Prove the Pythagorean identities
Vložit
- čas přidán 18. 06. 2016
- How to prove the pythagorean identities?
Prove sin^2(theta)+cos^2(theta)=1,
Prove tan^2(theta)+1=sec^2(theta),
Prove 1+cot^2(theta)=csc^2(theta),
Join my fb group: / blackpenredpen
bprp trig lessons...
blackpenredpen.com
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](http://i.ytimg.com/vi/p6j2nZKwf20/mqdefault.jpg)
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](/img/tr.png)
On an Unit circle:
Sine and Cosine build a right triangle with the radius which lenght is one. So sin²(θ)+cos²(θ)=1²=1
I've been through three pre-calc/trig textbooks looking for a proof on the three main identities. Still hadn't seen it until I ran across this video. Thank you very much!
that's the problem with modern education: it doesn't explain *how* things work. It just tells us what to do with what we're given
@@nostalgiafactor733 That feeling when you're told to memorize the values of cos and sin of 30, 45 and 60 because they come up often.
You put 1+cot(theta)=cot(theta) in the description and scared the sh!t outta me lol
thank you for pointing out my typo
Fun fact: Pythagoras, the creator of the Pythagorean Theorem, was also a cult leader. One time he and his members were locked in a room and needed to get out. Pythagoras had his members form a human ladder to the window at the top, and he climbed out, leaving his members alone in the aforementioned room to die.
What a dork
I needed this video , thank you very much for making it❤
4:30 , Well Proved. Was very helpful to me.
i was struggling with this problem for hours, thank you!!!
While watching the last few minutes, I realized you can get an interesting relationship csc^2 (theta) + sec^2(theta) = 1/(sin^2(theta)•cos^2(theta) ) and then use the double angle formula sin(2•theta) = 2•sin(theta)•cos(theta) to get an expression for the inverse of sin^2(2•theta) or the inverse of that to be 1/(sec^2 + csc^2)
Super late reply I know. Out of curiosity I did the algebra on notepad to see if you were right. However, I got the answer of sin^2(2x) = 4/(sec^(x) + csc^(x)). I think you forgot to carry a 4 from the sin^2(2x). We know that sin(2x) = 2sin(x)cos(x), therefore sin^2(2x) = 4sin^2(x)cos^2(x).
The Omniversal Gamer possibly... will check, Thanks!
You r such a life saver thank u so much
Consider the graph y = x
consider the distance from origin to the point some finite radius 'r'
let's say the angle it makes is ∆ (I know choosing ∆ as an angle is pretty weird but.. bruh my keyboard doesn't have many math symbols)
Now you know that sin∆ = y/r , cos∆ = x/r
=> x = rcos∆
y = rsin∆
Now using the Pythagorean theorem and plugging in for x and y
We get r²(sin²∆+cos²∆) = r²
Since r > 0 we can cancel it out, when r is 0, namely the distance from the origin is 0, or the side length for hypotenuse is not there, thus a right triangle can't be formed, we are talking about (x,y) € (0,0), trivial...
So we get : sin²∆ + cos²∆ = 1
Thank you so much 🙏🙏🙏
From Algeria Thank you ,sir 👏😄
Wow very nice videi
hypobtenuse
Re Tend hypopetuse?
Parenthesis ×
Prenesis √
Multiply?
Mutipie
Question: could we have proven this with a reference to the unit circle (i.e. c = 1)?
In this case you can just apply the Pythagorean theorem: the squared values of sine and cosine will add up to 1, which is the radius of said circle! I hope that was clear enough
awesome 🖤🖤🖤❤️❤️
Tomorrow is my exam and I got the solution thankyou very much 🙏
sin^2(theta)+cos^2(theta)=1 is known as the idiot formula in Denmark. i have no idea why.
Probably because any idiot can remember it... It's so simple😀
why should we divide C square on both sides
thank you very much
It can be proved using similar triangles
thanks a lot
Put a video about why sine is opposite over hypotenuse... Pls pls pls
bruh thank u so much
Nice Video
Thanks bro
This is a stretch since it's been a year, but hopefully someone will answer: how does this prove for any angle theta? If we use right triangles then theta is strictly between 0 and 90 degrees, how is this proof valid for angles greater than 90?
Daniel Dubnikov consider this: is the value of the sine or cosine of an angle dependent on what sort of triangle it's in? No! Cos (30) as part of a right angled triangle is the same as cos (30) as part of an isoceles. Both are (sqrt3)/2. Any one angle can be made into a right angled triangle, even angles over 90 degrees, so for any theta you can use Pythagoras to determine sin2theta+cos2theta =1.
Daniel Dubnikov for angles over 90 degrees, consider that you can calculate the sine function of any angle, that's why the sin graph goes on forever. It's just harder to calculate on paper, you need to work with negative numbers etc.
I love you bro
You are as good as if not better than PatrickJMT. You're definitely better than Salmon Kahn.
Hello there, I thank you for your nice comment. However, they are the ones who gave me the ideas to do CZcams math videos. So, they are still the legends!
The new guy Khan has is much better than Sal. Less stammering ,no mistakes, and no.. no repeating... no repeating himself... him... self... like.. repeating himself... like... this. Liiike.... like thiiiissss.
I like it
Dope
also, sin^2(2theta)+cos^2(2theta)=1 for the same reasons
actually, if both arguments are multiplied by the same coefficient, it is still 1
nice
I have a problem
Cos2thita-sin2thita=0
How is possible
I think you got a wrong info dude
how could you prove this identity? sin(2theta)=2sin(theta)cos(theta), by the way good video
sin(A+B) = sinAcosB + sinBcosA
therefore,
sin(2A) = sin(A+A) = sinAcosA+cosAsinA = 2sinAcosA
cot ?? WTF ? need ctg
99.99% of the world uses cot for cotangent. I've never even seen "ctg" before this. Where did you learn that?
@@VeritasEtAequitas in Russia use "tg, ctg, cosec"
in Japan we only use sin cos tan. no cosec, sec, cot
I don’t know why I know this but I’m in grade 6
Are you genios
We prove this from triangle but in triangle if angle in 0 then no triangle creat then how sin20+coc20=1
Cos 0 is 1 😂
sin 0° = 0
cos 0° = 1
sin² 0° + cos² 0°
= 0² + 1²
= 1
I prefer to see you write on the board. Feels more human than just watching the writing appear
hay boy
are you a chinese?
🐧🐧
Lol
I think you should meet a spoken English teacher
This is insanely obvious
why should we divide C square on both sides
That is what transforms a^2 + b^2 = c^2 into ratios for each, which is what trigonometric identities are. This is how the relationship between simple lengths of the sides becomes the relationship of trig ratios, or identities..