Is it legal to divide an equation by a trigonometric expression which may equal zero?
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I have the following function: $f(x) = cos(4x) + sin(4x)$ in the range of $0 < x < frac{pi}{2}$.
Obviously as you can see this function is defined for every $x$. Let's say I want to find the extremum points for this function. I derive it, and attempt to compare it to zero to find possible extrema:
$$f'(x) = 4cos(4x) - 4sin(4x) = 0$$
$$cos(4x) = sin(4x)$$
My question — am I allowed to do the following:
$$sin(4x) = cos(4x) quad/div cos(4x)$$
$$tan(4x) = 1$$
From here I can solve for $x$ and get possible extremum points. But notice that it is possible for the divisor to be equal to zero within the provided range, and both the function and derivative are still defined for those points. Is this illegal? Should I instead use trigonometric identities to break down these sort of problems? Alternatively, am I allowed to do this granted I provide a certain explanation?
real-analysis calculus trigonometry
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
$endgroup$
add a comment |
$begingroup$
I have the following function: $f(x) = cos(4x) + sin(4x)$ in the range of $0 < x < frac{pi}{2}$.
Obviously as you can see this function is defined for every $x$. Let's say I want to find the extremum points for this function. I derive it, and attempt to compare it to zero to find possible extrema:
$$f'(x) = 4cos(4x) - 4sin(4x) = 0$$
$$cos(4x) = sin(4x)$$
My question — am I allowed to do the following:
$$sin(4x) = cos(4x) quad/div cos(4x)$$
$$tan(4x) = 1$$
From here I can solve for $x$ and get possible extremum points. But notice that it is possible for the divisor to be equal to zero within the provided range, and both the function and derivative are still defined for those points. Is this illegal? Should I instead use trigonometric identities to break down these sort of problems? Alternatively, am I allowed to do this granted I provide a certain explanation?
real-analysis calculus trigonometry
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
$endgroup$
2
$begingroup$
You can always treat $cos(4x)=0$ as a separate case.
$endgroup$
– EuxhenH
Jan 8 at 22:21
2
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It is not legal in Kansas.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 22:24
1
$begingroup$
If $cos (4x) = 0$, then $|sin (4x)| = 1$ and so the equation cannot hold. Hence you may safely divide by $cos(4x)$. (Even if Dorothy disagrees.)
$endgroup$
– copper.hat
Jan 8 at 22:51
add a comment |
$begingroup$
I have the following function: $f(x) = cos(4x) + sin(4x)$ in the range of $0 < x < frac{pi}{2}$.
Obviously as you can see this function is defined for every $x$. Let's say I want to find the extremum points for this function. I derive it, and attempt to compare it to zero to find possible extrema:
$$f'(x) = 4cos(4x) - 4sin(4x) = 0$$
$$cos(4x) = sin(4x)$$
My question — am I allowed to do the following:
$$sin(4x) = cos(4x) quad/div cos(4x)$$
$$tan(4x) = 1$$
From here I can solve for $x$ and get possible extremum points. But notice that it is possible for the divisor to be equal to zero within the provided range, and both the function and derivative are still defined for those points. Is this illegal? Should I instead use trigonometric identities to break down these sort of problems? Alternatively, am I allowed to do this granted I provide a certain explanation?
real-analysis calculus trigonometry
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
$endgroup$
I have the following function: $f(x) = cos(4x) + sin(4x)$ in the range of $0 < x < frac{pi}{2}$.
Obviously as you can see this function is defined for every $x$. Let's say I want to find the extremum points for this function. I derive it, and attempt to compare it to zero to find possible extrema:
$$f'(x) = 4cos(4x) - 4sin(4x) = 0$$
$$cos(4x) = sin(4x)$$
My question — am I allowed to do the following:
$$sin(4x) = cos(4x) quad/div cos(4x)$$
$$tan(4x) = 1$$
From here I can solve for $x$ and get possible extremum points. But notice that it is possible for the divisor to be equal to zero within the provided range, and both the function and derivative are still defined for those points. Is this illegal? Should I instead use trigonometric identities to break down these sort of problems? Alternatively, am I allowed to do this granted I provide a certain explanation?
real-analysis calculus trigonometry
real-analysis calculus trigonometry
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
edited Jan 8 at 22:16
daedsidog
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
asked Jan 8 at 22:15
daedsidogdaedsidog
30417
30417
2
$begingroup$
You can always treat $cos(4x)=0$ as a separate case.
$endgroup$
– EuxhenH
Jan 8 at 22:21
2
$begingroup$
It is not legal in Kansas.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 22:24
1
$begingroup$
If $cos (4x) = 0$, then $|sin (4x)| = 1$ and so the equation cannot hold. Hence you may safely divide by $cos(4x)$. (Even if Dorothy disagrees.)
$endgroup$
– copper.hat
Jan 8 at 22:51
add a comment |
2
$begingroup$
You can always treat $cos(4x)=0$ as a separate case.
$endgroup$
– EuxhenH
Jan 8 at 22:21
2
$begingroup$
It is not legal in Kansas.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 22:24
1
$begingroup$
If $cos (4x) = 0$, then $|sin (4x)| = 1$ and so the equation cannot hold. Hence you may safely divide by $cos(4x)$. (Even if Dorothy disagrees.)
$endgroup$
– copper.hat
Jan 8 at 22:51
2
2
$begingroup$
You can always treat $cos(4x)=0$ as a separate case.
$endgroup$
– EuxhenH
Jan 8 at 22:21
$begingroup$
You can always treat $cos(4x)=0$ as a separate case.
$endgroup$
– EuxhenH
Jan 8 at 22:21
2
2
$begingroup$
It is not legal in Kansas.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 22:24
$begingroup$
It is not legal in Kansas.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 22:24
1
1
$begingroup$
If $cos (4x) = 0$, then $|sin (4x)| = 1$ and so the equation cannot hold. Hence you may safely divide by $cos(4x)$. (Even if Dorothy disagrees.)
$endgroup$
– copper.hat
Jan 8 at 22:51
$begingroup$
If $cos (4x) = 0$, then $|sin (4x)| = 1$ and so the equation cannot hold. Hence you may safely divide by $cos(4x)$. (Even if Dorothy disagrees.)
$endgroup$
– copper.hat
Jan 8 at 22:51
add a comment |
1 Answer
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Yes you can, because $$cos(4x)=0$$ cannot correspond to a solution (this would imply $sin(4x)=pm1ne0$).
The usual way to handle such situations is to perform a case study:
$cos(4x)=0to$ impossible;
$cos(4x)ne0totan(4x)=1$.
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
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add a comment |
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$begingroup$
Yes you can, because $$cos(4x)=0$$ cannot correspond to a solution (this would imply $sin(4x)=pm1ne0$).
The usual way to handle such situations is to perform a case study:
$cos(4x)=0to$ impossible;
$cos(4x)ne0totan(4x)=1$.
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
Yes you can, because $$cos(4x)=0$$ cannot correspond to a solution (this would imply $sin(4x)=pm1ne0$).
The usual way to handle such situations is to perform a case study:
$cos(4x)=0to$ impossible;
$cos(4x)ne0totan(4x)=1$.
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
Yes you can, because $$cos(4x)=0$$ cannot correspond to a solution (this would imply $sin(4x)=pm1ne0$).
The usual way to handle such situations is to perform a case study:
$cos(4x)=0to$ impossible;
$cos(4x)ne0totan(4x)=1$.
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
$endgroup$
Yes you can, because $$cos(4x)=0$$ cannot correspond to a solution (this would imply $sin(4x)=pm1ne0$).
The usual way to handle such situations is to perform a case study:
$cos(4x)=0to$ impossible;
$cos(4x)ne0totan(4x)=1$.
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
answered Jan 8 at 22:21
Yves DaoustYves Daoust
132k676229
132k676229
add a comment |
add a comment |
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2
$begingroup$
You can always treat $cos(4x)=0$ as a separate case.
$endgroup$
– EuxhenH
Jan 8 at 22:21
2
$begingroup$
It is not legal in Kansas.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 22:24
1
$begingroup$
If $cos (4x) = 0$, then $|sin (4x)| = 1$ and so the equation cannot hold. Hence you may safely divide by $cos(4x)$. (Even if Dorothy disagrees.)
$endgroup$
– copper.hat
Jan 8 at 22:51