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Isolating the Y variable in a Trigonometric Equation

  • Thread starter opus
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  • #1
opus
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Homework Statement


##4y=cos\left(4πx+\frac{3}{2}\right)##

Homework Equations




The Attempt at a Solution


In dividing both sides by 4, I got:

##y=\frac{1}{4}cos\left(πx+\frac{3}{8}\right)## But I am told this is incorrect.
Not sure if dividing everything by 4 here is an allowable technique, or if I did it wrong, but I'm not sure where else to go other than isolating the y like this.
 

Answers and Replies

  • #2
opus
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Actually I think I got it. I tried dividing the LHS by 4 and ONLY the coefficient of cos on the RHS and that seemed to work. My algebra seems to go out the window when I see trig words, so why wouldn't I divide all factors on the RHS by 4?
 
  • #3
ehild
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Homework Statement


##4y=cos\left(4πx+\frac{3}{2}\right)##

Homework Equations




The Attempt at a Solution


In dividing both sides by 4, I got:

##y=\frac{1}{4}cos\left(πx+\frac{3}{8}\right)## But I am told this is incorrect.
Not sure if dividing everything by 4 here is an allowable technique, or if I did it wrong, but I'm not sure where else to go other than isolating the y like this.
No, it is wrong. When dividing a product, you divide only one factor.
Think: what is y if 4y = 12? and what is y if 4y=3x4? Do you really think the two y values are different?
 
  • #4
opus
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Makes complete sense now. Thank you!
 
  • #5
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When dividing a product, you divide only one factor.
True, but ##\cos(4\pi x + \frac 3 2)## isn't a product.
Closer to the point, if ##4y = \sqrt{36}##, it would be incorrect to write ##y = \frac 1 4 \sqrt 9## (which happens to be ##\frac 3 4##). The corrected equation is ##y = \frac 1 4 \sqrt{36} = \frac 6 4 = \frac 3 2##.

Of course, with this equation, it would have been simpler to write it as 4y = 6, from which we get y = 3/2.
 
  • #6
opus
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Thank you Mark. So what would we call it if not a product? Just a typical function such as f(x) where we wouldnt be dividing the input of the function?
 
  • #7
Ray Vickson
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Actually I think I got it. I tried dividing the LHS by 4 and ONLY the coefficient of cos on the RHS and that seemed to work. My algebra seems to go out the window when I see trig words, so why wouldn't I divide all factors on the RHS by 4?
If the trig function is throwing you off, just call it something else (temporarily). If you let ##C = \cos(4 \pi x + \frac 3 2)##, your original equation is ##4y = C,## and the solution is ##y = \frac1 4 C.## Now put back the value of ##C## in terms of ##x##. When you do that it becomes absolutely clear that you do not pull the ##\frac{1}{4}## inside the ##\cos( \cdot) ## function.
 
Last edited:
  • #8
opus
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If the trig function is throwing you off, just call it something else (temporarily). If you let ##C = \cos(4 \pi x + \frac 3 2)##, your original equation is ##4y = C,## and the solution is ##y = \frac1 4 C.##. Now put back the value of ##C## in terms of ##x##. When you do that it becomes absolutely clear that you do not pull the ##\frac{1}{4}## inside the ##\cos( \cdot) ## function.
Im going to try that! Thats a great idea, thank you.
 
  • #9
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So what would we call it if not a product? Just a typical function such as f(x) where we wouldnt be dividing the input of the function?
Just call it a function. The expression "f(x)" does not mean "f times x", so it's definitely not a product. It's almost never the case that ##\frac 1 b f(x)## is equal to ##f(\frac 1 b x)##, so you generally can't "distribute" the factor ##\frac 1 b## to into the function.

Here are some example of what I mean:
##\frac 1 2 \sin(x) \ne \sin(\frac x 2)##
##\frac 1 3 \sqrt{x} \ne \sqrt{\frac x 3}##
##2e^x \ne e^{2x}##
and so on. In the above, ##\ne## should be interpreted as "not identically equal to." For some of my examples, there might be values of x that make the expressions on each side equal, but for most values of x, the paired expressions aren't equal.
 
  • #10
opus
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Thanks Mark!
 

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