Created an equation in trig class today

Created an equation in trig class today.

Prove it wrong.

>You cant

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wolframalpha.com/input/?i=sin(ax) + sin(bx) = 0
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>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/input/?i=sin(ax) + sin(bx) = 0

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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