Solving for $x$ in $sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$
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Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$
How do I find $x$?
I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
trigonometry inverse-function
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add a comment |
$begingroup$
Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$
How do I find $x$?
I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
trigonometry inverse-function
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$begingroup$
Possibly useful - math.stackexchange.com/questions/672575/…
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– Eevee Trainer
Jan 8 at 2:50
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But I don't know x^2+y^2
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– user5722540
Jan 8 at 2:52
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I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
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– D.B.
Jan 8 at 4:06
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Online open ended answer. Deadline of 150 seconds.
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– user5722540
Jan 8 at 4:09
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Well, it takes Maple less than a second to get ${frac {sqrt {2522-1164,sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer...
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– Robert Israel
Jan 8 at 4:23
add a comment |
$begingroup$
Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$
How do I find $x$?
I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
trigonometry inverse-function
$endgroup$
Given an equation: $$sin^{-1}(2x) + sin^{-1}(3x) = frac pi 4$$
How do I find $x$?
I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
trigonometry inverse-function
trigonometry inverse-function
edited Jan 8 at 4:22
Blue
47.9k870153
47.9k870153
asked Jan 8 at 2:46
user5722540user5722540
1638
1638
$begingroup$
Possibly useful - math.stackexchange.com/questions/672575/…
$endgroup$
– Eevee Trainer
Jan 8 at 2:50
$begingroup$
But I don't know x^2+y^2
$endgroup$
– user5722540
Jan 8 at 2:52
$begingroup$
I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
$endgroup$
– D.B.
Jan 8 at 4:06
$begingroup$
Online open ended answer. Deadline of 150 seconds.
$endgroup$
– user5722540
Jan 8 at 4:09
$begingroup$
Well, it takes Maple less than a second to get ${frac {sqrt {2522-1164,sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer...
$endgroup$
– Robert Israel
Jan 8 at 4:23
add a comment |
$begingroup$
Possibly useful - math.stackexchange.com/questions/672575/…
$endgroup$
– Eevee Trainer
Jan 8 at 2:50
$begingroup$
But I don't know x^2+y^2
$endgroup$
– user5722540
Jan 8 at 2:52
$begingroup$
I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
$endgroup$
– D.B.
Jan 8 at 4:06
$begingroup$
Online open ended answer. Deadline of 150 seconds.
$endgroup$
– user5722540
Jan 8 at 4:09
$begingroup$
Well, it takes Maple less than a second to get ${frac {sqrt {2522-1164,sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer...
$endgroup$
– Robert Israel
Jan 8 at 4:23
$begingroup$
Possibly useful - math.stackexchange.com/questions/672575/…
$endgroup$
– Eevee Trainer
Jan 8 at 2:50
$begingroup$
Possibly useful - math.stackexchange.com/questions/672575/…
$endgroup$
– Eevee Trainer
Jan 8 at 2:50
$begingroup$
But I don't know x^2+y^2
$endgroup$
– user5722540
Jan 8 at 2:52
$begingroup$
But I don't know x^2+y^2
$endgroup$
– user5722540
Jan 8 at 2:52
$begingroup$
I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
$endgroup$
– D.B.
Jan 8 at 4:06
$begingroup$
I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
$endgroup$
– D.B.
Jan 8 at 4:06
$begingroup$
Online open ended answer. Deadline of 150 seconds.
$endgroup$
– user5722540
Jan 8 at 4:09
$begingroup$
Online open ended answer. Deadline of 150 seconds.
$endgroup$
– user5722540
Jan 8 at 4:09
$begingroup$
Well, it takes Maple less than a second to get ${frac {sqrt {2522-1164,sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer...
$endgroup$
– Robert Israel
Jan 8 at 4:23
$begingroup$
Well, it takes Maple less than a second to get ${frac {sqrt {2522-1164,sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer...
$endgroup$
– Robert Israel
Jan 8 at 4:23
add a comment |
6 Answers
6
active
oldest
votes
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Or this way using
- $cos(a+b) = cos a cos b - sin a sin b$
- $cos a = sqrt{1-sin^2 a}$
begin{eqnarray*}
sin^{-1}(2x) + sin^{-1}(3x) & = & frac pi 4 \
sqrt{1-4x^2}sqrt{1-9x^2} - 6x^2 & = & frac{sqrt{2}}{2} \
(1-4x^2)(1-9x^2) &=& left( frac{sqrt{2}}{2} +6x^2 right)^2 \
frac{1}{2} & = & (6sqrt{2}+13)x^2 \
end{eqnarray*}
The positive solution gives:
$$boxed{x = frac{1}{sqrt{2(6sqrt{2}+13)}}}$$
$endgroup$
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
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– user5722540
Jan 8 at 7:48
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Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
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@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
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– trancelocation
Jan 8 at 7:55
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Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
add a comment |
$begingroup$
There is a useful identity that we can use in this case:
$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$
From here we can substitute:
$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$
We are then left with:
$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$
From here, you can solve for $x$.
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$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
add a comment |
$begingroup$
Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.
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When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
add a comment |
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I'm gonna derive the general function for $arcsin x$ then go from there.
Recall that
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that
$$u=ix+sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ to see that
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
and at this point I used Wolfram|Alpha to see that
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
I will update my answer once I figure out how this result is found
Edit:
Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)+6ix^2-isqrt{(1-4x^2)(1-9x^2)}=frac{1-i}{sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)=frac1{sqrt2}$$
Which @ClaudeLeibovici showed reduced to
$$97y^2-13y+frac14=0$$
with $y=x^2$. Using the quadratic formula, we see that
$$y=frac{13+sqrt{72}}{194}$$
which reduces to
$$y=frac{13}{194}+frac{3sqrt2}{97}$$
Taking $sqrt{cdot}$ on both sides,
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
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Is there a workaround to approach solution under 150seconds?
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– user5722540
Jan 8 at 7:49
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@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
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– clathratus
Jan 8 at 17:54
add a comment |
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We need $-1le3xle1$
But if $xle0,$ the left hand side $le0$
Now $3x=sin(pi/4-arcsin(2x))$
$3sqrt2x=sqrt{1-(2x)^2}-2x$
$sqrt{1-4x^2}=x(3sqrt2+2)$
Square both sides
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add a comment |
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Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
$$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+frac{1}{4}=0$$ which is simple.
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Or this way using
- $cos(a+b) = cos a cos b - sin a sin b$
- $cos a = sqrt{1-sin^2 a}$
begin{eqnarray*}
sin^{-1}(2x) + sin^{-1}(3x) & = & frac pi 4 \
sqrt{1-4x^2}sqrt{1-9x^2} - 6x^2 & = & frac{sqrt{2}}{2} \
(1-4x^2)(1-9x^2) &=& left( frac{sqrt{2}}{2} +6x^2 right)^2 \
frac{1}{2} & = & (6sqrt{2}+13)x^2 \
end{eqnarray*}
The positive solution gives:
$$boxed{x = frac{1}{sqrt{2(6sqrt{2}+13)}}}$$
$endgroup$
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
$endgroup$
– user5722540
Jan 8 at 7:48
$begingroup$
Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
$begingroup$
@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
$endgroup$
– trancelocation
Jan 8 at 7:55
$begingroup$
Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
add a comment |
$begingroup$
Or this way using
- $cos(a+b) = cos a cos b - sin a sin b$
- $cos a = sqrt{1-sin^2 a}$
begin{eqnarray*}
sin^{-1}(2x) + sin^{-1}(3x) & = & frac pi 4 \
sqrt{1-4x^2}sqrt{1-9x^2} - 6x^2 & = & frac{sqrt{2}}{2} \
(1-4x^2)(1-9x^2) &=& left( frac{sqrt{2}}{2} +6x^2 right)^2 \
frac{1}{2} & = & (6sqrt{2}+13)x^2 \
end{eqnarray*}
The positive solution gives:
$$boxed{x = frac{1}{sqrt{2(6sqrt{2}+13)}}}$$
$endgroup$
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
$endgroup$
– user5722540
Jan 8 at 7:48
$begingroup$
Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
$begingroup$
@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
$endgroup$
– trancelocation
Jan 8 at 7:55
$begingroup$
Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
add a comment |
$begingroup$
Or this way using
- $cos(a+b) = cos a cos b - sin a sin b$
- $cos a = sqrt{1-sin^2 a}$
begin{eqnarray*}
sin^{-1}(2x) + sin^{-1}(3x) & = & frac pi 4 \
sqrt{1-4x^2}sqrt{1-9x^2} - 6x^2 & = & frac{sqrt{2}}{2} \
(1-4x^2)(1-9x^2) &=& left( frac{sqrt{2}}{2} +6x^2 right)^2 \
frac{1}{2} & = & (6sqrt{2}+13)x^2 \
end{eqnarray*}
The positive solution gives:
$$boxed{x = frac{1}{sqrt{2(6sqrt{2}+13)}}}$$
$endgroup$
Or this way using
- $cos(a+b) = cos a cos b - sin a sin b$
- $cos a = sqrt{1-sin^2 a}$
begin{eqnarray*}
sin^{-1}(2x) + sin^{-1}(3x) & = & frac pi 4 \
sqrt{1-4x^2}sqrt{1-9x^2} - 6x^2 & = & frac{sqrt{2}}{2} \
(1-4x^2)(1-9x^2) &=& left( frac{sqrt{2}}{2} +6x^2 right)^2 \
frac{1}{2} & = & (6sqrt{2}+13)x^2 \
end{eqnarray*}
The positive solution gives:
$$boxed{x = frac{1}{sqrt{2(6sqrt{2}+13)}}}$$
answered Jan 8 at 7:42
trancelocationtrancelocation
10.2k1722
10.2k1722
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
$endgroup$
– user5722540
Jan 8 at 7:48
$begingroup$
Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
$begingroup$
@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
$endgroup$
– trancelocation
Jan 8 at 7:55
$begingroup$
Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
add a comment |
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
$endgroup$
– user5722540
Jan 8 at 7:48
$begingroup$
Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
$begingroup$
@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
$endgroup$
– trancelocation
Jan 8 at 7:55
$begingroup$
Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
$endgroup$
– user5722540
Jan 8 at 7:48
$begingroup$
The answer is not matching @clathratus solution. Please confirm.
$endgroup$
– user5722540
Jan 8 at 7:48
$begingroup$
Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
$begingroup$
Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here.
$endgroup$
– trancelocation
Jan 8 at 7:50
$begingroup$
@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
$endgroup$
– trancelocation
Jan 8 at 7:55
$begingroup$
@user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed.
$endgroup$
– trancelocation
Jan 8 at 7:55
$begingroup$
Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
$begingroup$
Got that. Thanks.
$endgroup$
– user5722540
Jan 8 at 8:05
add a comment |
$begingroup$
There is a useful identity that we can use in this case:
$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$
From here we can substitute:
$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$
We are then left with:
$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$
From here, you can solve for $x$.
$endgroup$
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
add a comment |
$begingroup$
There is a useful identity that we can use in this case:
$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$
From here we can substitute:
$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$
We are then left with:
$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$
From here, you can solve for $x$.
$endgroup$
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
add a comment |
$begingroup$
There is a useful identity that we can use in this case:
$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$
From here we can substitute:
$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$
We are then left with:
$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$
From here, you can solve for $x$.
$endgroup$
There is a useful identity that we can use in this case:
$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$
From here we can substitute:
$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$
We are then left with:
$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$
From here, you can solve for $x$.
answered Jan 8 at 3:02
GnumbertesterGnumbertester
37919
37919
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
add a comment |
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:18
add a comment |
$begingroup$
Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.
$endgroup$
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
add a comment |
$begingroup$
Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.
$endgroup$
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
add a comment |
$begingroup$
Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.
$endgroup$
Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.
edited Jan 8 at 3:18
answered Jan 8 at 3:02
D.B.D.B.
1,2128
1,2128
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
add a comment |
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
$begingroup$
When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes?
$endgroup$
– user5722540
Jan 8 at 3:19
add a comment |
$begingroup$
I'm gonna derive the general function for $arcsin x$ then go from there.
Recall that
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that
$$u=ix+sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ to see that
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
and at this point I used Wolfram|Alpha to see that
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
I will update my answer once I figure out how this result is found
Edit:
Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)+6ix^2-isqrt{(1-4x^2)(1-9x^2)}=frac{1-i}{sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)=frac1{sqrt2}$$
Which @ClaudeLeibovici showed reduced to
$$97y^2-13y+frac14=0$$
with $y=x^2$. Using the quadratic formula, we see that
$$y=frac{13+sqrt{72}}{194}$$
which reduces to
$$y=frac{13}{194}+frac{3sqrt2}{97}$$
Taking $sqrt{cdot}$ on both sides,
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
$endgroup$
$begingroup$
Is there a workaround to approach solution under 150seconds?
$endgroup$
– user5722540
Jan 8 at 7:49
$begingroup$
@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
$endgroup$
– clathratus
Jan 8 at 17:54
add a comment |
$begingroup$
I'm gonna derive the general function for $arcsin x$ then go from there.
Recall that
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that
$$u=ix+sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ to see that
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
and at this point I used Wolfram|Alpha to see that
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
I will update my answer once I figure out how this result is found
Edit:
Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)+6ix^2-isqrt{(1-4x^2)(1-9x^2)}=frac{1-i}{sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)=frac1{sqrt2}$$
Which @ClaudeLeibovici showed reduced to
$$97y^2-13y+frac14=0$$
with $y=x^2$. Using the quadratic formula, we see that
$$y=frac{13+sqrt{72}}{194}$$
which reduces to
$$y=frac{13}{194}+frac{3sqrt2}{97}$$
Taking $sqrt{cdot}$ on both sides,
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
$endgroup$
$begingroup$
Is there a workaround to approach solution under 150seconds?
$endgroup$
– user5722540
Jan 8 at 7:49
$begingroup$
@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
$endgroup$
– clathratus
Jan 8 at 17:54
add a comment |
$begingroup$
I'm gonna derive the general function for $arcsin x$ then go from there.
Recall that
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that
$$u=ix+sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ to see that
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
and at this point I used Wolfram|Alpha to see that
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
I will update my answer once I figure out how this result is found
Edit:
Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)+6ix^2-isqrt{(1-4x^2)(1-9x^2)}=frac{1-i}{sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)=frac1{sqrt2}$$
Which @ClaudeLeibovici showed reduced to
$$97y^2-13y+frac14=0$$
with $y=x^2$. Using the quadratic formula, we see that
$$y=frac{13+sqrt{72}}{194}$$
which reduces to
$$y=frac{13}{194}+frac{3sqrt2}{97}$$
Taking $sqrt{cdot}$ on both sides,
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
$endgroup$
I'm gonna derive the general function for $arcsin x$ then go from there.
Recall that
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that
$$u=ix+sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ to see that
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=frac{1+i}{sqrt2}$$
and at this point I used Wolfram|Alpha to see that
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
I will update my answer once I figure out how this result is found
Edit:
Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)+6ix^2-isqrt{(1-4x^2)(1-9x^2)}=frac{1-i}{sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)=frac1{sqrt2}$$
Which @ClaudeLeibovici showed reduced to
$$97y^2-13y+frac14=0$$
with $y=x^2$. Using the quadratic formula, we see that
$$y=frac{13+sqrt{72}}{194}$$
which reduces to
$$y=frac{13}{194}+frac{3sqrt2}{97}$$
Taking $sqrt{cdot}$ on both sides,
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
edited Jan 8 at 4:48
answered Jan 8 at 3:57
clathratusclathratus
3,740333
3,740333
$begingroup$
Is there a workaround to approach solution under 150seconds?
$endgroup$
– user5722540
Jan 8 at 7:49
$begingroup$
@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
$endgroup$
– clathratus
Jan 8 at 17:54
add a comment |
$begingroup$
Is there a workaround to approach solution under 150seconds?
$endgroup$
– user5722540
Jan 8 at 7:49
$begingroup$
@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
$endgroup$
– clathratus
Jan 8 at 17:54
$begingroup$
Is there a workaround to approach solution under 150seconds?
$endgroup$
– user5722540
Jan 8 at 7:49
$begingroup$
Is there a workaround to approach solution under 150seconds?
$endgroup$
– user5722540
Jan 8 at 7:49
$begingroup$
@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
$endgroup$
– clathratus
Jan 8 at 17:54
$begingroup$
@user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig
$endgroup$
– clathratus
Jan 8 at 17:54
add a comment |
$begingroup$
We need $-1le3xle1$
But if $xle0,$ the left hand side $le0$
Now $3x=sin(pi/4-arcsin(2x))$
$3sqrt2x=sqrt{1-(2x)^2}-2x$
$sqrt{1-4x^2}=x(3sqrt2+2)$
Square both sides
$endgroup$
add a comment |
$begingroup$
We need $-1le3xle1$
But if $xle0,$ the left hand side $le0$
Now $3x=sin(pi/4-arcsin(2x))$
$3sqrt2x=sqrt{1-(2x)^2}-2x$
$sqrt{1-4x^2}=x(3sqrt2+2)$
Square both sides
$endgroup$
add a comment |
$begingroup$
We need $-1le3xle1$
But if $xle0,$ the left hand side $le0$
Now $3x=sin(pi/4-arcsin(2x))$
$3sqrt2x=sqrt{1-(2x)^2}-2x$
$sqrt{1-4x^2}=x(3sqrt2+2)$
Square both sides
$endgroup$
We need $-1le3xle1$
But if $xle0,$ the left hand side $le0$
Now $3x=sin(pi/4-arcsin(2x))$
$3sqrt2x=sqrt{1-(2x)^2}-2x$
$sqrt{1-4x^2}=x(3sqrt2+2)$
Square both sides
answered Jan 8 at 4:13
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
$$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+frac{1}{4}=0$$ which is simple.
$endgroup$
add a comment |
$begingroup$
Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
$$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+frac{1}{4}=0$$ which is simple.
$endgroup$
add a comment |
$begingroup$
Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
$$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+frac{1}{4}=0$$ which is simple.
$endgroup$
Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2=frac 12$$ that is to say
$$72 x^4-13x^2+frac 12=12 sqrt{1-9 x^2} sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+frac{1}{4}=0$$ which is simple.
answered Jan 8 at 4:17
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
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$begingroup$
Possibly useful - math.stackexchange.com/questions/672575/…
$endgroup$
– Eevee Trainer
Jan 8 at 2:50
$begingroup$
But I don't know x^2+y^2
$endgroup$
– user5722540
Jan 8 at 2:52
$begingroup$
I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question?
$endgroup$
– D.B.
Jan 8 at 4:06
$begingroup$
Online open ended answer. Deadline of 150 seconds.
$endgroup$
– user5722540
Jan 8 at 4:09
$begingroup$
Well, it takes Maple less than a second to get ${frac {sqrt {2522-1164,sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer...
$endgroup$
– Robert Israel
Jan 8 at 4:23