When does y = sin⁡ (x + a) and y = sin⁡ (x + b) have identical graphs?

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Aion

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In what way can the function [MATH]y = sin⁡ (x + a)[/MATH] and [MATH]y = sin⁡ (x + b)[/MATH] have identical graphs even if a ≠ b?
 
LCKurtz said:
What do you think the answer is, and why?
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I think its possible since you are only laterally displacing the sine graph. I've tried to solve the equation but I think its beyond me

[MATH]Sin(x+a)=Sin(x+b)[/MATH] Let [MATH]\alpha = x+a[/MATH] and [MATH]\beta = x+b[/MATH][MATH]Sin(\alpha)=Sin(\beta)[/MATH][MATH]\alpha_{1}=\beta+n2\pi[/MATH][MATH]\alpha_{2}=\pi - \beta +n2\pi[/MATH][MATH]x+a_1=x+b+n2\pi[/MATH][MATH]x+a_2=\pi -(x+b)+n2\pi[/MATH][MATH]a_1=b+n2\pi[/MATH][MATH]a_2=\pi-2x-b+n2\pi[/MATH]
 
A more relevant one would be the periodicity of the sine: [MATH]\sin(x + 2\pi) = \sin(x)[/MATH].
 
[MATH]y_1=Sin(x+a)=Sin(x)Cos(a)+Sin(a)Cos(x)[/MATH][MATH]y_2=Sin(x+b)=Sin(x)Cos(b)+Sin(b)Cos(x)[/MATH]
We can see that these functions are identical if
[MATH]Sin(a)=Sin(b)[/MATH] and
[MATH]Cos(a)=Cos(b)[/MATH], if we divide the first equation by the second we obtain
[MATH]Tan(a)=Tan(b)[/MATH][MATH]a=b+n\pi[/MATH]
Is this the solution? I'm guessing there are several solutions.
 
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Aion said:
[MATH]y_1=Sin(x+a)=Sin(x)Cos(a)+Sin(a)Cos(x)[/MATH][MATH]y_2=Sin(x+b)=Sin(x)Cos(b)+Sin(b)Cos(x)[/MATH]
We can see that these functions are identical if
[MATH]Sin(a)=Sin(b)[/MATH] and
[MATH]Cos(a)=Cos(b)[/MATH], if we divide the first equation by the second we obtain
[MATH]Tan(a)=Tan(b)[/MATH][MATH]a=b+n\pi[/MATH]
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Not quite. tan(x) is has a period of [math]\pi[/math] but sin(x) has a period of [math]2 \pi[/math]. The correct solution is [math]a = b + 2 \pi n[/math].

-Dan
 
topsquark said:
Not quite. tan(x) is has a period of [math]\pi[/math] but sin(x) has a period of [math]2 \pi[/math]. The correct solution is [math]a = b + 2 \pi n[/math].

-Dan
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Is that the only solution? I need to cover all solutions:d
 
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Aion said:
Is that the only solution? I need to cover all solutions:d
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That's not one solution; it's infinitely many. (n can be any integer.)

Tell us what you're thinking, please. Is there a reason you think that might not be the complete answer?
 
Dr.Peterson said:
That's not one solution; it's infinitely many. (n can be any integer.)

Tell us what you're thinking, please. Is there a reason you think that might not be the complete answer?
Click to expand...

Well no I don't know. I've been told that a sinefunction generally has two sets of solutions. If [MATH]Sin(x)=k[/MATH] where [MATH]-1\leq k \leq1[/MATH] then the solutions for x is given by the following

[MATH]x_1=arcsin(k)+n2\pi[/MATH][MATH]x_2=\pi-arcsin(k)+n2\pi[/MATH]
 
Aion said:
Well no I don't know. I've been told that a sinefunction generally has two sets of solutions. If [MATH]Sin(x)=k[/MATH] where [MATH]-1\leq k \leq1[/MATH] then the solutions for x is given by the following

[MATH]x_1=arcsin(k)+n2\pi[/MATH][MATH]x_2=\pi-arcsin(k)+n2\pi[/MATH]
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But sin(x) = k isn't the equation you are trying to solve!

-Dan
 
If the two angles differ by 2pi or a multiply of 2pi then the sine of those two angles would b equal.

For example sin(30) = sin(30 + 2pi) = sin(30 + 4pi) = sin(30 + 6pi) = ... = sin(30 - 2pi) = sin(30 - 4pi) = sin(30-6pi)...

So we want (x-a) - (x-b) = b-a = 2kpi where k is any integer other than 0. Why can't k=0? Solving b-a=2kpi or b yields that b= a+ 2kpi.
 
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