Sir now Ill completely forget memorizing these formula Because from Now on I can derive these formula anytime , anywhere Long Live BlackPenRedPen Yeeay!!!
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.
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!
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) ] ?
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!
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
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. :/
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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!
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.
this is probably the best teacher I have in terms of all the trigonometic identities. Very simple and consice! Love it!
I am so happy you put this up 2 YEARS AGO!!! thank you so much, never investigated these relationships
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?
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!
I have been looking for a proof just like this and found it. Thank you so much!
This proof videos are my favorite, thanks! Love u ❤
That demostrative videos are amazing!!!
Thank you!!
Thank you! I have been find the proofs for this, since the precalculus lesson I attended didn't prove this for us
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.
THIS IS SO FUN! please keep on proving stuff and do more videos with this OG style!!
San Samman OK!!!!!!
You make me so happy now, thanks a lot!
After 11 months, still helpful!
Wanted to say Thank you! Learnt a lot from you till date :)
Pretty easy identity to prove but still a useful one
Thank you very much!!! It becomes so much easier to memorise now that I know how it works, video very much appreciated!
beautiful magnificently explained.... 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
That's just great! 😯
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) ]
?
Great video, thx.
Really helped me out!
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 th-cam.com/video/2SlvKnlVx7U/w-d-xo.html
this helped me understand! thank you
thank you so much!!!
I like the video so much.
Thank you!!!
Cool job!!!
trigonometric proofs are beautiful
bless ur soul
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!
Can you racionalize 1/[cuberoot(a)+cuberoot(b)+cuberoot(c)] please? Love your videos
amazing
can you derive this formula using euler's formula ? (without subtituting alpha+beta = A and alpha-beta = B ?)
thank you so much
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
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
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. :/
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.
thanks brother proof of sum to product identities is not in my book for some reason
For those who looking for video link mentioned in the video i.e formula for Sum of angles
here it is: th-cam.com/video/2SlvKnlVx7U/w-d-xo.html
Does there exists something like that for cosine
Yes! Use the sum and difference formulas for cosine and you can get the results.
But be clear with video clarity it's somehow blurrrr
谢谢
I wonder who thumbs down
Link for those pens please lmao
First I was scared 😬💀 f d channel but it was very useful
The point P ≒ P
solve pls sin(3x)/cos(x)=39/41
老哥，听不懂啊
超
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