If $sin^-1x+sin^-1y+sin^-1z = pi$, then $xsqrt1-x^2 + ysqrt1-y^2 +zsqrt1-z^2 = 2xyz$.
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If $sin^-1x+sin^-1y+sin^-1z = pi$, prove that
$$xsqrt1-x^2 + ysqrt1-y^2 +zsqrt1-z^2 = 2xyz$$
I tried rewriting $z$ in terms of $x$ and $y$. I got
$$z=xsqrt1-y^2 + ysqrt1-x^2$$
I further tried simplifying this term, but I just couldn't get the desired result.
trigonometry
edited Aug 27 at 16:08
Blue
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asked Aug 27 at 14:03
S.Nep
424
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up vote
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If $sin^-1x+sin^-1y+sin^-1z = pi$, prove that
$$xsqrt1-x^2 + ysqrt1-y^2 +zsqrt1-z^2 = 2xyz$$
I tried rewriting $z$ in terms of $x$ and $y$. I got
$$z=xsqrt1-y^2 + ysqrt1-x^2$$
I further tried simplifying this term, but I just couldn't get the desired result.
trigonometry
edited Aug 27 at 16:08
Blue
44.1k868141
asked Aug 27 at 14:03
S.Nep
424
1
Hi, welcome! Posts which are just a problem pasted in don't get a very good reception. You'll get better answers if you include the work you've done on the problem and where you got stuck.
– Matthew Leingang
Aug 27 at 14:07
What have you tried? Are you stuck somewhere? If you don't show your work/thoughts, the question will be downvoted/closed.
– Andrei
Aug 27 at 14:07
3
I take it you know the compound angle formula. You may want to try and write $x=sinalpha, y=sinbeta, z=singamma$. Can you try to express $z$ in terms of $x$ and $y$?
– daruma
Aug 27 at 14:10
Have you tried spherical coordinates?
– Henry Lee
Aug 27 at 14:31
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $sin^-1x+sin^-1y+sin^-1z = pi$, prove that
$$xsqrt1-x^2 + ysqrt1-y^2 +zsqrt1-z^2 = 2xyz$$
I tried rewriting $z$ in terms of $x$ and $y$. I got
$$z=xsqrt1-y^2 + ysqrt1-x^2$$
I further tried simplifying this term, but I just couldn't get the desired result.
trigonometry
edited Aug 27 at 16:08
Blue
44.1k868141
asked Aug 27 at 14:03
S.Nep
424
If $sin^-1x+sin^-1y+sin^-1z = pi$, prove that
$$xsqrt1-x^2 + ysqrt1-y^2 +zsqrt1-z^2 = 2xyz$$
I tried rewriting $z$ in terms of $x$ and $y$. I got
$$z=xsqrt1-y^2 + ysqrt1-x^2$$
I further tried simplifying this term, but I just couldn't get the desired result.
trigonometry
edited Aug 27 at 16:08
Blue
44.1k868141
asked Aug 27 at 14:03
S.Nep
424
edited Aug 27 at 16:08
Blue
44.1k868141
edited Aug 27 at 16:08
Blue
44.1k868141
edited Aug 27 at 16:08
Blue
44.1k868141
44.1k868141
asked Aug 27 at 14:03
S.Nep
424
asked Aug 27 at 14:03
S.Nep
424
asked Aug 27 at 14:03
S.Nep
424
424
1
Hi, welcome! Posts which are just a problem pasted in don't get a very good reception. You'll get better answers if you include the work you've done on the problem and where you got stuck.
– Matthew Leingang
Aug 27 at 14:07
What have you tried? Are you stuck somewhere? If you don't show your work/thoughts, the question will be downvoted/closed.
– Andrei
Aug 27 at 14:07
3
I take it you know the compound angle formula. You may want to try and write $x=sinalpha, y=sinbeta, z=singamma$. Can you try to express $z$ in terms of $x$ and $y$?
– daruma
Aug 27 at 14:10
Have you tried spherical coordinates?
– Henry Lee
Aug 27 at 14:31
add a comment |Â
1
Hi, welcome! Posts which are just a problem pasted in don't get a very good reception. You'll get better answers if you include the work you've done on the problem and where you got stuck.
– Matthew Leingang
Aug 27 at 14:07
What have you tried? Are you stuck somewhere? If you don't show your work/thoughts, the question will be downvoted/closed.
– Andrei
Aug 27 at 14:07
3
I take it you know the compound angle formula. You may want to try and write $x=sinalpha, y=sinbeta, z=singamma$. Can you try to express $z$ in terms of $x$ and $y$?
– daruma
Aug 27 at 14:10
Have you tried spherical coordinates?
– Henry Lee
Aug 27 at 14:31
1
1
Hi, welcome! Posts which are just a problem pasted in don't get a very good reception. You'll get better answers if you include the work you've done on the problem and where you got stuck.
– Matthew Leingang
Aug 27 at 14:07
Hi, welcome! Posts which are just a problem pasted in don't get a very good reception. You'll get better answers if you include the work you've done on the problem and where you got stuck.
– Matthew Leingang
Aug 27 at 14:07
What have you tried? Are you stuck somewhere? If you don't show your work/thoughts, the question will be downvoted/closed.
– Andrei
Aug 27 at 14:07
What have you tried? Are you stuck somewhere? If you don't show your work/thoughts, the question will be downvoted/closed.
– Andrei
Aug 27 at 14:07
3
3
I take it you know the compound angle formula. You may want to try and write $x=sinalpha, y=sinbeta, z=singamma$. Can you try to express $z$ in terms of $x$ and $y$?
– daruma
Aug 27 at 14:10
I take it you know the compound angle formula. You may want to try and write $x=sinalpha, y=sinbeta, z=singamma$. Can you try to express $z$ in terms of $x$ and $y$?
– daruma
Aug 27 at 14:10
Have you tried spherical coordinates?
– Henry Lee
Aug 27 at 14:31
Have you tried spherical coordinates?
– Henry Lee
Aug 27 at 14:31
add a comment |Â
3 Answers
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Let $$sin^-1x=A,sin^-1y=B,sin^-1z=C$$and $A+B+C=pi$
It is enough to prove that $$sin2A+sin2B+sin2C=4sin Asin Bsin C$$
Let us take $LHS$$$implies2sin(A+B)cos(A-B)+2sin Ccos C$$$$=2sin Ccos(A-B)+2sin Ccos C$$$$=2sin C[cos(A-B)+cos C]$$$$=2sin C[cos(A-B)-cos(A+B)]$$$$=2sin C 2sin B sin A$$$$=4sin Asin Bsin C$$$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=2sin Asin Bsin C=2xyz$$
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
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Proof.$blacktriangleleft$ Let $A = arcsin(x), B = arcsin(y), C = arcsin(z)$, then $A,B,C in [-pi/2, pi/2]$, thus $cos(A), cos(B), cos(C) geqslant 0$. Therefore we are requested to prove
$$
sin(A)cos(A) + sin(B)cos(B) + sin(C) cos(C) = 2sin(A) sin(B) sin(C)quad [A + B + C = pi].
$$
Now calculate:
beginalign*
mathrm LHS &= sin(A)cos(A) + sin(B)cos(B) - sin(A+B) cos(A+B) \
&= frac 12 (sin(2A) + sin(2B)) - sin(A+B) cos(A+B)\
&= sin(A+B) cos(A-B) - sin(A+B) cos(A+B)\
&= sin(A+B) (cos(A-B) - cos(A+B))\
&= 2sin(C)sin(A) sin(B) = mathrm RHS.
endalign*
Hence the equation. $blacktriangleright$
answered Aug 27 at 15:27
xbh
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Good start. Indeed, on the one hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin^-1x=pi-(sin^-1y+sin^-1z) iff \
sin (sin^-1x)=sin(sin^-1y+sin^-1z) iff \
x=ycos(sin^-1z)+zcos (sin^-1y) =ysqrt1-z^2+zsqrt1-y^2.$$
Similarly:
$$y=xsqrt1-z^2+zsqrt1-x^2;\
z=xsqrt1-y^2+ysqrt1-x^2.$$
Multiply the three equations by $sqrt1-x^2,sqrt1-y^2,sqrt1-z^2$, respectively, and add up:
$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=\
2left(colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2)right)$$
On the other hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin(sin^-1x+sin^-1y+sin^-1z)=0 iff \
sin(sin^-1x)cdot cos(sin^-1y+sin^-1z)colorred+\
cos(sin^-1x)cdot sin(sin^-1y+sin^-1z)=0 iff \
xcdot (cos(sin^-1y)cos(sin^-1z)-sin(sin^-1y)sin(sin^-1z))colorred+\
sqrt1-x^2cdot (ycos(sin^-1z)+zcos(sin^-1y))=0 iff \
xcdot (sqrt(1-y^2)(1-z^2)-yz)colorred+sqrt1-x^2cdot (ysqrt1-z^2+zsqrt1-y^2)=0 iff \
xyz=colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2).$$
answered Aug 27 at 15:39
farruhota
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $$sin^-1x=A,sin^-1y=B,sin^-1z=C$$and $A+B+C=pi$
It is enough to prove that $$sin2A+sin2B+sin2C=4sin Asin Bsin C$$
Let us take $LHS$$$implies2sin(A+B)cos(A-B)+2sin Ccos C$$$$=2sin Ccos(A-B)+2sin Ccos C$$$$=2sin C[cos(A-B)+cos C]$$$$=2sin C[cos(A-B)-cos(A+B)]$$$$=2sin C 2sin B sin A$$$$=4sin Asin Bsin C$$$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=2sin Asin Bsin C=2xyz$$
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
add a comment |Â
up vote
0
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Let $$sin^-1x=A,sin^-1y=B,sin^-1z=C$$and $A+B+C=pi$
It is enough to prove that $$sin2A+sin2B+sin2C=4sin Asin Bsin C$$
Let us take $LHS$$$implies2sin(A+B)cos(A-B)+2sin Ccos C$$$$=2sin Ccos(A-B)+2sin Ccos C$$$$=2sin C[cos(A-B)+cos C]$$$$=2sin C[cos(A-B)-cos(A+B)]$$$$=2sin C 2sin B sin A$$$$=4sin Asin Bsin C$$$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=2sin Asin Bsin C=2xyz$$
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $$sin^-1x=A,sin^-1y=B,sin^-1z=C$$and $A+B+C=pi$
It is enough to prove that $$sin2A+sin2B+sin2C=4sin Asin Bsin C$$
Let us take $LHS$$$implies2sin(A+B)cos(A-B)+2sin Ccos C$$$$=2sin Ccos(A-B)+2sin Ccos C$$$$=2sin C[cos(A-B)+cos C]$$$$=2sin C[cos(A-B)-cos(A+B)]$$$$=2sin C 2sin B sin A$$$$=4sin Asin Bsin C$$$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=2sin Asin Bsin C=2xyz$$
Let $$sin^-1x=A,sin^-1y=B,sin^-1z=C$$and $A+B+C=pi$
It is enough to prove that $$sin2A+sin2B+sin2C=4sin Asin Bsin C$$
Let us take $LHS$$$implies2sin(A+B)cos(A-B)+2sin Ccos C$$$$=2sin Ccos(A-B)+2sin Ccos C$$$$=2sin C[cos(A-B)+cos C]$$$$=2sin C[cos(A-B)-cos(A+B)]$$$$=2sin C 2sin B sin A$$$$=4sin Asin Bsin C$$$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=2sin Asin Bsin C=2xyz$$
answered Aug 27 at 15:19
user587574
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
add a comment |Â
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
Very nice approach. I too thought the same but ended up doing a silly mistake. Thanks.
– S.Nep
Aug 27 at 16:10
add a comment |Â
up vote
0
down vote
Proof.$blacktriangleleft$ Let $A = arcsin(x), B = arcsin(y), C = arcsin(z)$, then $A,B,C in [-pi/2, pi/2]$, thus $cos(A), cos(B), cos(C) geqslant 0$. Therefore we are requested to prove
$$
sin(A)cos(A) + sin(B)cos(B) + sin(C) cos(C) = 2sin(A) sin(B) sin(C)quad [A + B + C = pi].
$$
Now calculate:
beginalign*
mathrm LHS &= sin(A)cos(A) + sin(B)cos(B) - sin(A+B) cos(A+B) \
&= frac 12 (sin(2A) + sin(2B)) - sin(A+B) cos(A+B)\
&= sin(A+B) cos(A-B) - sin(A+B) cos(A+B)\
&= sin(A+B) (cos(A-B) - cos(A+B))\
&= 2sin(C)sin(A) sin(B) = mathrm RHS.
endalign*
Hence the equation. $blacktriangleright$
answered Aug 27 at 15:27
xbh
3,062219
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up vote
0
down vote
Proof.$blacktriangleleft$ Let $A = arcsin(x), B = arcsin(y), C = arcsin(z)$, then $A,B,C in [-pi/2, pi/2]$, thus $cos(A), cos(B), cos(C) geqslant 0$. Therefore we are requested to prove
$$
sin(A)cos(A) + sin(B)cos(B) + sin(C) cos(C) = 2sin(A) sin(B) sin(C)quad [A + B + C = pi].
$$
Now calculate:
beginalign*
mathrm LHS &= sin(A)cos(A) + sin(B)cos(B) - sin(A+B) cos(A+B) \
&= frac 12 (sin(2A) + sin(2B)) - sin(A+B) cos(A+B)\
&= sin(A+B) cos(A-B) - sin(A+B) cos(A+B)\
&= sin(A+B) (cos(A-B) - cos(A+B))\
&= 2sin(C)sin(A) sin(B) = mathrm RHS.
endalign*
Hence the equation. $blacktriangleright$
answered Aug 27 at 15:27
xbh
3,062219
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Proof.$blacktriangleleft$ Let $A = arcsin(x), B = arcsin(y), C = arcsin(z)$, then $A,B,C in [-pi/2, pi/2]$, thus $cos(A), cos(B), cos(C) geqslant 0$. Therefore we are requested to prove
$$
sin(A)cos(A) + sin(B)cos(B) + sin(C) cos(C) = 2sin(A) sin(B) sin(C)quad [A + B + C = pi].
$$
Now calculate:
beginalign*
mathrm LHS &= sin(A)cos(A) + sin(B)cos(B) - sin(A+B) cos(A+B) \
&= frac 12 (sin(2A) + sin(2B)) - sin(A+B) cos(A+B)\
&= sin(A+B) cos(A-B) - sin(A+B) cos(A+B)\
&= sin(A+B) (cos(A-B) - cos(A+B))\
&= 2sin(C)sin(A) sin(B) = mathrm RHS.
endalign*
Hence the equation. $blacktriangleright$
answered Aug 27 at 15:27
xbh
3,062219
Proof.$blacktriangleleft$ Let $A = arcsin(x), B = arcsin(y), C = arcsin(z)$, then $A,B,C in [-pi/2, pi/2]$, thus $cos(A), cos(B), cos(C) geqslant 0$. Therefore we are requested to prove
$$
sin(A)cos(A) + sin(B)cos(B) + sin(C) cos(C) = 2sin(A) sin(B) sin(C)quad [A + B + C = pi].
$$
Now calculate:
beginalign*
mathrm LHS &= sin(A)cos(A) + sin(B)cos(B) - sin(A+B) cos(A+B) \
&= frac 12 (sin(2A) + sin(2B)) - sin(A+B) cos(A+B)\
&= sin(A+B) cos(A-B) - sin(A+B) cos(A+B)\
&= sin(A+B) (cos(A-B) - cos(A+B))\
&= 2sin(C)sin(A) sin(B) = mathrm RHS.
endalign*
Hence the equation. $blacktriangleright$
answered Aug 27 at 15:27
xbh
3,062219
answered Aug 27 at 15:27
xbh
3,062219
answered Aug 27 at 15:27
xbh
3,062219
answered Aug 27 at 15:27
xbh
3,062219
3,062219
add a comment |Â
add a comment |Â
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0
down vote
Good start. Indeed, on the one hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin^-1x=pi-(sin^-1y+sin^-1z) iff \
sin (sin^-1x)=sin(sin^-1y+sin^-1z) iff \
x=ycos(sin^-1z)+zcos (sin^-1y) =ysqrt1-z^2+zsqrt1-y^2.$$
Similarly:
$$y=xsqrt1-z^2+zsqrt1-x^2;\
z=xsqrt1-y^2+ysqrt1-x^2.$$
Multiply the three equations by $sqrt1-x^2,sqrt1-y^2,sqrt1-z^2$, respectively, and add up:
$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=\
2left(colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2)right)$$
On the other hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin(sin^-1x+sin^-1y+sin^-1z)=0 iff \
sin(sin^-1x)cdot cos(sin^-1y+sin^-1z)colorred+\
cos(sin^-1x)cdot sin(sin^-1y+sin^-1z)=0 iff \
xcdot (cos(sin^-1y)cos(sin^-1z)-sin(sin^-1y)sin(sin^-1z))colorred+\
sqrt1-x^2cdot (ycos(sin^-1z)+zcos(sin^-1y))=0 iff \
xcdot (sqrt(1-y^2)(1-z^2)-yz)colorred+sqrt1-x^2cdot (ysqrt1-z^2+zsqrt1-y^2)=0 iff \
xyz=colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2).$$
answered Aug 27 at 15:39
farruhota
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Good start. Indeed, on the one hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin^-1x=pi-(sin^-1y+sin^-1z) iff \
sin (sin^-1x)=sin(sin^-1y+sin^-1z) iff \
x=ycos(sin^-1z)+zcos (sin^-1y) =ysqrt1-z^2+zsqrt1-y^2.$$
Similarly:
$$y=xsqrt1-z^2+zsqrt1-x^2;\
z=xsqrt1-y^2+ysqrt1-x^2.$$
Multiply the three equations by $sqrt1-x^2,sqrt1-y^2,sqrt1-z^2$, respectively, and add up:
$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=\
2left(colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2)right)$$
On the other hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin(sin^-1x+sin^-1y+sin^-1z)=0 iff \
sin(sin^-1x)cdot cos(sin^-1y+sin^-1z)colorred+\
cos(sin^-1x)cdot sin(sin^-1y+sin^-1z)=0 iff \
xcdot (cos(sin^-1y)cos(sin^-1z)-sin(sin^-1y)sin(sin^-1z))colorred+\
sqrt1-x^2cdot (ycos(sin^-1z)+zcos(sin^-1y))=0 iff \
xcdot (sqrt(1-y^2)(1-z^2)-yz)colorred+sqrt1-x^2cdot (ysqrt1-z^2+zsqrt1-y^2)=0 iff \
xyz=colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2).$$
answered Aug 27 at 15:39
farruhota
14.9k2633
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up vote
0
down vote
up vote
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down vote
Good start. Indeed, on the one hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin^-1x=pi-(sin^-1y+sin^-1z) iff \
sin (sin^-1x)=sin(sin^-1y+sin^-1z) iff \
x=ycos(sin^-1z)+zcos (sin^-1y) =ysqrt1-z^2+zsqrt1-y^2.$$
Similarly:
$$y=xsqrt1-z^2+zsqrt1-x^2;\
z=xsqrt1-y^2+ysqrt1-x^2.$$
Multiply the three equations by $sqrt1-x^2,sqrt1-y^2,sqrt1-z^2$, respectively, and add up:
$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=\
2left(colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2)right)$$
On the other hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin(sin^-1x+sin^-1y+sin^-1z)=0 iff \
sin(sin^-1x)cdot cos(sin^-1y+sin^-1z)colorred+\
cos(sin^-1x)cdot sin(sin^-1y+sin^-1z)=0 iff \
xcdot (cos(sin^-1y)cos(sin^-1z)-sin(sin^-1y)sin(sin^-1z))colorred+\
sqrt1-x^2cdot (ycos(sin^-1z)+zcos(sin^-1y))=0 iff \
xcdot (sqrt(1-y^2)(1-z^2)-yz)colorred+sqrt1-x^2cdot (ysqrt1-z^2+zsqrt1-y^2)=0 iff \
xyz=colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2).$$
answered Aug 27 at 15:39
farruhota
14.9k2633
Good start. Indeed, on the one hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin^-1x=pi-(sin^-1y+sin^-1z) iff \
sin (sin^-1x)=sin(sin^-1y+sin^-1z) iff \
x=ycos(sin^-1z)+zcos (sin^-1y) =ysqrt1-z^2+zsqrt1-y^2.$$
Similarly:
$$y=xsqrt1-z^2+zsqrt1-x^2;\
z=xsqrt1-y^2+ysqrt1-x^2.$$
Multiply the three equations by $sqrt1-x^2,sqrt1-y^2,sqrt1-z^2$, respectively, and add up:
$$xsqrt1-x^2+ysqrt1-y^2+zsqrt1-z^2=\
2left(colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2)right)$$
On the other hand:
$$sin^-1x+sin^-1y+sin^-1z = pi iff \
sin(sin^-1x+sin^-1y+sin^-1z)=0 iff \
sin(sin^-1x)cdot cos(sin^-1y+sin^-1z)colorred+\
cos(sin^-1x)cdot sin(sin^-1y+sin^-1z)=0 iff \
xcdot (cos(sin^-1y)cos(sin^-1z)-sin(sin^-1y)sin(sin^-1z))colorred+\
sqrt1-x^2cdot (ycos(sin^-1z)+zcos(sin^-1y))=0 iff \
xcdot (sqrt(1-y^2)(1-z^2)-yz)colorred+sqrt1-x^2cdot (ysqrt1-z^2+zsqrt1-y^2)=0 iff \
xyz=colorbluexsqrt(1-y^2)(1-z^2)+ysqrt(1-x^2)(1-z^2)+zsqrt(1-x^2)(1-y^2).$$
answered Aug 27 at 15:39
farruhota
14.9k2633
answered Aug 27 at 15:39
farruhota
14.9k2633
answered Aug 27 at 15:39
farruhota
14.9k2633
answered Aug 27 at 15:39
farruhota
14.9k2633
14.9k2633
add a comment |Â
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1
Hi, welcome! Posts which are just a problem pasted in don't get a very good reception. You'll get better answers if you include the work you've done on the problem and where you got stuck.
– Matthew Leingang
Aug 27 at 14:07
What have you tried? Are you stuck somewhere? If you don't show your work/thoughts, the question will be downvoted/closed.
– Andrei
Aug 27 at 14:07
3
I take it you know the compound angle formula. You may want to try and write $x=sinalpha, y=sinbeta, z=singamma$. Can you try to express $z$ in terms of $x$ and $y$?
– daruma
Aug 27 at 14:10
Have you tried spherical coordinates?
– Henry Lee
Aug 27 at 14:31