A Trigonometric Reduction
$begingroup$
I am trying to prove an equation is always positive given certain constraints for the variables, the equation I've come up with is as follows
$ 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)$
In my previous post it is commented that this expression can be simplified to the following equation
$ sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$
Is this correct and if so, how did they do it? Like I said, I'm not a mathematician so I haven't been able to recreate it but if this is true it could prove useful.
algebra-precalculus trigonometry
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
$endgroup$
add a comment |
$begingroup$
I am trying to prove an equation is always positive given certain constraints for the variables, the equation I've come up with is as follows
$ 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)$
In my previous post it is commented that this expression can be simplified to the following equation
$ sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$
Is this correct and if so, how did they do it? Like I said, I'm not a mathematician so I haven't been able to recreate it but if this is true it could prove useful.
algebra-precalculus trigonometry
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
$endgroup$
add a comment |
$begingroup$
I am trying to prove an equation is always positive given certain constraints for the variables, the equation I've come up with is as follows
$ 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)$
In my previous post it is commented that this expression can be simplified to the following equation
$ sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$
Is this correct and if so, how did they do it? Like I said, I'm not a mathematician so I haven't been able to recreate it but if this is true it could prove useful.
algebra-precalculus trigonometry
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
$endgroup$
I am trying to prove an equation is always positive given certain constraints for the variables, the equation I've come up with is as follows
$ 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)$
In my previous post it is commented that this expression can be simplified to the following equation
$ sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$
Is this correct and if so, how did they do it? Like I said, I'm not a mathematician so I haven't been able to recreate it but if this is true it could prove useful.
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
edited Jan 20 at 15:24
Lee David Chung Lin
4,47841242
4,47841242
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
asked Dec 24 '18 at 8:24
RoryHectorRoryHector
79214
79214
add a comment |
add a comment |
1 Answer
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$begingroup$
Pretty sure the RHS is zero right? The coefficient seems to be canceled out.
begin{align}
& 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)\
&= 2 b [cos(t+2r)- cos(t)] sin(a t) -2[cos(a t)-cos(a t + 2 b (pi - r - t))]sin(t)\
&= -4bsin(r)sin(r+t)sin(at) -4 sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)\
&=-4left[bsin(r)sin(r+t)sin(at) + sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)right]
end{align}
Discarding the $(-4)$, you get your expression. This is just using a standard product/sum formula.
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
$endgroup$
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Pretty sure the RHS is zero right? The coefficient seems to be canceled out.
begin{align}
& 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)\
&= 2 b [cos(t+2r)- cos(t)] sin(a t) -2[cos(a t)-cos(a t + 2 b (pi - r - t))]sin(t)\
&= -4bsin(r)sin(r+t)sin(at) -4 sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)\
&=-4left[bsin(r)sin(r+t)sin(at) + sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)right]
end{align}
Discarding the $(-4)$, you get your expression. This is just using a standard product/sum formula.
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
$endgroup$
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
add a comment |
$begingroup$
Pretty sure the RHS is zero right? The coefficient seems to be canceled out.
begin{align}
& 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)\
&= 2 b [cos(t+2r)- cos(t)] sin(a t) -2[cos(a t)-cos(a t + 2 b (pi - r - t))]sin(t)\
&= -4bsin(r)sin(r+t)sin(at) -4 sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)\
&=-4left[bsin(r)sin(r+t)sin(at) + sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)right]
end{align}
Discarding the $(-4)$, you get your expression. This is just using a standard product/sum formula.
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
$endgroup$
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
add a comment |
$begingroup$
Pretty sure the RHS is zero right? The coefficient seems to be canceled out.
begin{align}
& 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)\
&= 2 b [cos(t+2r)- cos(t)] sin(a t) -2[cos(a t)-cos(a t + 2 b (pi - r - t))]sin(t)\
&= -4bsin(r)sin(r+t)sin(at) -4 sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)\
&=-4left[bsin(r)sin(r+t)sin(at) + sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)right]
end{align}
Discarding the $(-4)$, you get your expression. This is just using a standard product/sum formula.
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
$endgroup$
Pretty sure the RHS is zero right? The coefficient seems to be canceled out.
begin{align}
& 2 b cos(t+2r) sin(a t) - 2 sin(t) cos(a t + 2 b (pi - r - t)) + 2 sin(t) cos(a t) - 2 b cos(t) sin(a t)\
&= 2 b [cos(t+2r)- cos(t)] sin(a t) -2[cos(a t)-cos(a t + 2 b (pi - r - t))]sin(t)\
&= -4bsin(r)sin(r+t)sin(at) -4 sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)\
&=-4left[bsin(r)sin(r+t)sin(at) + sin(b (-r - t + π)) sin(a t + b (-r - t + π))sin(t)right]
end{align}
Discarding the $(-4)$, you get your expression. This is just using a standard product/sum formula.
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
answered Dec 24 '18 at 8:32
Karn WatcharasupatKarn Watcharasupat
3,9642526
3,9642526
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
add a comment |
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
$begingroup$
Thank you very much but I think there is a mistake on the second line should it be $2 b [ cos(t+2 r) - cos(t)] sin(a t) + 2[cos(a t) - cos(a t + 2 b(pi - r - t))] sin(t)$ which I think would cause the final result to be $sin(b(pi-r-t)) sin(t) sin(b(pi - r - t) + a t) - b sin(r) sin(a t) sin(r+t)$ after getting rid of a factor of 4
$endgroup$
– RoryHector
Dec 24 '18 at 9:44
add a comment |
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