Mixing
$int_theta^theta+pir(u)cos u, du=-lambdasintheta$ and $int_theta^theta+pir(u)sin u;du=lambdacostheta$ for certain $r(u)$
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We are given some definitions about a function $r(u)$:
1) $r(u)>0$
2) $r(u)= r(u+2pi)$
3) $r(u)+r(u+pi)= lambda$
4) $r(u)$ is continuous
5) $int_0^pir(u)costheta=0$ and $int_0^pir(u)sintheta=lambda$
The following identities are to be proven:
The first proceeds as follows:
I have tried over and over to prove the second identity, and it seems like it should be easy. However, I keep getting $lambda +lambdacostheta$ instead of $lambdacostheta$ ... what am I missing here? And sorry for the awful formatting, as you can see I'm fairly new to MSE.
calculus algebra-precalculus trigonometry
edited Jul 14 at 21:18
Blue
43.7k868141
asked Jul 14 at 19:43
Math Enthusiast
576
add a comment |Â
up vote
4
down vote
favorite
We are given some definitions about a function $r(u)$:
1) $r(u)>0$
2) $r(u)= r(u+2pi)$
3) $r(u)+r(u+pi)= lambda$
4) $r(u)$ is continuous
5) $int_0^pir(u)costheta=0$ and $int_0^pir(u)sintheta=lambda$
The following identities are to be proven:
The first proceeds as follows:
I have tried over and over to prove the second identity, and it seems like it should be easy. However, I keep getting $lambda +lambdacostheta$ instead of $lambdacostheta$ ... what am I missing here? And sorry for the awful formatting, as you can see I'm fairly new to MSE.
calculus algebra-precalculus trigonometry
edited Jul 14 at 21:18
Blue
43.7k868141
asked Jul 14 at 19:43
Math Enthusiast
576
2
You should be ending up with $lambda +lambda(costheta-cos0)$ which will give you the desired result since $cos0=1$.
– John Wayland Bales
Jul 14 at 20:09
Aha! Thank you so much, I just ran into so much mental gridlock there.
– Math Enthusiast
Jul 14 at 20:13
You're welcome.
– John Wayland Bales
Jul 14 at 20:14
@MathEnthusiast The approach below is positive.
– Salahamam_ Fatima
Jul 14 at 23:53
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
We are given some definitions about a function $r(u)$:
1) $r(u)>0$
2) $r(u)= r(u+2pi)$
3) $r(u)+r(u+pi)= lambda$
4) $r(u)$ is continuous
5) $int_0^pir(u)costheta=0$ and $int_0^pir(u)sintheta=lambda$
The following identities are to be proven:
The first proceeds as follows:
I have tried over and over to prove the second identity, and it seems like it should be easy. However, I keep getting $lambda +lambdacostheta$ instead of $lambdacostheta$ ... what am I missing here? And sorry for the awful formatting, as you can see I'm fairly new to MSE.
calculus algebra-precalculus trigonometry
edited Jul 14 at 21:18
Blue
43.7k868141
asked Jul 14 at 19:43
Math Enthusiast
576
We are given some definitions about a function $r(u)$:
1) $r(u)>0$
2) $r(u)= r(u+2pi)$
3) $r(u)+r(u+pi)= lambda$
4) $r(u)$ is continuous
5) $int_0^pir(u)costheta=0$ and $int_0^pir(u)sintheta=lambda$
The following identities are to be proven:
The first proceeds as follows:
I have tried over and over to prove the second identity, and it seems like it should be easy. However, I keep getting $lambda +lambdacostheta$ instead of $lambdacostheta$ ... what am I missing here? And sorry for the awful formatting, as you can see I'm fairly new to MSE.
calculus algebra-precalculus trigonometry
edited Jul 14 at 21:18
Blue
43.7k868141
asked Jul 14 at 19:43
Math Enthusiast
576
edited Jul 14 at 21:18
Blue
43.7k868141
edited Jul 14 at 21:18
Blue
43.7k868141
edited Jul 14 at 21:18
Blue
43.7k868141
43.7k868141
asked Jul 14 at 19:43
Math Enthusiast
576
asked Jul 14 at 19:43
Math Enthusiast
576
asked Jul 14 at 19:43
Math Enthusiast
576
576
2
You should be ending up with $lambda +lambda(costheta-cos0)$ which will give you the desired result since $cos0=1$.
– John Wayland Bales
Jul 14 at 20:09
Aha! Thank you so much, I just ran into so much mental gridlock there.
– Math Enthusiast
Jul 14 at 20:13
You're welcome.
– John Wayland Bales
Jul 14 at 20:14
@MathEnthusiast The approach below is positive.
– Salahamam_ Fatima
Jul 14 at 23:53
add a comment |Â
2
You should be ending up with $lambda +lambda(costheta-cos0)$ which will give you the desired result since $cos0=1$.
– John Wayland Bales
Jul 14 at 20:09
Aha! Thank you so much, I just ran into so much mental gridlock there.
– Math Enthusiast
Jul 14 at 20:13
You're welcome.
– John Wayland Bales
Jul 14 at 20:14
@MathEnthusiast The approach below is positive.
– Salahamam_ Fatima
Jul 14 at 23:53
2
2
You should be ending up with $lambda +lambda(costheta-cos0)$ which will give you the desired result since $cos0=1$.
– John Wayland Bales
Jul 14 at 20:09
You should be ending up with $lambda +lambda(costheta-cos0)$ which will give you the desired result since $cos0=1$.
– John Wayland Bales
Jul 14 at 20:09
Aha! Thank you so much, I just ran into so much mental gridlock there.
– Math Enthusiast
Jul 14 at 20:13
Aha! Thank you so much, I just ran into so much mental gridlock there.
– Math Enthusiast
Jul 14 at 20:13
You're welcome.
– John Wayland Bales
Jul 14 at 20:14
You're welcome.
– John Wayland Bales
Jul 14 at 20:14
@MathEnthusiast The approach below is positive.
– Salahamam_ Fatima
Jul 14 at 23:53
@MathEnthusiast The approach below is positive.
– Salahamam_ Fatima
Jul 14 at 23:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $$F(theta)=int_theta^theta+pir(u)sin(u)du.$$
The derivative is
$$F'(theta)=$$
$$r(theta+pi)sin(theta+pi)-r(theta)sin(theta)=$$
$$-sin(theta)Bigl(r(theta+pi)+r(theta)Bigr)=$$
$$-lambda sin(theta)$$
and after integration,
$$F(theta)=lambda cos(theta)+C$$
with $theta=0$, it becomes
$$int_0^pi r(u)sin(u)du=lambda=lambda+C$$
thus $$C=0$$
Done.
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $$F(theta)=int_theta^theta+pir(u)sin(u)du.$$
The derivative is
$$F'(theta)=$$
$$r(theta+pi)sin(theta+pi)-r(theta)sin(theta)=$$
$$-sin(theta)Bigl(r(theta+pi)+r(theta)Bigr)=$$
$$-lambda sin(theta)$$
and after integration,
$$F(theta)=lambda cos(theta)+C$$
with $theta=0$, it becomes
$$int_0^pi r(u)sin(u)du=lambda=lambda+C$$
thus $$C=0$$
Done.
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
add a comment |Â
up vote
2
down vote
accepted
Let $$F(theta)=int_theta^theta+pir(u)sin(u)du.$$
The derivative is
$$F'(theta)=$$
$$r(theta+pi)sin(theta+pi)-r(theta)sin(theta)=$$
$$-sin(theta)Bigl(r(theta+pi)+r(theta)Bigr)=$$
$$-lambda sin(theta)$$
and after integration,
$$F(theta)=lambda cos(theta)+C$$
with $theta=0$, it becomes
$$int_0^pi r(u)sin(u)du=lambda=lambda+C$$
thus $$C=0$$
Done.
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $$F(theta)=int_theta^theta+pir(u)sin(u)du.$$
The derivative is
$$F'(theta)=$$
$$r(theta+pi)sin(theta+pi)-r(theta)sin(theta)=$$
$$-sin(theta)Bigl(r(theta+pi)+r(theta)Bigr)=$$
$$-lambda sin(theta)$$
and after integration,
$$F(theta)=lambda cos(theta)+C$$
with $theta=0$, it becomes
$$int_0^pi r(u)sin(u)du=lambda=lambda+C$$
thus $$C=0$$
Done.
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
Let $$F(theta)=int_theta^theta+pir(u)sin(u)du.$$
The derivative is
$$F'(theta)=$$
$$r(theta+pi)sin(theta+pi)-r(theta)sin(theta)=$$
$$-sin(theta)Bigl(r(theta+pi)+r(theta)Bigr)=$$
$$-lambda sin(theta)$$
and after integration,
$$F(theta)=lambda cos(theta)+C$$
with $theta=0$, it becomes
$$int_0^pi r(u)sin(u)du=lambda=lambda+C$$
thus $$C=0$$
Done.
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
edited Jul 14 at 21:20
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
answered Jul 14 at 21:04
Salahamam_ Fatima
33.6k21229
33.6k21229
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
add a comment |Â
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
Thank you very much for answering :) Another user was helpful in pointing out my original error in the comments above, but I will mark this as the correct answer as well because I think it's a very cool approach! In fact, this method is somewhat more direct than the one I was following in my paper...
– Math Enthusiast
Jul 15 at 2:06
add a comment |Â
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2
You should be ending up with $lambda +lambda(costheta-cos0)$ which will give you the desired result since $cos0=1$.
– John Wayland Bales
Jul 14 at 20:09
Aha! Thank you so much, I just ran into so much mental gridlock there.
– Math Enthusiast
Jul 14 at 20:13
You're welcome.
– John Wayland Bales
Jul 14 at 20:14
@MathEnthusiast The approach below is positive.
– Salahamam_ Fatima
Jul 14 at 23:53