simplifies to [math]x r^2[/math], and in the last term [math]a = r - r \cos x[/math] and [math]b = r \tan x - r \sin x[/math].

>>7846621
[math]A = \frac{1}{2}\left(r^2\left(\tan{x} - \frac{x}{\pi}\right)- ab\right)[/math]

>>7846621
>>7846633
>see this >>7846632
But the first term is not divided by 2.

>>7846638
but it's a triangle

File: Area.png (13KB, 318x508px) Image search: [Google]

13KB, 318x508px

>>7845615
I'm not checking your work. I'm too fucking lazy. All you have to do is define two areas (pic related) and subtract.

>>7846683
[math]\frac{r^2}{2}((1 + \cos x) \sin x - x)[/math]
This seems to be better, thanks.

>>7846747
[math]\frac{r^2}{2}(\cos x \sin x - x)[/math]
This is even better.

>>7846756
>[math]\frac{r^2}{2}(2 \sin x - \cos x \sin x - x)[/math]
> Oh well, that's the same I got before, I think this is as small as it can get.

op i hope you realize that this obsession you seem to have with "simplifying" trigonometric functions "as much as possible" is a total waste of time. if you want to understand the behavior as a function of the angle just plot it on a computer -- it's not like a bunch of random sines will give you any more intuitive understanding than a bunch of cosines, sines, and tangents.

You were finished by the 4th line in the OP, I don't understand the point of this thread.

>>7846797

What program ?

>>7846797
I need to use it in other formulae and in integrals, so the simpler the better.
And i think that using only sines makes it more elegant.

>>7846892
I usually use wolfram alpha, works great.
You could also use mathematica or geogebra (maybe even matlab?).

(cosx+tanx)/2-(a+b)/2-(xr/2)