Created an equation in trig class today.
Prove it wrong.
>You cant
Created an equation in trig class today.
Prove it wrong.
>You cant
Other urls found in this thread:
wolframalpha.com
twitter.com
>Assumes radians
First, tell us why this isn't trivial.
Please pick out this equation from a list of trig identites.
>it isnt there
New trig identity
Explain what is n and a and b and your stupid equation should be simplified.
Bring the 2 up
Its official.
I am the next Isaac Newton
It follows from the sum of two sines expressed as a product of a sine and a cosine, which is listed there.
Only you get half the roots wrong. It should be a+b n the denominator for even n but a-b for odd n.
Have you ever studied trig?
N represents the number of half rotations around a circle, in radians (pi = half-circle)
>explain a and b
sin (ax) + sin (bx) = 0
a=55 b =67
sin 55x + sin 67x = 0
Or actually, I missed the factor of 1/2 in the denominator. The roots are correct as they are but that's only half of them. There's also the roots from the cosine term which are odd multiples of pi/(a-b)
what if a=0.999... and b=-1
Nope, Wolfram alpha has it.
You probably took it from them.
All you did was divide by two the numerator and denominator, you fuckretard cuckbastard.
>Junior high general
Works for both even and odd n
>Wolfram has it
Proof
see
wolframalpha.com
Here you retard.
lololololololol
>user asked wolfram alpha to solve my equation
>It proved I was right.
Thanks, user
For the record, wolfram does not have this equation.
It will only solve it on the equation calculator. User must first type in my first sinusoidal equation.
Wolfram solves the equation with the same answer I did.
>Proof Im right.
Veeky Forumsfags just got owned by a /pol/ack
sin(ax) + sin(bx) = 0
sin(ax + 2pi*n) = - sin(bx), sin(ax + 2pi*n) = sin(-bx) = sin(pi + bx)
ax + 2pi*n = pi + bx
(2n-1)*pi = (b-a)x
(2n - 1)*pi/(b-a) = x.
>(2n - 1)*pi/(b-a) = x.
This equation is accurate, although -1 is uneeded
>>(2n - 2031 + 777)*pi/(b-a) = x.
This is still true. But the -2031 and +777 are uneccesary...
>(2n*pi)/(b-a) == (n*pi)/2(b-a)
This equation holds true for all n's. adding or subtracting 1 doesnt make a difference
this shit boggles my mind
why does it work for arbitrary reals?
Why don't you show us the proof, nigger
Is sci legitimately retarded? This is grade school shit. Jesus christ just kill yourselves already
Fermat couldnt prove his final theorem, i cant prove this shit, I only know it works
>this shit boggles my mind
>why does it work for arbitrary reals?
Because I am the next Isaac Newton.
>/thread
>x=(1+2n)pi/(a+b)
Your 1 is uneccesary. It does absolutely nothing in this case. you may as well have x=(1000+2n)pi/(a+b)
>x=2n*pi/(a+b)
is the same as
>x=n*pi/[(a+b)*2]
This is retarded.
>This is retarded
Your just mad its 100% accurate
>All real solutions may be found by using the following equation
Demonstrably false.
Take the equation sin(5x) + sin(x).
You claim all real solutions are found with the equation x = n * pi / 3
Based on your assertion, there should be only four real solutions on the accompanying graph: 0, pi/3, 2pi/3, and pi. However, as you can plainly see, there are -six- real solutions. Your equation has neglected two of them.
The other equation required in order to obtain all real solutions is x = pi*(1+2n) / 4
As stated, these two equations are obtained by using the sum identity for two sines and solving for when sin( x+y /2) and cos( x-y /2) equal 0. The first one solves to the equation in your picture. The second one solves to x = pi(1+2n) / (a-b).
Maybe, but no one is more autistically driven to bite the bait like a /v/irgin.
Can't you use a taylor series to prove it wrong ?
>owned
You have to be eighteen to use this board
So what you're saying is that you stole this proof?
simple:
sin(ax) + sin(bx) = 0
sin(ax) = -sin(bx)
a*x = b*x + n*(2*k + 1)*pi with k in Z.
-> x = n*(2k+1)*pi/(a-b)
yes he had proof it was just to large to write in the margin
>Prove it wrong
a=0, b=0
You fucking retard.
BAHAHAHA
omg i can't breath
LOOOOOOOOOOL KEKEKEKEK
OP BTFO
Easy. Let n = 0. 0/0 is any number you want
>samefag brainlet btfo once again
oh shit. what are you doing posting this on Veeky Forums? publish it in a journal and claim your small monetary prize
sin(ax) + sin(bx) = 0
sin(ax) = -sin(bx)
sin(ax) = sin(-bx)
ax = -bx or (a + b)x = 0
If a = -b, any x in R works.
If a != -b, and WLOG a is 0, then solution is all x such that = (pi * k)/a for some integer k.
If a != -b, and neither are 0, we must have x = (pi * k)/(a + b) for some integer k.
sorry, for the last condition, the numerator should read 2*pi*k since ax = -bx modulo 2*pi
>0/0 is any number you want
0/0 is fucking undefined wtf
0/0=0 faggot
Maybe in Bizarromathland, but not in this Earth
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