My professor didn't have time to derive these equations for the class and we had to settle for memorizing them. It feels so good to know where they come from, now!
That description of cosine as an "even" function and sine as an "odd" function really cleared up some other things for me, let alone these identities. Thanks a lot.
This video is truly a timeless classic. It was puzzling to me why the sum/diff formula is the way it is, and so I was doing it mindlessly. Solving trig problems make more sense now. Thanks!
dont know if you guys cares but if you guys are stoned like me during the covid times then you can stream pretty much all the latest series on Instaflixxer. Have been watching with my brother for the last months :)
You are my hero. Like- I'm supposed to do a report on this topic and I was so lost until I found this vid. Thank you. Thank you soo much. Like seriously, you saved me.
This video was so much helpful! I really can't thank you enough, sir. By the way sir, why have you stopped making videos?Your videos are very intuitive
Based on the sketch in the video, point S has angle A in standard position, and point R has angle A-B in standard position. The cosine of an angle in the second quadrant yields a negative ratio since cosine is defined as the adjacent side (negative), over hypotenuse (one). The cosine ratios of A and A-B both lie somewhere in the open interval (-1,0) since cos(90)=0 and cos(180)=-1. Angles A, B, and A-B are arbitrary; no matter where you draw them, this derivation is carried out in the same way.
Thank you so much for this informative video. My books have 2 proofs ( one using a rectangle, other uses 2 triangles) but I have no idea why they used the geometric proofs since it's only limited to acute angles and also kinda hard to understand imo. They should have included these proofs, not only are they easy to understand, you can also see why the domains are (-inf, inf) instead of just acute positive angles.
Like the vid. Thanks. You should check out a program called omnidazzle. I use the bullseye feature from it when I am projecting in class and it puts a highlight around the cursor. It would really bring attention to your cursor. Just a suggestion.
Shouldn't the points S and R have an x co-ordinate -cosA and -cos(A) as the x co-ordinate is in the second quadrant where x is negative, therefore cos should be negative?
i also reccomend watching 3b1b’s lockdown math lecture on imaginary number fundamentals because it culminates to intuitively understanding this identity
Thanks for the vid sir. You've done a very good job with your explanation. I'm still confused a bit with the coordinates though. I get points P (1,0) & Q (cosB, sinB) , I'm confused with pt S (cosA, sinA) and R (cos (A-B), sin(A-B)). Shouldn't pt S' coords. be (Cos 180-A, Sin 180-A)? I beg you to explain the coordinates sir. Pls and thanks Everything else, the algebra and etc., are clear though .
Thank you very much for the explanation! Really nice and clear. One question, I noticed some question in the comment about the S and R point being on the 2nd quadrant but you didn't write the cosine to be negative. Am I correct to assume that it doesn't matter? Since most of the sum and difference identities problem will divide the one big angle to 2 acute one hence it'll only require the first quadrant. Also, the identities will still stand regardless of where you draw the line, you can even draw them all to be on the 1st quadrant. Am I correct?
Hello Anrico. I'm glad you found the video helpful. If angle A is in the second quadrant, then yes, cos(A) will be negative. But we write it as cos(A). Think of it as cos(3pi/4). We don't "see the negative sign" until we write cos(3pi/4) as -sqrt(2)/2. Until then, it's just cos(3pi/4). Same idea in the proof. If we knew what A was then we could get the actual cosine value of it and it would indeed be a negative number. But we don't need to know the value of A for the proof. Thus, it's just kept as cos(A). And yes, this proof is in general. The angles A and B could be in any quadrant.
@@RandyAnderson Thank you so much for the response! You're right, why didn't I think of that. I really appreciate you taking the time to answer my question. Cheers.
Yes it is negative if the angle brings it into the second quadrant. If you are referring to 9:20 where cos(-B)=cos(B), it is correct because the angle -B isn't in the 2nd quadrant, it is in the fourth quadrant. So both are positive and cos(B)=cos(-B)
Ok you changed the distance formula into the addition substraction identities but you never changed them into the same equation to prove that they're equal. As in you never show cos(a-b) = cos(a-b)
You're purely just manipulating equations algebraically. For the most part, algebraic manipulation isn't meant to be understood step-by-step on an intuitive level. The equations themselves however are based on observation and intuition. If a + 2 = 5, then we can algebraically manipulate this equation by solving for a: a = 5 - 3. Furthermore, if we have a statement/argument like Secx, you can also represent that statement/argument as 1/Cosx. This means that Secx = 1/Cosx and that they are fundamentally the same, it's just that they're written in different forms. Remember, this is a PROOF; this is establishing that no matter what inputs you have, you'll always get the same equations. This is the same as entering variables and seeing if you can manipulate the the functions into something else but in a more general sense. Proofs do not establish intuition or explain why it works, it just proves that with enough algebraic manipulation, we can get the end result.
My professor didn't have time to derive these equations for the class and we had to settle for memorizing them. It feels so good to know where they come from, now!
That description of cosine as an "even" function and sine as an "odd" function really cleared up some other things for me, let alone these identities. Thanks a lot.
How is life now? After 12years
This video is truly a timeless classic. It was puzzling to me why the sum/diff formula is the way it is, and so I was doing it mindlessly. Solving trig problems make more sense now. Thanks!
You are welcome. I'm glad you found the video helpful. Study well!
This was beautiful! You actually know what you are talking about.
dont know if you guys cares but if you guys are stoned like me during the covid times then you can stream pretty much all the latest series on Instaflixxer. Have been watching with my brother for the last months :)
@Jaxon Peyton Yup, I've been using InstaFlixxer for since november myself :)
really helpful, helped me understand the equations I've been learning in school.
Excellent, a rare resource that is truly needed.
I'm glad to hear you found the video helpful. Study well.
You are my hero. Like- I'm supposed to do a report on this topic and I was so lost until I found this vid. Thank you. Thank you soo much. Like seriously, you saved me.
You're welcome. I'm glad you found the video helpful. I wish you the best on your report. Study well.
Thank you. I was looking to prove the sum and difference identities. My book does it the same way but hearing it really helped!
I really appreciate the video man. I understand so much more about the subtraction/addition identities. Thanks!
This video was so much helpful! I really can't thank you enough, sir. By the way sir, why have you stopped making videos?Your videos are very intuitive
Thank you. I'm glad you found the video helpful. I haven't made any new videos in a while because I haven't needed to for my classes. Study well.
@@RandyAnderson , ok sir. Hope you have a wonderful day. Take care
Beautifully demonstrated.
Thank you. Study well.
Wonderful video! Very helpful!
I'm glad you found the video helpful! Study well!
great video and clear explanation like khan academy even better
Now I understand the derivation, thanks!
Absolutely BEAUTIFUL.
Very clear explanation of such complex identity !!!. Thanks a ton.
Absolutely awesome and clear video. Thanks a lot!!!!
Beautiful Awesome
Based on the sketch in the video, point S has angle A in standard position, and point R has angle A-B in standard position. The cosine of an angle in the second quadrant yields a negative ratio since cosine is defined as the adjacent side (negative), over hypotenuse (one). The cosine ratios of A and A-B both lie somewhere in the open interval (-1,0) since cos(90)=0 and cos(180)=-1. Angles A, B, and A-B are arbitrary; no matter where you draw them, this derivation is carried out in the same way.
Thank you! Super clear.
Thank you so much for this informative video. My books have 2 proofs ( one using a rectangle, other uses 2 triangles) but I have no idea why they used the geometric proofs since it's only limited to acute angles and also kinda hard to understand imo. They should have included these proofs, not only are they easy to understand, you can also see why the domains are (-inf, inf) instead of just acute positive angles.
Thank you for your explanation sir
You're welcome. Study well.
Like the vid. Thanks. You should check out a program called omnidazzle. I use the bullseye feature from it when I am projecting in class and it puts a highlight around the cursor. It would really bring attention to your cursor. Just a suggestion.
thanks man this is the real math dont memorize formulas
You're welcome. Study well.
Thank you, that was a very helpful video!
Awesome! Thanks
In my opinion this is one of the toughest things in High School Math
it's not difficult,just do not explain it clearly,i have the best proof of it.
Shouldn't the points S and R have an x co-ordinate -cosA and -cos(A) as the x co-ordinate is in the second quadrant where x is negative, therefore cos should be negative?
i also reccomend watching 3b1b’s lockdown math lecture on imaginary number fundamentals because it culminates to intuitively understanding this identity
Thanks for the vid sir. You've done a very good job with your explanation. I'm still confused a bit with the coordinates though. I get points P (1,0) & Q (cosB, sinB) , I'm confused with pt S (cosA, sinA) and R (cos (A-B), sin(A-B)). Shouldn't pt S' coords. be (Cos 180-A, Sin 180-A)? I beg you to explain the coordinates sir. Pls and thanks
Everything else, the algebra and etc., are clear though .
Thank you very much for the explanation! Really nice and clear. One question, I noticed some question in the comment about the S and R point being on the 2nd quadrant but you didn't write the cosine to be negative. Am I correct to assume that it doesn't matter? Since most of the sum and difference identities problem will divide the one big angle to 2 acute one hence it'll only require the first quadrant. Also, the identities will still stand regardless of where you draw the line, you can even draw them all to be on the 1st quadrant. Am I correct?
Hello Anrico. I'm glad you found the video helpful. If angle A is in the second quadrant, then yes, cos(A) will be negative. But we write it as cos(A). Think of it as cos(3pi/4). We don't "see the negative sign" until we write cos(3pi/4) as -sqrt(2)/2. Until then, it's just cos(3pi/4). Same idea in the proof. If we knew what A was then we could get the actual cosine value of it and it would indeed be a negative number. But we don't need to know the value of A for the proof. Thus, it's just kept as cos(A). And yes, this proof is in general. The angles A and B could be in any quadrant.
@@RandyAnderson Thank you so much for the response! You're right, why didn't I think of that. I really appreciate you taking the time to answer my question. Cheers.
great proof
Why do we have to draw circles here? Just derive the identities
What happened to the square root?
Ian Gunay he squared both sides to get rid of it
isn't cosine in the 2nd quadrant negative??
Yes it is negative if the angle brings it into the second quadrant.
If you are referring to 9:20 where cos(-B)=cos(B), it is correct because the angle -B isn't in the 2nd quadrant, it is in the fourth quadrant. So both are positive and cos(B)=cos(-B)
Night b4 test, prob gonna save my @$$. Thanks!
Wow
Ok you changed the distance formula into the addition substraction identities but you never changed them into the same equation to prove that they're equal. As in you never show cos(a-b) = cos(a-b)
I still don’t understand this and I feel very stupid
You're purely just manipulating equations algebraically. For the most part, algebraic manipulation isn't meant to be understood step-by-step on an intuitive level. The equations themselves however are based on observation and intuition. If a + 2 = 5, then we can algebraically manipulate this equation by solving for a: a = 5 - 3. Furthermore, if we have a statement/argument like Secx, you can also represent that statement/argument as 1/Cosx. This means that Secx = 1/Cosx and that they are fundamentally the same, it's just that they're written in different forms. Remember, this is a PROOF; this is establishing that no matter what inputs you have, you'll always get the same equations. This is the same as entering variables and seeing if you can manipulate the the functions into something else but in a more general sense. Proofs do not establish intuition or explain why it works, it just proves that with enough algebraic manipulation, we can get the end result.
it's not the best proof,best proof is symmetry breaking.