Possible to prove that a particular trigonometric expression is always positive?
0
$begingroup$
Is it possible to prove
$$begin{align}
&phantom{+}sin(t) sinleft(frac{1}{4} ; epsilon ; (2 pi -2 r - t) + rright) sinleft(frac{1}{2} ; (2-epsilon) (pi-r-t)right) \
&+ frac{1}{2} ; (epsilon-2) sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
is always positive given the following information?
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
trigonometry
edited Jan 7 at 20:17
Blue
47.7k870151
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
$endgroup$
|
show 3 more comments
0
$begingroup$
Is it possible to prove
$$begin{align}
&phantom{+}sin(t) sinleft(frac{1}{4} ; epsilon ; (2 pi -2 r - t) + rright) sinleft(frac{1}{2} ; (2-epsilon) (pi-r-t)right) \
&+ frac{1}{2} ; (epsilon-2) sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
is always positive given the following information?
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
trigonometry
edited Jan 7 at 20:17
Blue
47.7k870151
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
$endgroup$
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
@Neal This is the point where I couldn't see another way to reduce the equation to make it simpler and the ($epsilon-2$) term makes it so that I can't just say everything is positive. I'm not a mathematician or anything so I've had a difficult time with this one.
$endgroup$
– RoryHector
Jan 7 at 20:24
1
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
@Blue, I tried to explain on the last post that the answer is complicated but I’ll give it another go. We have a triangle with a “base” side and two non-base sides. Two additional triangles are placed on top of the non-base sides of the original triangle, where each non-base side of the original triangle is the base side of a subsequent triangle. So there are 3 triangles, the original, and the two stacked on top. If you fix one base angle of the original, and take the partial derivative of the height of each triangle with respect to the other base angle,
$endgroup$
– RoryHector
Jan 7 at 22:04
|
show 3 more comments
0
0
0
$begingroup$
Is it possible to prove
$$begin{align}
&phantom{+}sin(t) sinleft(frac{1}{4} ; epsilon ; (2 pi -2 r - t) + rright) sinleft(frac{1}{2} ; (2-epsilon) (pi-r-t)right) \
&+ frac{1}{2} ; (epsilon-2) sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
is always positive given the following information?
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
trigonometry
edited Jan 7 at 20:17
Blue
47.7k870151
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
$endgroup$
Is it possible to prove
$$begin{align}
&phantom{+}sin(t) sinleft(frac{1}{4} ; epsilon ; (2 pi -2 r - t) + rright) sinleft(frac{1}{2} ; (2-epsilon) (pi-r-t)right) \
&+ frac{1}{2} ; (epsilon-2) sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
is always positive given the following information?
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
trigonometry
trigonometry
edited Jan 7 at 20:17
Blue
47.7k870151
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
edited Jan 7 at 20:17
Blue
47.7k870151
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
edited Jan 7 at 20:17
Blue
47.7k870151
edited Jan 7 at 20:17
Blue
47.7k870151
edited Jan 7 at 20:17
Blue
47.7k870151
47.7k870151
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
asked Jan 7 at 20:09
RoryHectorRoryHector
10412
10412
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
@Neal This is the point where I couldn't see another way to reduce the equation to make it simpler and the ($epsilon-2$) term makes it so that I can't just say everything is positive. I'm not a mathematician or anything so I've had a difficult time with this one.
$endgroup$
– RoryHector
Jan 7 at 20:24
1
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
@Blue, I tried to explain on the last post that the answer is complicated but I’ll give it another go. We have a triangle with a “base” side and two non-base sides. Two additional triangles are placed on top of the non-base sides of the original triangle, where each non-base side of the original triangle is the base side of a subsequent triangle. So there are 3 triangles, the original, and the two stacked on top. If you fix one base angle of the original, and take the partial derivative of the height of each triangle with respect to the other base angle,
$endgroup$
– RoryHector
Jan 7 at 22:04
|
show 3 more comments
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
@Neal This is the point where I couldn't see another way to reduce the equation to make it simpler and the ($epsilon-2$) term makes it so that I can't just say everything is positive. I'm not a mathematician or anything so I've had a difficult time with this one.
$endgroup$
– RoryHector
Jan 7 at 20:24
1
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
@Blue, I tried to explain on the last post that the answer is complicated but I’ll give it another go. We have a triangle with a “base” side and two non-base sides. Two additional triangles are placed on top of the non-base sides of the original triangle, where each non-base side of the original triangle is the base side of a subsequent triangle. So there are 3 triangles, the original, and the two stacked on top. If you fix one base angle of the original, and take the partial derivative of the height of each triangle with respect to the other base angle,
$endgroup$
– RoryHector
Jan 7 at 22:04
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
@Neal This is the point where I couldn't see another way to reduce the equation to make it simpler and the ($epsilon-2$) term makes it so that I can't just say everything is positive. I'm not a mathematician or anything so I've had a difficult time with this one.
$endgroup$
– RoryHector
Jan 7 at 20:24
$begingroup$
@Neal This is the point where I couldn't see another way to reduce the equation to make it simpler and the ($epsilon-2$) term makes it so that I can't just say everything is positive. I'm not a mathematician or anything so I've had a difficult time with this one.
$endgroup$
– RoryHector
Jan 7 at 20:24
1
1
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
@Blue, I tried to explain on the last post that the answer is complicated but I’ll give it another go. We have a triangle with a “base” side and two non-base sides. Two additional triangles are placed on top of the non-base sides of the original triangle, where each non-base side of the original triangle is the base side of a subsequent triangle. So there are 3 triangles, the original, and the two stacked on top. If you fix one base angle of the original, and take the partial derivative of the height of each triangle with respect to the other base angle,
$endgroup$
– RoryHector
Jan 7 at 22:04
$begingroup$
@Blue, I tried to explain on the last post that the answer is complicated but I’ll give it another go. We have a triangle with a “base” side and two non-base sides. Two additional triangles are placed on top of the non-base sides of the original triangle, where each non-base side of the original triangle is the base side of a subsequent triangle. So there are 3 triangles, the original, and the two stacked on top. If you fix one base angle of the original, and take the partial derivative of the height of each triangle with respect to the other base angle,
$endgroup$
– RoryHector
Jan 7 at 22:04
|
show 3 more comments
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$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
@Neal This is the point where I couldn't see another way to reduce the equation to make it simpler and the ($epsilon-2$) term makes it so that I can't just say everything is positive. I'm not a mathematician or anything so I've had a difficult time with this one.
$endgroup$
– RoryHector
Jan 7 at 20:24
1
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
@Blue, I tried to explain on the last post that the answer is complicated but I’ll give it another go. We have a triangle with a “base” side and two non-base sides. Two additional triangles are placed on top of the non-base sides of the original triangle, where each non-base side of the original triangle is the base side of a subsequent triangle. So there are 3 triangles, the original, and the two stacked on top. If you fix one base angle of the original, and take the partial derivative of the height of each triangle with respect to the other base angle,
$endgroup$
– RoryHector
Jan 7 at 22:04