# Isolating the Y variable in a Trigonometric Equation

Gold Member

## Homework Statement

$4y=cos\left(4πx+\frac{3}{2}\right)$

## 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.

Related Precalculus Mathematics Homework Help News on Phys.org
Gold Member
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?

ehild
Homework Helper

## Homework Statement

$4y=cos\left(4πx+\frac{3}{2}\right)$

## 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?

• opus
Gold Member
Makes complete sense now. Thank you!

Mark44
Mentor
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.

• opus
Gold Member
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?

Ray Vickson
Homework Helper
Dearly Missed
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:
• opus
Gold Member
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.

Mark44
Mentor
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.

• opus
Gold Member
Thanks Mark!