sin(15 degrees) = ?
HTML-код
- Опубликовано: 2 июн 2024
- We will evaluate the exact value of sin(15 degrees), i.e. sin(pi/12) but there are three ways of doing so. We will first use the angle-difference formula for sine, then the half-angle formula for sine, and finally a 15-75-90 special triangle. This tutorial will help you with your trigonometry and precalculus classes!
Shop my math t-shirts & hoodies on Amazon: 👉 amzn.to/3qBeuw6
-----------------------------
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 / blackpenredpen . Every bit of support means the world to me and motivates me to keep bringing you the best math lessons! Thank you!
-----------------------------
#math #algebra #mathbasics
0:00 3 ways of evaluating sin(15 degrees)
0:21 using angle difference formula
3:56 using half-angle identity
7:17 using15-75-90 special triangle
I just memorized that the 15-75-90 triangle has legs root6-root2 and root6+root2, and hypotenuse 4.
ok
@@savitatawade2403 It makes sense. We can remember 1 - root3 - 2 easily enough. It's not that far of a leap.
I will grant you, other than the 45-45-90 the 30-60-90 and the 15-75-90 triangle, there are no other really good candidates for easy memorization. The Pythagorean right triangles have easy trig functions but irrational angle measures. The 36-54-90 and 18-72-90 are not hard to derive, but the nested radicals makes it really confusing to remember.
@@Qermaq ok
@@savitatawade2403ok
I like it, stealing
this is the first video ive seen of these that I was able to solve instinctively! I just finished taking trig, I really enjoyed it
Layin' down the knowledge in this one! I am going to steal some of this for my IB math course I teach, use it as a little seasoning to keep things spicy in class.
*The third one is 🥶*
We can reach to choice (C) by multiplying of choice (B) by
sqrt(2+sqrt(3))
I liked the last method. I used the expansion for Cos(A +B) or Cos(2X) when A=B.
Does Amazon sell that whiteboard that erases with a tap
Too bad they don't : )
math teachers have special powers
Waiting for the next video - love the expression of the result of different approaches as non-decimal values. Glad it is a next video so I have time to think about the results.
Great job, teacher!
Third one is just amazing 😮
This video is amazing.
As are all your videos doing everything by first principles and explaining in detail, your students are really lucky. :)
As an engineer for years the objectives were a quick functional answer. Your presentation reveals a hidden world of wonders. This is really cool. Thanks.
Funny, I did it by a 4th way. sin(30) = 1/2 = sin(2*15) = 2 sin(15)cos(15) = 2 sin(15)sqrt(1 - sin(15)^2)
Let s = sin(15), s*sqrt(1 - s^2) = 1/4, s^2(1 - s^2) = 1/16. Let x = s^2, x(1-x) = 1/16
x^2 - x + 1/16 = 0, x = (1 +/- sqrt(1 - 4/16))/2 = (1 +/- sqrt(12/16))/2
choose the -, x = (1 - sqrt(3/4))/2 = (2 - sqrt(3))/4 = s^2
s = sin(15) = sqrt(2 - sqrt(3))/2 = 1/(2*sqrt(2 + sqrt(3)))
I know, this is a very stupid way to do it.
Simply wonderful 👍👍👏👏
How to consrtruct 90-75-15 triangle
1. Draw line through chosen point A and B
2. Draw circle in center at A and radius AB
3. Draw circle in center at B and radius AB
Let intersection of those two circles be C and D
4. Draw line through points C and D
5. Draw circle in center C and radius AC
Let point E be the intersection of line CD and circle at center in C and radius AC , C(C,AC)
Let point M be the intersection of lines AB and CD
Triangle AME will be 90-75-15 triangle
Yes, I tried it in GeoGebra and it works! Did you come up with that yourself or did someone teach you that?
@@volodymyrgandzhuk361 I came to this construction with trigonometry
tan(75) = 2+sqrt(3) = (1+sqrt(3)/2)/(1/2)
Segment 1/2 length of original segment we can get from construction of equilateral triangle , sqrt(3)/2 is the length of height of equilateral triangle so i built this 90-75-15 triangle on equilateral triangle
fun easy video!
How do you calculate the sin/cos of an arbitrary angle?
The easiest way to explain how you could (in concept) calculate the sine or cosine of any angle, just with arithmetic operations, is to use Taylor series.
First, you'd preprocess the angle to get it between -pi/4 and +pi/4, so that it is as close to zero as practical (the midpoint of the Taylor series). If the angle after preprocessing is >pi/4, you'd find the opposite trig function of its complimentary angle. For instance, given 255 degrees, this is 180 degrees away from 75 degrees, and a related angle is 15 degrees. The negative sine of 15 degrees will give you cos(255 deg), and the negative cosine of 15 degrees will give you sine of 255 deg. So we calculate the sine and cosine of 15 deg, which will indirectly tell us the opposite functions for 255 deg.
Next, you'd translate to radians, and evaluate the Taylor series of the corresponding trig functions. Using Lagrange's error formula, you can determine how many terms you need to get the desired tolerance on the output of the trig function.
Evaluating the Taylor series for sin(15 deg) & cos(15 deg):
sin(15 deg) = sin(pi/12) = (pi/12) - (pi/12)^3/3! + (pi/12)^5/5! - (pi/12)^7/7! + ... = 0.258819, within a tolerance of 11 digits
cos(15 deg) = cos(pi/12) = 1 - (pi/12)^2/2! + (pi/12)^4/4! - (pi/12)^6/6! + ... = 0.965926, within a tolerance of 10 digits
This isn't the way calculators actually do it, since the CORDIC algorithm is more efficient. I'll leave that to you to search for.
The beauty of that half-angle identity.
At least for the double angle and 15-75-90 method, you can easily see that both answers are the same if you realise that the latter denominator is (half the) conjugate of the former's numerator.
The triangle method is closely related to the half-angle method, because it can be generalized to derive the half-angle formula. Just replace 30° with an arbitrary angle θ.
Now that I think about it, this really gives you the half-angle formula for the _cosine,_ not the sine. (Just rationalize the denominator, and factor to simplify.) Then you can get the half-angle formula for the sine by applying a Pythagorean identity. But if you get a half-angle formula for the sine _directly_ from the triangle, then it's not the usual one, and it's not as nice.
Funny you mentioned it, I made a community post about this not too long ago
thanks alot for teaching us these tricks, i am currently doing the chapter basic maths and these type of questions will help me alot in calculation for physics and maths
a lot is two words.
Could someone tell me if sin(x) [x-acute] is an rational number only for x=30 degrees? Is it possible to prove that for example sin(1) is irrational number without calculate te value of it? thx;)
As x increases from 0 to 90, sin(x) passes through every number between 0 and 1. So there are infinitely many acute angles for which sin(x) is a rational number.
Me after watching this🗿🗿
Answer within seconds √3-1/2√2 just easy😂😂😂😂
most people like me only remember pi/6, pi/4, and pi/3, so this is basically answering sin(pi/4 - pi/6) which u gotta find on paper but yeah
hey, kid, 'prove that the three are equivalent' was the homework, not 'what is sin(15°)'. quit pretending like you're a genius for knowing basic trigonometry and maybe people will actually like you instead of tolerating your presence.
@@GirishManjunathMusic I hope your genius , I am so dumb that I only remembered sin basic value ok . I hope you to be good in the comments but don't be rude saying your are genius then I have a question why don't you find sin(43) and tell me with proof then your are genius
@@gineethr7099 i never said I'm a genius, kid.
@@GirishManjunathMusicage?
It's just class 11 th concept of maths in india (way to easy😒)
Cool, but the video is more
Yh no shit, this is the basic maths channel.