Sir now Ill completely forget memorizing these formula Because from Now on I can derive these formula anytime , anywhere Long Live BlackPenRedPen Yeeay!!!
5*: Everything I have found on the Internet so far has been a proof, not a derivation; meaning that they start with the answer and then simplify the answer on the RHS to equal the LHS. Yours is the first real derivation I've found. Love it!
My textbook did a terrible job explaining this proof, and most proofs I found involve proving itself. Thanks for making this proof crystal clear. Well done.
an informal way to derive alpha in terms of A and B: Say for example we take sin(40) + sin(60). how would we determine alpha and beta? well, it's not too hard to see it'll be 50-10 and 50+10. alpha is fifty, because it's the average, and beta is 10, because its the difference between each term and the average.
Can you make videos explaining how to solve equations involving the floor function? An example of such equation would be floor(x)-2floor(x/2) = 1. Great videos by the way!
this is a very elegant proof. I on the other hand, started from the 2sin((a+b)/2)cos((a-b)/2) and arrive at sin(a)+sin(b): 2sin((a+b)/2)cos((a-b)/2) =2[cos(a/2)sin(b/2) + cos(b/2)sin(a/2)][cos(a/2)cos(-b/2) -sin(a/2)sin(-b/2) ] =2[cos(a/2)sin(b/2) + cos(b/2)sin(a/2)][cos(a/2)cos(b/2) +sin(a/2)sin(b/2) ] =2[ cos(a/2)sin(a/2)(cos^2(b/2)+sin^2(b/2)) + cos(b/2)sin(b/2)(cos^2(a/2)+sin^2(a/2)) ] =2[ cos(a/2)sin(a/2) + cos(b/2)sin(b/2) ] =2cos(a/2)sin(a/2) + 2cos(b/2)sin(b/2) =sin(2a/2) + sin(2b/2) =sin(a) + sin(b)
How did you get from =2[ cos(a/2)sin(a/2)(cos^2(b/2)+sin^2(b/2)) + cos(b/2)sin(b/2)(cos^2(a/2)+sin^2(a/2)) ] =2[ cos(a/2)sin(a/2) + cos(b/2)sin(b/2) ] ?
this identity makes it not circular reasoning to use lhopitals rule on lim_{h->0} (1-cos(h))/h, because you can derive the cosine versions of this identity by replacing a with a+π/2, and b with b+π/2, and for proving the derivitives of sine and cosine, just use these identities instead of expanding out via the sum and difference identities, d/dx(sin(x))=lim_{h->0} (sin(x+h)-sin(x))/h=lim_{h->0} (2cos((x+h+x)/2)sin((x+h-x)/2))/h=lim_{h->0} (2cos(x+h/2)sin(h/2))/h, and do the same thing for cosine
Very curious at 3:33 . . . Aww, why bother with the clumsy step of multiplying alpha - beta = B by negative one at all? Come on, simply SUBTRACT the whole thing from alpha + beta = A, and you IMMEDIATELY get 2beta = A - B Also, at 4:30 . . . No parentheses will be necessary for single-variable arguments in trigonometric functions, thus it is perfectly ok to write sinA + sinB rather than the, again very clumsy, sin(A) + sin(B) . . . especially that you were already writing in black and red ^_^ Finally, are you also on Facebook? I'd love to join you if you happen to be there!
e^xcosz-(1/3)e^(3x)cos(3z)+(1/5)e^(5x)cos(5z)-.... Please help me with this series.I am asked to find the infinite sum of this series.Got this from a complex variable book. :/
Sir now Ill completely forget memorizing these formula
Because from Now on I can derive these formula anytime , anywhere
Long Live BlackPenRedPen Yeeay!!!
My pleasure!
this is probably the best teacher I have in terms of all the trigonometic identities. Very simple and consice! Love it!
Blackpenredpen white paper. You are uplifting the channel name. Good job!
Yup!!!! Thanks!!!!
@@blackpenredpen love this vid. kinda irrelevant, but what pens do you use?
I am so happy you put this up 2 YEARS AGO!!! thank you so much, never investigated these relationships
This proof videos are my favorite, thanks! Love u ❤
I have been looking for a proof just like this and found it. Thank you so much!
5*: Everything I have found on the Internet so far has been a proof, not a derivation; meaning that they start with the answer and then simplify the answer on the RHS to equal the LHS. Yours is the first real derivation I've found. Love it!
You make me so happy now, thanks a lot!
That demostrative videos are amazing!!!
Thank you!!
Wanted to say Thank you! Learnt a lot from you till date :)
Thank you very much!!! It becomes so much easier to memorise now that I know how it works, video very much appreciated!
Thank you! I have been find the proofs for this, since the precalculus lesson I attended didn't prove this for us
beautiful magnificently explained.... thank you so much
My textbook did a terrible job explaining this proof, and most proofs I found involve proving itself. Thanks for making this proof crystal clear. Well done.
After 11 months, still helpful!
Really helped me out!
That's just great! 😯
Great video, thx.
THIS IS SO FUN! please keep on proving stuff and do more videos with this OG style!!
San Samman OK!!!!!!
this helped me understand! thank you
Pretty easy identity to prove but still a useful one
thank you so much!!!
an informal way to derive alpha in terms of A and B:
Say for example we take sin(40) + sin(60). how would we determine alpha and beta? well, it's not too hard to see it'll be 50-10 and 50+10. alpha is fifty, because it's the average, and beta is 10, because its the difference between each term and the average.
Cool job!!!
Can you make videos explaining how to solve equations involving the floor function? An example of such equation would be floor(x)-2floor(x/2) = 1.
Great videos by the way!
bless ur soul
I like the video so much.
Thank you!!!
amazing
thank you so much
Everybody know this. your are my favourite teacher and i hoped that it will be a geometric explain.
Thank you!!!
Where’s the dabbing man? Love the vids👌
no please
this is a very elegant proof.
I on the other hand, started from the 2sin((a+b)/2)cos((a-b)/2) and arrive at sin(a)+sin(b):
2sin((a+b)/2)cos((a-b)/2)
=2[cos(a/2)sin(b/2) + cos(b/2)sin(a/2)][cos(a/2)cos(-b/2) -sin(a/2)sin(-b/2) ]
=2[cos(a/2)sin(b/2) + cos(b/2)sin(a/2)][cos(a/2)cos(b/2) +sin(a/2)sin(b/2) ]
=2[ cos(a/2)sin(a/2)(cos^2(b/2)+sin^2(b/2)) + cos(b/2)sin(b/2)(cos^2(a/2)+sin^2(a/2)) ]
=2[ cos(a/2)sin(a/2) + cos(b/2)sin(b/2) ]
=2cos(a/2)sin(a/2) + 2cos(b/2)sin(b/2)
=sin(2a/2) + sin(2b/2)
=sin(a) + sin(b)
How did you get from
=2[ cos(a/2)sin(a/2)(cos^2(b/2)+sin^2(b/2)) + cos(b/2)sin(b/2)(cos^2(a/2)+sin^2(a/2)) ]
=2[ cos(a/2)sin(a/2) + cos(b/2)sin(b/2) ]
?
For the difference, it is sin(A)-sin(B)=2*cos((A+B)/2)*sin((A-B)/2)
I love these so much. All just connects together. I lvoe math.
trigonometric proofs are beautiful
Can you racionalize 1/[cuberoot(a)+cuberoot(b)+cuberoot(c)] please? Love your videos
Could you please do int_0^1 int_0^1 [1/(1-xy)] dx dy = zeta(2)?
Thank you and of course great channel ;D
谢谢
can you derive this formula using euler's formula ? (without subtituting alpha+beta = A and alpha-beta = B ?)
The sin (a+b)
Vid isn't in the description
yah, that's what I looked for too
Boypig24 sorry I forgot. It's here ua-cam.com/video/2SlvKnlVx7U/v-deo.html
thanks brother proof of sum to product identities is not in my book for some reason
this identity makes it not circular reasoning to use lhopitals rule on lim_{h->0} (1-cos(h))/h, because you can derive the cosine versions of this identity by replacing a with a+π/2, and b with b+π/2, and for proving the derivitives of sine and cosine, just use these identities instead of expanding out via the sum and difference identities, d/dx(sin(x))=lim_{h->0} (sin(x+h)-sin(x))/h=lim_{h->0} (2cos((x+h+x)/2)sin((x+h-x)/2))/h=lim_{h->0} (2cos(x+h/2)sin(h/2))/h, and do the same thing for cosine
First I was scared 😬💀 f d channel but it was very useful
Did your school tell you that you can't use their classroom white board for videos any longer?
Snarky Mark I live 42 miles away from my school.
Ouch, quite the commute.
Very curious at 3:33 . . .
Aww, why bother with the clumsy step of
multiplying alpha - beta = B by negative one at all?
Come on, simply SUBTRACT the whole thing from alpha + beta = A,
and you IMMEDIATELY get 2beta = A - B
Also, at 4:30 . . .
No parentheses will be necessary for single-variable arguments in trigonometric functions, thus it is perfectly ok to write sinA + sinB rather than the, again very clumsy, sin(A) + sin(B) . . . especially that you were already writing in black and red ^_^
Finally, are you also on Facebook? I'd love to join you if you happen to be there!
But be clear with video clarity it's somehow blurrrr
For those who looking for video link mentioned in the video i.e formula for Sum of angles
here it is: ua-cam.com/video/2SlvKnlVx7U/v-deo.html
Does there exists something like that for cosine
Yes! Use the sum and difference formulas for cosine and you can get the results.
The point P ≒ P
I wonder who thumbs down
e^xcosz-(1/3)e^(3x)cos(3z)+(1/5)e^(5x)cos(5z)-....
Please help me with this series.I am asked to find the infinite sum of this series.Got this from a complex variable book. :/
Link for those pens please lmao
超
solve pls sin(3x)/cos(x)=39/41
老哥，听不懂啊
Can you solve
Z^3 - 4j = 0
What did you ment by w and j? Are they random variables?if they are so you need at lest one more equation to solve them