I finally understand trigonometry

[MagicBoy]

I love that guy's YouTube channel. The dude touches everything from mechanics to chemistry. And I finally understand trigonometry (too bad his speech is too monotonic, but the content is good):

https://youtu.be/yV5QykhwSPY

[Natacha]

I'm happy for your break through. Trigonometri is actually kinda cool.

[abcdeqwe]

Happy for you dude yeah there’s a lot of great math channels out there I recommend blackpenredpen and 3blie1brown although their content is a bit more advanced than trig they are still really good at communicating

[Alexnbin]

I might need this
Thank you for sharing

[MagicBoy]

So I watched the 3rd part of the tutorial, and at some point he says that the length of the chord can be easily related to the sine of the half-angle, and leaves it to us to show it. But I couldn't figure it out.

Any idea?

[Natacha]

So I watched the 3rd part of the tutorial, and at some point he says that the length of the chord can be easily related to the sine of the half-angle, and leaves it to us to show it. But I couldn't figure it out.

Any idea?

Hey :-) If you an AT before the name you are refering to, it's likely you will get an straight answer quickly. And I guessyour question might be refered to RyanIsGay24

[abcdeqwe]

I love that guy's YouTube channel. The dude touches everything from mechanics to chemistry. And I finally understand trigonometry (too bad his speech is too monotonic, but the content is good):

https://youtu.be/yV5QykhwSPY

So if drop a perpendicular from the point where the hypotenuse touches the circle you will break the triangle into two right triangles. The bigger triangle has side lengths rcosx, rsinx, and r. The smaller triangle has side lengths r-rcosx, rsinx, and c (length of chord). Since the smaller triangle is a right triangle we can use the Pythagorean theorem, which means that (r-rcosx)^2 + (rsinx)^2 = c^2. Expanding the left hand side gives 2(r^2)-2(r^2)cosx=c^2. Now multiply numerator and denominator by 2/2 and factor out the numerator, which leads to 4(r^2)((1-cosx)/2) = c^2. Now square root both sides to get 2r*sqrt((1-cosx)/2). Then it is a trig identity that sqrt((1-cosx)/2) is equal to sin(x/2) so all in all we have chord of x = 2r*sin(x/2).

[MagicBoy]

So if drop a perpendicular from the point where the hypotenuse touches the circle you will break the triangle into two right triangles. The bigger triangle has side lengths rcosx, rsinx, and r. The smaller triangle has side lengths r-rcosx, rsinx, and c (length of chord). Since the smaller triangle is a right triangle we can use the Pythagorean theorem, which means that (r-rcosx)^2 + (rsinx)^2 = c^2. Expanding the left hand side gives 2(r^2)-2(r^2)cosx=c^2. Now multiply numerator and denominator by 2/2 and factor out the numerator, which leads to 4(r^2)((1-cosx)/2) = c^2. Now square root both sides to get 2r*sqrt((1-cosx)/2). Then it is a trig identity that sqrt((1-cosx)/2) is equal to sin(x/2) so all in all we have chord of x = 2r*sin(x/2).

Yes, I actually just figured it out and made this figure to explain it:

https://i.ibb.co/p4hgCL9/Screenshot-from-2020-10-08-10-33-45.png (https://ibb.co/0YskPGd)

Thanks!

[foxwolf]

I recommend you khan academy site which gives taste to learn more maths to me