Integral inequality for sin function
Let $0<r<1$ and $tgeq 0$ real numbers. Is it true that $$int_t^{t+r} sin(x), dx leq int_{frac{pi}{2}-frac{r}{2}}^{frac{pi}{2}+frac{r}{2}}sin(x), dx ,? $$
I suspect that yes, since both integration intervals have length $r$ and $sin$ has maximum in second one (RHS).
Hints are welcome.
integral-inequality
asked Nov 27 '18 at 21:11
Sigur
4,48811736
add a comment |
Let $0<r<1$ and $tgeq 0$ real numbers. Is it true that $$int_t^{t+r} sin(x), dx leq int_{frac{pi}{2}-frac{r}{2}}^{frac{pi}{2}+frac{r}{2}}sin(x), dx ,? $$
I suspect that yes, since both integration intervals have length $r$ and $sin$ has maximum in second one (RHS).
Hints are welcome.
integral-inequality
asked Nov 27 '18 at 21:11
Sigur
4,48811736
add a comment |
Let $0<r<1$ and $tgeq 0$ real numbers. Is it true that $$int_t^{t+r} sin(x), dx leq int_{frac{pi}{2}-frac{r}{2}}^{frac{pi}{2}+frac{r}{2}}sin(x), dx ,? $$
I suspect that yes, since both integration intervals have length $r$ and $sin$ has maximum in second one (RHS).
Hints are welcome.
integral-inequality
asked Nov 27 '18 at 21:11
Sigur
4,48811736
Let $0<r<1$ and $tgeq 0$ real numbers. Is it true that $$int_t^{t+r} sin(x), dx leq int_{frac{pi}{2}-frac{r}{2}}^{frac{pi}{2}+frac{r}{2}}sin(x), dx ,? $$
I suspect that yes, since both integration intervals have length $r$ and $sin$ has maximum in second one (RHS).
Hints are welcome.
integral-inequality
integral-inequality
asked Nov 27 '18 at 21:11
Sigur
4,48811736
asked Nov 27 '18 at 21:11
Sigur
4,48811736
edited Nov 27 '18 at 21:18
asked Nov 27 '18 at 21:11
Sigur
4,48811736
asked Nov 27 '18 at 21:11
Sigur
4,48811736
asked Nov 27 '18 at 21:11
Sigur
4,48811736
4,48811736
add a comment |
add a comment |
1 Answer
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Well, one way to do it (though unelegant) would be to write out the solutions to the integrals explicitly. So, you are asking us to prove that:
$$cos(t)-cos(r+t)leq cos(pi/2-r/2)-cos(pi/2+r/2)$$
This is true, since by one of the Prosthaphaeresis Formulas, namely:
$$cos(alpha)-cos(beta)=-2sinleft[frac{1}{2}(alpha+beta)right]sinleft[frac{1}{2}(alpha-beta)right]$$
We have that:
$$cos(t)-cos(r+t)=2sin(r/2)sin(r/2+t)$$
And:
$$cos(pi/2-r/2)-cos(pi/2+r/2)=2sin(r/2)$$
And fortunately, since $0<r<1$, we have that $sin(r/2)> 0$ and since $r/2+t$ is a real number, $sin(r/2+t)leq 1$.
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
1
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
add a comment |
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1 Answer
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Well, one way to do it (though unelegant) would be to write out the solutions to the integrals explicitly. So, you are asking us to prove that:
$$cos(t)-cos(r+t)leq cos(pi/2-r/2)-cos(pi/2+r/2)$$
This is true, since by one of the Prosthaphaeresis Formulas, namely:
$$cos(alpha)-cos(beta)=-2sinleft[frac{1}{2}(alpha+beta)right]sinleft[frac{1}{2}(alpha-beta)right]$$
We have that:
$$cos(t)-cos(r+t)=2sin(r/2)sin(r/2+t)$$
And:
$$cos(pi/2-r/2)-cos(pi/2+r/2)=2sin(r/2)$$
And fortunately, since $0<r<1$, we have that $sin(r/2)> 0$ and since $r/2+t$ is a real number, $sin(r/2+t)leq 1$.
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
1
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
add a comment |
Well, one way to do it (though unelegant) would be to write out the solutions to the integrals explicitly. So, you are asking us to prove that:
$$cos(t)-cos(r+t)leq cos(pi/2-r/2)-cos(pi/2+r/2)$$
This is true, since by one of the Prosthaphaeresis Formulas, namely:
$$cos(alpha)-cos(beta)=-2sinleft[frac{1}{2}(alpha+beta)right]sinleft[frac{1}{2}(alpha-beta)right]$$
We have that:
$$cos(t)-cos(r+t)=2sin(r/2)sin(r/2+t)$$
And:
$$cos(pi/2-r/2)-cos(pi/2+r/2)=2sin(r/2)$$
And fortunately, since $0<r<1$, we have that $sin(r/2)> 0$ and since $r/2+t$ is a real number, $sin(r/2+t)leq 1$.
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
1
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
add a comment |
Well, one way to do it (though unelegant) would be to write out the solutions to the integrals explicitly. So, you are asking us to prove that:
$$cos(t)-cos(r+t)leq cos(pi/2-r/2)-cos(pi/2+r/2)$$
This is true, since by one of the Prosthaphaeresis Formulas, namely:
$$cos(alpha)-cos(beta)=-2sinleft[frac{1}{2}(alpha+beta)right]sinleft[frac{1}{2}(alpha-beta)right]$$
We have that:
$$cos(t)-cos(r+t)=2sin(r/2)sin(r/2+t)$$
And:
$$cos(pi/2-r/2)-cos(pi/2+r/2)=2sin(r/2)$$
And fortunately, since $0<r<1$, we have that $sin(r/2)> 0$ and since $r/2+t$ is a real number, $sin(r/2+t)leq 1$.
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
Well, one way to do it (though unelegant) would be to write out the solutions to the integrals explicitly. So, you are asking us to prove that:
$$cos(t)-cos(r+t)leq cos(pi/2-r/2)-cos(pi/2+r/2)$$
This is true, since by one of the Prosthaphaeresis Formulas, namely:
$$cos(alpha)-cos(beta)=-2sinleft[frac{1}{2}(alpha+beta)right]sinleft[frac{1}{2}(alpha-beta)right]$$
We have that:
$$cos(t)-cos(r+t)=2sin(r/2)sin(r/2+t)$$
And:
$$cos(pi/2-r/2)-cos(pi/2+r/2)=2sin(r/2)$$
And fortunately, since $0<r<1$, we have that $sin(r/2)> 0$ and since $r/2+t$ is a real number, $sin(r/2+t)leq 1$.
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
edited Nov 27 '18 at 21:45
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
answered Nov 27 '18 at 21:38
projectilemotion
11.4k62041
11.4k62041
1
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
add a comment |
1
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
1
1
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Fortunately, $sinleq 1$!!!
– Sigur
Nov 27 '18 at 21:40
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals.
– projectilemotion
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks.
– Sigur
Nov 27 '18 at 21:42
add a comment |
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