I wanted to learn these for so long. thank you so much

Wow

## it has been 3 yearss!! I'm wondering 🤔 what happened to him 😢... I wish Nothin'' bad✓✓

Hats off❤

Sin comes from the 👹

As a lay man,no struggled to learn this shit at school. I don't use it anywhere in my life

Guys,guys,guys 6/2(1+2) =6/(2+4) = 6/6 =1 OR According To BODMAS(ik it’s basic stuff) =6/2*3 =3*3 =1

Thanks a lot

I guess you got sucked into another dimension? Where did you go?

this is the best explaination ever!!!

My great, great,… grandpa invented compound interest. i should tell the story of what lead to this invention. it is actually interesting.

wish i had youtube when i was going to school, would have got much better results then with Ms Parkes teaching me, it sounded and looked way more complicated back in school

I use BOMDAS

Nearly 60 years old, got O'level and A'Level in maths and used it in my engineering courses. Never have I seen trigonometry explained so well

Teachers say "do this" but rarely do they say, "here's the application" or "this is what's going on behind the scenes". Doing math with only the how and no why or when feels like biking uphill with the brakes on.

Why are they called sin, cos and tan?

Simple, excellent explanation. Thank you. Vetri South Africa 🙏🇿🇦🇿🇦🙏

This channel should have more subscribers as its explanation is so realistic and clear. Thank sir

thanks bro so much your videos make me understand math not just memorize it just like the school thanks❤

This video literally cracked the code of teaching

If you take any expression and replace the entire thing with a single variable, algebra says the smallest tree you can make is this: * / \ 1 X 1x, where x is a function in parentheses. All variables have coefficients. There is ALWAYS a multiplication happening after the function in parentheses is evaluated. Which makes sense because can't have something having a higher priority unless there is another function happening afterwards. You have to follow this pattern, otherwise algebra must be wrong. So no, other operators and coefficients of a parentheses don't have the same priority. Coefficients have the SAME priority as the contents of the parentheses. They are BOTH needed to complete the multiplication at the top of the simplest tree, and should be considered to be part of the first step.

I was always curious. Knowing the steps is not the same as understanding the logic of it. Once you do start to understand.. it is like a whole new world of possibilities opens up for you.. too bad semesters don't last that long. We'll have to carry a little bit of knowledge and carry it with us, hoping we can put it to work without the aid of technology

Higest priority is the paranthesis, so 6 : 2 * 3 divide and multiply has same priority and are solved left to right so, 6 : 2 = 3, and 3 * 3 = 9 I wrote a recursive descent expression evaluator, based on math operator priority, and it says the same. 9

Area of unit circle is π.

I never understood the store behind Sin, cos...etc. until I watch this amazing video Thank you

Years ago when I was studying I made yp this phrase to remind me ... Some Old Humans Can Also Have The Original Alphabet. Its that simple .. ha ha.

Been years since I had trig in high school. This video helped me understand more than my many years of studying.

Very helpful in understanding how each of these are calculated. However I was under the impression I was gonna see an explanation of where the names Sinus, Cosinus and Tangens came from.

Well done ,not some high school know it all .

What the flip was that about and wher'es the peanuts .

If you follow PEMDAS or BODMAS rules, the answer is 9. End of story.

Interesting, makes me now wonder how I ever got anything right at all.

The origin of Sin ;)

And part 2 never came...

Somehow, I got a B in Trig, but I never understood what I was doing, until now.

I think the problem with teaching math many times has a twofold handicap: 1.) most brilliant math teachers can't understand how to teach because it came so naturally and easy 2.) many math teachers are more interested in appearing brilliant than teaching!

Excellent video!

You explained what none of my math teachers ever did. I need to understand the origin of rge math to solve the problem. All my teachers could never explain this to us. Therefore I list interest. Can you explain why PEMDAS came about and who did it?

Mr Michael you claim to be a programmer but you don't know simple grade 5 maths. There's no ambiguity. There is no need to use your own parenthesis. India sent a rocket to the moon. They should have sent you and changed you from a programmer to Moonsamy

And I always ask myself where did the Pi come from and what is the relation between the triangle and circle, this presentation gives really an awesome explanation, thanks dude.

Although this answer is late and few will read this I hope to answer your question. To do this I want you to go to Gobekle Tepe and study the geometries of the buildings. Also go to CatalHoyuk. We can see that there are attempts here for rectilinear structures, but these structures fall short of being rectilinear. The first sort of relevant rectilinear structure was the E-Absu at Eridu whose first constructions are from 7300 years ago. The Mesopotamians would come to master rectilinear structures in the constructions of their cities. They also invented the Ox-cart which is essentially 8 legs pulling 4 rectilinear positioned wheels. In these processes the Sumerians and Babylonians produced our basic geometry tool kit. The circle has 360°. The also discovered the Pythagorean triangles (but for them the rectangle was more important). If you discover more of these eventually you get nearly isosceles right triangles, which mathematically can be used to estimate the chord of 90°. But why were these used. The perfect right triangles can be used to determine squareness of an angle in two dimensions. If you measure 3 units in one direction and 4 units in another direction you get 5 between the two endpoints if it’s square. So let’s start with this triangle 3-4-5. Let’s do this, let’s mirror 3-4-5 over 4, this gives an isosceles triangle of two sides of 5 and one side of 6. Now let’s divide by 5. This gives us one side of side 1.2. This is the chord of the double angle opposite side 3 in the original triangle. In this rescaled triangle the height is 4/5 or 0.8, the chord is 1.2 If we continue doing this for more nearly isosceles right triangles as we approach equity in the orthogonal sides the chord approaches 2x height. This side values of the two composite right triangles approach the SQRT (1/2). And the chord would be thus 2 * SQRT(1/2) = SQRT(4)* SQRT(1/2) = SQRT(4/2) = SQRT(2) But could we mathematically predict this using known values. I will solve your problem without using an infinite series. So geometrically 360° divides by 2,3,4,5,6,8,10,12,15,18,20,24,30,36, 40,45,60,72,90,120,180, 360 to give the chords of 180°, 120°, 90°, 72°, 60°, 45°, 36°, 30°, 24°, 20°, 18°, 15°, 12°, 10°, 8°, 6°, 5°, 4°, 3°, 2° and 1° What if we divided the 360 by 1, well that just gives a chord of 0 2, that gives the diameter 2 3, unknown 4, SQRT(2) 5, unknown 6, 1 8, unknown, and so on. So if chord of 180 degrees what is the two right triangles that compose the isosceles triangle? see 3-4-5 example Those mysterious triangles have a halfchord of 1 and a bisector of zero. So can we use these mysterious triangles to predict the chord of half the angle (90°). So we bisect the chord 180°. We then project the zero length bisector to intercept the circle. Since the bisector is zero length the distance is r-0 and on the unit circle this is 1. Ok so we have two orthogonal sides of something. So here’s the theory, if you draw any chord on a circle. The chord and the two radii always from and isosceles triangle. If you bisect the isosceles you get two right triangles, so it’s not just orthogonal because we bisected 180° which is 90°, but and bisection of the chord results in two right triangles to the intersection of the bisector and chord, BUT ALSO two right triangles to form the chord of the half angle. So the chord of the half angle of 180° is chord 90°= C and A^2 + B^2 = C^2 = 1^2 + 1^2 = 2 C = positive +/- SQRT (C^2) = + SQRT(2). I want you to note something mysterious here. If the chord of 180° = 2, the polygon of sides 2 has a perimeter of 4. If the polygon has 4 equal sides then the inscribed polygon has sides 1.4142x 4 = 5.657 So let’s solve one more. Let’s get the chord of 45°. The chord of 90° is 1.41428 Then the halfchord and bisector are 0.70714 because this is the only isosceles right triangle. Thus the thus the exterior triangle has a height of is 1-.70714 = 0.29285 and a base of 0.70714 SQRT(1/2) ^2 + 0.29285 ~ 0.5 + 0.0857 = 0.5857. Thus the chord of 45° = SQRT(0.587) = 0.76536686473018 Now if you drawn the actual chord of this out on a circle and draw a line from the one point to the opposite side of the other point we should see that we created the chord of 135. Now remember what I said above the chord of 360° is 0, that means the halfchord is also zero, which means the bisector is 1. If we draw such a line to the opposite side we create the chord of 180°, it’s halfchord is 1 and is bisector is Zero. If we take the chord of 90° we have a bisector os SQRT(1/2) and a halfchord of SQRT(1/2). If we repeat the same process, draw a line from one end point of the chord to the point opposite the other endpoint we get a new chord identical to the first with same halfchord and bisector. The bisector in the case of Chord of 360°(0°) = halfchord of 180° the halfchord 0° = bisector 180°. There seems to be a reciprocal relationship between the halfchords and bisectors of angles related by 180-x. In the case of 45° the new chord we created is 135°. The base is a diameter thus 180°, the chord 45° takes 45 from 180 leaving the chord 135°. We need to investigate this. The chord 135° is bisected, it forms a triangle with hypotenuse 1 and angles of 67.5. Remember the bisector always forms a right angle with the chord. 180 - 67.5 = 22.5° So this triangle shares the hypotenuse with right triangle the bisector of 45. That triangle bisects to form 2 22.5° angles. The adjacent angle is 67.5°. If you draw this out the right triangle created by the bisection of 135 creates a rectangle with the adjacent triangle bisected of chord 45°. Thus the halfchord of 45° = bisector of 135° and vice versa. We need the halfchord of 135° This is simply SQRT(1-chord^2 45°/4) We are not done yet. This 8 sided polygon created by concatenation 7 more chords is 0.7654 = 0.0032 + 0.0400 0.4800 5.6000 = 6.1232 If we repeat the process that we performed on the angle we can get the chords of 22.5,11.25 5.625 and so on. More interestingly the chords begin to approach Angle * pi/180° if we multiply this by the 2^divisions, we get essentially pi. This is the basis of trigonometry before sines and cosines. however knowing the chords and bisectors of 180, 90, 45, 135, 22.5, 157.5, 11.25 is not all that useful. We have five tricks cards up our sleeve. The first is the chord of 60°. Well we know it’s 1, so it’s half chord is 1/2, and thus its bisector is SQRT(1-1/4) = SQRT(3)/2. We can relate this to 120°, because it is 180-60 and thus the chord is the 2*SQRT(3)/2 = SQRT(3). But it’s a funny thing, chord 120° is also the chord of the double angle of 60°. Is there another way that is generally useful. Actually there is. The chord of the double angle is 2 Chord x * bisector x . So let’s check it out 2 * (Chord 60° = 1) * (bisector = SQRT(3)/2) = 2* SQRT(3)/2 = SQRT(3). Thus now we can double chords. What’s the double angle chord 120° = 2* SQRT(3) * 0.5 = SQRT(3), also correct, chord 240° = Chord (360°-240°=120°). The first trick is we can determine the chord of any angle with a known chord on the unit circle. Thus if we know the chord and the radius or the angle we can determine this. From this we can get the chords of 30°, 15°, 165°, 150° Our second trick Ptolemy’s gives to us, we cannot solve the chord of the pentagon, but we remember that the side of pentagon is the chord (360°/5 = 72°). This means the two other angles are 180°-72° = 2 * 108°/2 But the radius splits the interior angle of the pentagon, so that angle is 108°. We can use ptolemies theory on quadrilaterals to solve it. ABCD such that inscribed a circle the AB*CD + BC*DA = AC*BD were A,B,C,D,A is the order on the circle and AC and BD are diagonals. This results in 1 + X = X^2 for four of five points on a perfect pentagon. X solves to “The Magic Number” 0.5 + SQRT(5)/2. This is the chord of 108°. The halfchord is 0.25 + SQRT(5)/4, and from this we can get the bisector. Isn’t life grand. 180 - 108° = 72° so we can solve the pentagon, and we can get the chords of 36°, 144°, 18°,162°, 9°, 171°,81°. The next thing is that we can find the value a chord of the average angle of two known chords. This is chord (a+b)/2 = SQRT((halfcord a + halfchord b)^2 + (bisector a - bisector b)^2. Using this we are able to derive the chords of every third degree angle on the unit circle. Our forth trick is to find the chord of a nonagon. This is 360/9 or the chord of 40°. We have a triangle so we know all we really need to do is resolve the arc over one chord into three chords. This can be done with ptolemies formula and some intense algebra. But once this is done we can use all our tricks to get all the chords of all degrees angles on a circle. Our final trick is this, it’s going to be painful, it’s going to hurt. Very intense. This sine of any angle is the halfchord of double the angle. The cosine of any angle is the bisector of the chord of any angle. IOW in making a list of chords, halfchords and bisectors, as the Greeks did, all you need is another column with the angle divided by two and you instantly know the sine and cosine. Sin x = (Chord 2x)/2 cos x = SQRT (1- (chord 2x)^2/4) and so if you find your self in second century BCE starring down a list of chords, know this, to what ever resolution they provide you can get sine of any angle you so desire by finding the closest double angle, getting the halfchord and you have the sin (desired angle +/- resolution/4). The Babylonians and Greeks were not dunces. All the hard work later mathematicians claim they have done, already done 2200 years ago. What changed is the way we solve problems.

Thank you so much for this video. Youre the onl one explaining this in the whole internet❤❤❤

Now, if only they can explain why it is "maths" when they also say "science" and not sciences...even Google thinks maths is a misspelling. Math was used by the speakers of the Kings/Queens language all the way up until the last century. Much like Association Football being shortened to soccer, then they up and changed it to football. The big explanation I see is mathematics is a mass noun. Well, ok. Why then do you take a science class and not a sciences class? Science is a mass noun too, encompassing many disciplines. Sure, you can study the sciences, but mathematics would be the term for studying that field. I think that some folks just wanted to add a lisp sound. You take a math class, you apply math, you work on your math homework. If you are a math major, you will study different fields of mathematics. Mass noun and plural noun are not the same thing. The originators of the English language should know this. Even the English would never say, "I am taking a sciences class".

Wow, awesome video and very clear explanation. Where is part2?

Still waiting for part 2 ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

The Commutative Property only works in Addition and Multiplication.

6/2(1+2) = 9. This problem takes the form of a/b *c. a/b*c and a/bc are not equal to each other. We can flip the order of operands around and still maintain the solution (Commutative Property). It's a binary operation. The problem is separated by multiplication. Any time you see a number right next to the parentheses, it means multiplication.

Best teacher ❤🙏

Why we only can take decimal digit at opp. Side ??

Anyone in 2024?