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Help: trigonometric equation

by physicsRookie
Tags: equation, trigonometric
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physicsRookie
#1
Jan19-06, 05:16 PM
P: 5
[tex]x_1(cos\alpha-1) + x_2sin\alpha = 0 [/tex]
[tex]x_1sin\alpha + x_2(-cos\alpha-1) = 0 [/tex]
How to solve this equation? Can anyone help me?
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cepheid
#2
Jan20-06, 03:29 AM
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PF Gold
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It's a system of equations: 2 equations in 2 unknowns. That means you can solve it. Just solve for one unknown in terms of the other using the first equation, and then subsitute that into the second.
physicsRookie
#3
Jan20-06, 05:24 AM
P: 5
Let me try...

Solve equation 1:
[tex]x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}[/tex]

Substitute it to the second:
[tex]x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0[/tex]

[tex]x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0[/tex]

[tex]2x_1sin\alpha = 0[/tex]

What is the solutions for [tex]2x_1sin\alpha = 0[/tex]?
Obviously one is [tex]x_1=0[/tex], but if [tex]sin\alpha = 0[/tex], then...

HallsofIvy
#4
Jan20-06, 05:46 AM
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Thanks
PF Gold
P: 39,682
Help: trigonometric equation

Quote Quote by physicsRookie
Let me try...

Solve equation 1:
[tex]x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}[/tex]

Substitute it to the second:
[tex]x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0[/tex]

[tex]x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0[/tex]

[tex]2x_1sin\alpha = 0[/tex]

What is the solutions for [tex]2x_1sin\alpha = 0[/tex]?
Obviously one is [tex]x_1=0[/tex], but if [tex]sin\alpha = 0[/tex], then...
Well done. The point is, of course, that [tex]\alpha[/tex] is a number (not one of the variables) so these can be solved like any pair of equations for x1 and x2.
Notice, by the way, that if [tex]sin\alpha= 0[/tex], your first step, dividing by that, would be invalid. You have to look at this case separately.
If [tex]sin\alpha= 0[/tex], then [tex]cos\alpha[/tex] is either 1 or -1.

What do your equations look like if [tex]sin\alpha= 0[/tex] and [tex]cos\alpha= 1[/tex]?

What do your equations look like if [tex]sin\alpha= 0[/tex] and [tex]cos\alpha= -1[/tex]?
physicsRookie
#5
Jan20-06, 06:36 AM
P: 5
HallsofIvy, thanks.
Quote Quote by HallsofIvy
What do your equations look like if [tex]sin\alpha= 0[/tex] and [tex]cos\alpha= 1[/tex]?
[tex]2x_1=0 and 0=0 => x_1=0, x_2 [/tex] could be any number
Quote Quote by HallsofIvy
What do your equations look like if [tex]sin\alpha= 0[/tex] and [tex]cos\alpha= -1[/tex]?
[tex] 0=0 and -2x_2=0 => x_2=0, x_1 [/tex] could be any number

I just try another solution.
Rewrite the equations:
[tex](x_1cos\alpha + x_2sin\alpha) - x_1 = 0 [/tex]
[tex](x_1sin\alpha - x_2cos\alpha) - x_2 = 0 [/tex]

Suppose [tex]x_1 = cos\frac{\alpha}{2}, x_2 = sin\frac{\alpha}{2}[/tex], then

[tex]cos\frac{\alpha}{2}cos\alpha + sin\frac{\alpha}{2}sin\alpha - cos\frac{\alpha}{2} = 0 [/tex]

[tex]cos\frac{\alpha}{2}sin\alpha - sin\frac{\alpha}{2}cos\alpha - sin\frac{\alpha}{2} = 0 [/tex]

It works!

I am wondering whether there is some general method to solve [tex]x_1, x_2[/tex] depending on [tex]\alpha[/tex] or not.


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