Mixing
Solve: $arctan(2x)+arctan(3x) = frac{pi}{4}$
$begingroup$
Solve: $arctan(2x)+arctan(3x) = frac{pi}{4}$
I started by applying tan on both sides,
$frac{2x+3x}{1-2xtimes3x}=tanfrac{pi}{4}$
This yields $x=frac16,$ $x=-1$
But -1 doesn't satisfy the equation! Can you help me identify my mistake? Thanks.
P.S. I saw some similar questions posted on this same problem. But it didn't address how to correctly solve this equation.
trigonometry inverse-function
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
asked Jan 15 at 4:10
emilemil
431410
$endgroup$
add a comment |
$begingroup$
Solve: $arctan(2x)+arctan(3x) = frac{pi}{4}$
I started by applying tan on both sides,
$frac{2x+3x}{1-2xtimes3x}=tanfrac{pi}{4}$
This yields $x=frac16,$ $x=-1$
But -1 doesn't satisfy the equation! Can you help me identify my mistake? Thanks.
P.S. I saw some similar questions posted on this same problem. But it didn't address how to correctly solve this equation.
trigonometry inverse-function
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
asked Jan 15 at 4:10
emilemil
431410
$endgroup$
$begingroup$
Use math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 15 at 4:13
add a comment |
$begingroup$
Solve: $arctan(2x)+arctan(3x) = frac{pi}{4}$
I started by applying tan on both sides,
$frac{2x+3x}{1-2xtimes3x}=tanfrac{pi}{4}$
This yields $x=frac16,$ $x=-1$
But -1 doesn't satisfy the equation! Can you help me identify my mistake? Thanks.
P.S. I saw some similar questions posted on this same problem. But it didn't address how to correctly solve this equation.
trigonometry inverse-function
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
asked Jan 15 at 4:10
emilemil
431410
$endgroup$
Solve: $arctan(2x)+arctan(3x) = frac{pi}{4}$
I started by applying tan on both sides,
$frac{2x+3x}{1-2xtimes3x}=tanfrac{pi}{4}$
This yields $x=frac16,$ $x=-1$
But -1 doesn't satisfy the equation! Can you help me identify my mistake? Thanks.
P.S. I saw some similar questions posted on this same problem. But it didn't address how to correctly solve this equation.
trigonometry inverse-function
trigonometry inverse-function
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
asked Jan 15 at 4:10
emilemil
431410
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
asked Jan 15 at 4:10
emilemil
431410
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
edited Jan 15 at 19:31
Michael Rozenberg
104k1891196
104k1891196
asked Jan 15 at 4:10
emilemil
431410
asked Jan 15 at 4:10
emilemil
431410
asked Jan 15 at 4:10
emilemil
431410
431410
$begingroup$
Use math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 15 at 4:13
add a comment |
$begingroup$
Use math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 15 at 4:13
$begingroup$
Use math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 15 at 4:13
$begingroup$
Use math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 15 at 4:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$-1$ is not valid because $arctan(-2)+arctan(-3)=-135^{circ}neq45^{circ}.$
We got this value because $tan45^{circ}=tan(-135^{circ})$
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
add a comment |
$begingroup$
Your argument only tells you that if $x$ is a solution of the equation then $x$ must be one of the two values you have obtained. It doesn't not follow that both the values actually satisfy the original equation. (If you try to retrace your steps you will have to use that fact $tan$ is a one-to-one function, but this is not true). After finding possible values of $x$ you should go back to the original equation and keep only those values that actually satisfy it.
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
Adding angles corresponds to multiplying complex numbers.
$arctan(theta)$ is the angle (argument) of the complex number $1+theta i.$
Then
$(1+2xi)(1+3xi) = (1-6x^2)+ 5xi$
is the complex number with angle
$arctan(2x)+arctan(3x)$.
Therefore we want to find $x$ such that $(1-6x^2)+5xi$ is a complex number with angle $pi / 4$. In other words we want it to be a real multiple of $1+i$, or equivalently a complex number with equal real and imaginary part. However, not all such numbers have angle $pi/4$ so we can obtain false solutions.
This leads to the same equation that you have, although now it may be easier to see that for $x < 0$, you are adding angles in the clockwise direction, i.e. your angles are negative. Therefore $x = -1$ gives a false solution, and corresponds to the complex number $-5-5i$ that has angle
$arctan(-2)+arctan(-3) = -3pi/4$.
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$-1$ is not valid because $arctan(-2)+arctan(-3)=-135^{circ}neq45^{circ}.$
We got this value because $tan45^{circ}=tan(-135^{circ})$
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
add a comment |
$begingroup$
$-1$ is not valid because $arctan(-2)+arctan(-3)=-135^{circ}neq45^{circ}.$
We got this value because $tan45^{circ}=tan(-135^{circ})$
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
add a comment |
$begingroup$
$-1$ is not valid because $arctan(-2)+arctan(-3)=-135^{circ}neq45^{circ}.$
We got this value because $tan45^{circ}=tan(-135^{circ})$
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$-1$ is not valid because $arctan(-2)+arctan(-3)=-135^{circ}neq45^{circ}.$
We got this value because $tan45^{circ}=tan(-135^{circ})$
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 6:09
Michael RozenbergMichael Rozenberg
104k1891196
104k1891196
add a comment |
add a comment |
$begingroup$
Your argument only tells you that if $x$ is a solution of the equation then $x$ must be one of the two values you have obtained. It doesn't not follow that both the values actually satisfy the original equation. (If you try to retrace your steps you will have to use that fact $tan$ is a one-to-one function, but this is not true). After finding possible values of $x$ you should go back to the original equation and keep only those values that actually satisfy it.
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
Your argument only tells you that if $x$ is a solution of the equation then $x$ must be one of the two values you have obtained. It doesn't not follow that both the values actually satisfy the original equation. (If you try to retrace your steps you will have to use that fact $tan$ is a one-to-one function, but this is not true). After finding possible values of $x$ you should go back to the original equation and keep only those values that actually satisfy it.
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
Your argument only tells you that if $x$ is a solution of the equation then $x$ must be one of the two values you have obtained. It doesn't not follow that both the values actually satisfy the original equation. (If you try to retrace your steps you will have to use that fact $tan$ is a one-to-one function, but this is not true). After finding possible values of $x$ you should go back to the original equation and keep only those values that actually satisfy it.
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
Your argument only tells you that if $x$ is a solution of the equation then $x$ must be one of the two values you have obtained. It doesn't not follow that both the values actually satisfy the original equation. (If you try to retrace your steps you will have to use that fact $tan$ is a one-to-one function, but this is not true). After finding possible values of $x$ you should go back to the original equation and keep only those values that actually satisfy it.
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Jan 15 at 6:07
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
add a comment |
add a comment |
$begingroup$
Adding angles corresponds to multiplying complex numbers.
$arctan(theta)$ is the angle (argument) of the complex number $1+theta i.$
Then
$(1+2xi)(1+3xi) = (1-6x^2)+ 5xi$
is the complex number with angle
$arctan(2x)+arctan(3x)$.
Therefore we want to find $x$ such that $(1-6x^2)+5xi$ is a complex number with angle $pi / 4$. In other words we want it to be a real multiple of $1+i$, or equivalently a complex number with equal real and imaginary part. However, not all such numbers have angle $pi/4$ so we can obtain false solutions.
This leads to the same equation that you have, although now it may be easier to see that for $x < 0$, you are adding angles in the clockwise direction, i.e. your angles are negative. Therefore $x = -1$ gives a false solution, and corresponds to the complex number $-5-5i$ that has angle
$arctan(-2)+arctan(-3) = -3pi/4$.
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
$endgroup$
add a comment |
$begingroup$
Adding angles corresponds to multiplying complex numbers.
$arctan(theta)$ is the angle (argument) of the complex number $1+theta i.$
Then
$(1+2xi)(1+3xi) = (1-6x^2)+ 5xi$
is the complex number with angle
$arctan(2x)+arctan(3x)$.
Therefore we want to find $x$ such that $(1-6x^2)+5xi$ is a complex number with angle $pi / 4$. In other words we want it to be a real multiple of $1+i$, or equivalently a complex number with equal real and imaginary part. However, not all such numbers have angle $pi/4$ so we can obtain false solutions.
This leads to the same equation that you have, although now it may be easier to see that for $x < 0$, you are adding angles in the clockwise direction, i.e. your angles are negative. Therefore $x = -1$ gives a false solution, and corresponds to the complex number $-5-5i$ that has angle
$arctan(-2)+arctan(-3) = -3pi/4$.
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
$endgroup$
add a comment |
$begingroup$
Adding angles corresponds to multiplying complex numbers.
$arctan(theta)$ is the angle (argument) of the complex number $1+theta i.$
Then
$(1+2xi)(1+3xi) = (1-6x^2)+ 5xi$
is the complex number with angle
$arctan(2x)+arctan(3x)$.
Therefore we want to find $x$ such that $(1-6x^2)+5xi$ is a complex number with angle $pi / 4$. In other words we want it to be a real multiple of $1+i$, or equivalently a complex number with equal real and imaginary part. However, not all such numbers have angle $pi/4$ so we can obtain false solutions.
This leads to the same equation that you have, although now it may be easier to see that for $x < 0$, you are adding angles in the clockwise direction, i.e. your angles are negative. Therefore $x = -1$ gives a false solution, and corresponds to the complex number $-5-5i$ that has angle
$arctan(-2)+arctan(-3) = -3pi/4$.
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
$endgroup$
Adding angles corresponds to multiplying complex numbers.
$arctan(theta)$ is the angle (argument) of the complex number $1+theta i.$
Then
$(1+2xi)(1+3xi) = (1-6x^2)+ 5xi$
is the complex number with angle
$arctan(2x)+arctan(3x)$.
Therefore we want to find $x$ such that $(1-6x^2)+5xi$ is a complex number with angle $pi / 4$. In other words we want it to be a real multiple of $1+i$, or equivalently a complex number with equal real and imaginary part. However, not all such numbers have angle $pi/4$ so we can obtain false solutions.
This leads to the same equation that you have, although now it may be easier to see that for $x < 0$, you are adding angles in the clockwise direction, i.e. your angles are negative. Therefore $x = -1$ gives a false solution, and corresponds to the complex number $-5-5i$ that has angle
$arctan(-2)+arctan(-3) = -3pi/4$.
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
edited Jan 15 at 6:21
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
answered Jan 15 at 5:44
Gunnar SveinssonGunnar Sveinsson
6115
6115
add a comment |
add a comment |
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$begingroup$
Use math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 15 at 4:13