Mixing
Help explaining the simplification of an integral
$begingroup$
I am trying to understand the following steps my teacher did in class (from top to bottom). I tried to look up different trigonometric identities but couldn't figure out where the arrival of cosine squared came from, and the other sine values arrived to the right of the plus sign.
Thanks!
$$ begin{align}
pi &= frac 1 T int_0^t F cos omega t frac{F}{|{underline{z}_m}|} cos(omega t - theta) dt \
&= frac{F^2}{T |underline{z}_m|} int_0^t left[
cos^2 omega t cos theta + cos omega t sin omega t sin theta
right] dt
end{align}$$
calculus integration trigonometry definite-integrals
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
asked Jan 22 at 0:24
articatarticat
64
$endgroup$
add a comment |
$begingroup$
I am trying to understand the following steps my teacher did in class (from top to bottom). I tried to look up different trigonometric identities but couldn't figure out where the arrival of cosine squared came from, and the other sine values arrived to the right of the plus sign.
Thanks!
$$ begin{align}
pi &= frac 1 T int_0^t F cos omega t frac{F}{|{underline{z}_m}|} cos(omega t - theta) dt \
&= frac{F^2}{T |underline{z}_m|} int_0^t left[
cos^2 omega t cos theta + cos omega t sin omega t sin theta
right] dt
end{align}$$
calculus integration trigonometry definite-integrals
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
asked Jan 22 at 0:24
articatarticat
64
$endgroup$
1
$begingroup$
You can move the constant factors from inside the integral to outside. Then expand $cos(a+b)$ using the standard trig identity. There's a $cos^2$ because you have $cos(a)cos(a+b)$.
$endgroup$
– Ethan Bolker
Jan 22 at 0:41
$begingroup$
It will help to note that $$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$$
$endgroup$
– clathratus
Jan 22 at 1:33
add a comment |
$begingroup$
I am trying to understand the following steps my teacher did in class (from top to bottom). I tried to look up different trigonometric identities but couldn't figure out where the arrival of cosine squared came from, and the other sine values arrived to the right of the plus sign.
Thanks!
$$ begin{align}
pi &= frac 1 T int_0^t F cos omega t frac{F}{|{underline{z}_m}|} cos(omega t - theta) dt \
&= frac{F^2}{T |underline{z}_m|} int_0^t left[
cos^2 omega t cos theta + cos omega t sin omega t sin theta
right] dt
end{align}$$
calculus integration trigonometry definite-integrals
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
asked Jan 22 at 0:24
articatarticat
64
$endgroup$
I am trying to understand the following steps my teacher did in class (from top to bottom). I tried to look up different trigonometric identities but couldn't figure out where the arrival of cosine squared came from, and the other sine values arrived to the right of the plus sign.
Thanks!
$$ begin{align}
pi &= frac 1 T int_0^t F cos omega t frac{F}{|{underline{z}_m}|} cos(omega t - theta) dt \
&= frac{F^2}{T |underline{z}_m|} int_0^t left[
cos^2 omega t cos theta + cos omega t sin omega t sin theta
right] dt
end{align}$$
calculus integration trigonometry definite-integrals
calculus integration trigonometry definite-integrals
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
asked Jan 22 at 0:24
articatarticat
64
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
asked Jan 22 at 0:24
articatarticat
64
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
edited Jan 22 at 0:39
Eevee Trainer
7,61621338
7,61621338
asked Jan 22 at 0:24
articatarticat
64
asked Jan 22 at 0:24
articatarticat
64
asked Jan 22 at 0:24
articatarticat
64
64
1
$begingroup$
You can move the constant factors from inside the integral to outside. Then expand $cos(a+b)$ using the standard trig identity. There's a $cos^2$ because you have $cos(a)cos(a+b)$.
$endgroup$
– Ethan Bolker
Jan 22 at 0:41
$begingroup$
It will help to note that $$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$$
$endgroup$
– clathratus
Jan 22 at 1:33
add a comment |
1
$begingroup$
You can move the constant factors from inside the integral to outside. Then expand $cos(a+b)$ using the standard trig identity. There's a $cos^2$ because you have $cos(a)cos(a+b)$.
$endgroup$
– Ethan Bolker
Jan 22 at 0:41
$begingroup$
It will help to note that $$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$$
$endgroup$
– clathratus
Jan 22 at 1:33
1
1
$begingroup$
You can move the constant factors from inside the integral to outside. Then expand $cos(a+b)$ using the standard trig identity. There's a $cos^2$ because you have $cos(a)cos(a+b)$.
$endgroup$
– Ethan Bolker
Jan 22 at 0:41
$begingroup$
You can move the constant factors from inside the integral to outside. Then expand $cos(a+b)$ using the standard trig identity. There's a $cos^2$ because you have $cos(a)cos(a+b)$.
$endgroup$
– Ethan Bolker
Jan 22 at 0:41
$begingroup$
It will help to note that $$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$$
$endgroup$
– clathratus
Jan 22 at 1:33
$begingroup$
It will help to note that $$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$$
$endgroup$
– clathratus
Jan 22 at 1:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: The cosine angle subtraction identity says that $cos(a-b)=cos acos b+sin asin b$ for all $a,b$. What does this tell you for $a=omega t$, $b=theta$? Further, $F$ and $z_m$ are (presumably) constant with respect to $t$, so they can be brought out of the integral sign. Can you proceed?
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
$endgroup$
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
add a comment |
$begingroup$
Knowing the sum and difference trig cosine formulas would help:
$$cos(apm b)=cos(a)cos(b)mpsin(a)sin(b)$$
This applies here with $begin{bmatrix} a \ bend{bmatrix}=begin{bmatrix} omega t \ thetaend{bmatrix}$. Plugging this information into the integrand in your statement gives you the required result.
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
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add a comment |
$begingroup$
Assuming that $F$ and $left| z_m right|$ are constants, they can come all the way outside the integral.
This leaves you with:
$$
cos (omega t) cos (omega t - theta)
$$
as the integrand.
Note the trig identity:
$$
cos(a - b) = cos(a) cos(b) + sin(a) * sin(b)
$$
substituting
$$
a = omega t qquad and qquad b = theta,
$$
it follows
$$
cos(omega t - theta) = cos(omega t) cos(theta) + sin(omega t) sin(theta)
$$
From this, it follows:
$$
cos(omega t)cos(omega t - theta) = cos^2(omega t) cos(theta) + cos(omega t) sin(omega t) sin(theta)
$$
which is the integrand on your second line.
answered Jan 22 at 1:12
themathochistthemathochist
13
$endgroup$
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: The cosine angle subtraction identity says that $cos(a-b)=cos acos b+sin asin b$ for all $a,b$. What does this tell you for $a=omega t$, $b=theta$? Further, $F$ and $z_m$ are (presumably) constant with respect to $t$, so they can be brought out of the integral sign. Can you proceed?
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
$endgroup$
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
add a comment |
$begingroup$
Hint: The cosine angle subtraction identity says that $cos(a-b)=cos acos b+sin asin b$ for all $a,b$. What does this tell you for $a=omega t$, $b=theta$? Further, $F$ and $z_m$ are (presumably) constant with respect to $t$, so they can be brought out of the integral sign. Can you proceed?
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
$endgroup$
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
add a comment |
$begingroup$
Hint: The cosine angle subtraction identity says that $cos(a-b)=cos acos b+sin asin b$ for all $a,b$. What does this tell you for $a=omega t$, $b=theta$? Further, $F$ and $z_m$ are (presumably) constant with respect to $t$, so they can be brought out of the integral sign. Can you proceed?
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
$endgroup$
Hint: The cosine angle subtraction identity says that $cos(a-b)=cos acos b+sin asin b$ for all $a,b$. What does this tell you for $a=omega t$, $b=theta$? Further, $F$ and $z_m$ are (presumably) constant with respect to $t$, so they can be brought out of the integral sign. Can you proceed?
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
answered Jan 22 at 1:22
YiFanYiFan
4,5261727
4,5261727
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
add a comment |
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
I can thank you very much! Need to study my trig identities!!
$endgroup$
– articat
Jan 22 at 3:43
add a comment |
$begingroup$
Knowing the sum and difference trig cosine formulas would help:
$$cos(apm b)=cos(a)cos(b)mpsin(a)sin(b)$$
This applies here with $begin{bmatrix} a \ bend{bmatrix}=begin{bmatrix} omega t \ thetaend{bmatrix}$. Plugging this information into the integrand in your statement gives you the required result.
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
$endgroup$
add a comment |
$begingroup$
Knowing the sum and difference trig cosine formulas would help:
$$cos(apm b)=cos(a)cos(b)mpsin(a)sin(b)$$
This applies here with $begin{bmatrix} a \ bend{bmatrix}=begin{bmatrix} omega t \ thetaend{bmatrix}$. Plugging this information into the integrand in your statement gives you the required result.
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
$endgroup$
add a comment |
$begingroup$
Knowing the sum and difference trig cosine formulas would help:
$$cos(apm b)=cos(a)cos(b)mpsin(a)sin(b)$$
This applies here with $begin{bmatrix} a \ bend{bmatrix}=begin{bmatrix} omega t \ thetaend{bmatrix}$. Plugging this information into the integrand in your statement gives you the required result.
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
$endgroup$
Knowing the sum and difference trig cosine formulas would help:
$$cos(apm b)=cos(a)cos(b)mpsin(a)sin(b)$$
This applies here with $begin{bmatrix} a \ bend{bmatrix}=begin{bmatrix} omega t \ thetaend{bmatrix}$. Plugging this information into the integrand in your statement gives you the required result.
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
answered Jan 23 at 18:17
Paras KhoslaParas Khosla
1,732219
1,732219
add a comment |
add a comment |
$begingroup$
Assuming that $F$ and $left| z_m right|$ are constants, they can come all the way outside the integral.
This leaves you with:
$$
cos (omega t) cos (omega t - theta)
$$
as the integrand.
Note the trig identity:
$$
cos(a - b) = cos(a) cos(b) + sin(a) * sin(b)
$$
substituting
$$
a = omega t qquad and qquad b = theta,
$$
it follows
$$
cos(omega t - theta) = cos(omega t) cos(theta) + sin(omega t) sin(theta)
$$
From this, it follows:
$$
cos(omega t)cos(omega t - theta) = cos^2(omega t) cos(theta) + cos(omega t) sin(omega t) sin(theta)
$$
which is the integrand on your second line.
answered Jan 22 at 1:12
themathochistthemathochist
13
$endgroup$
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
add a comment |
$begingroup$
Assuming that $F$ and $left| z_m right|$ are constants, they can come all the way outside the integral.
This leaves you with:
$$
cos (omega t) cos (omega t - theta)
$$
as the integrand.
Note the trig identity:
$$
cos(a - b) = cos(a) cos(b) + sin(a) * sin(b)
$$
substituting
$$
a = omega t qquad and qquad b = theta,
$$
it follows
$$
cos(omega t - theta) = cos(omega t) cos(theta) + sin(omega t) sin(theta)
$$
From this, it follows:
$$
cos(omega t)cos(omega t - theta) = cos^2(omega t) cos(theta) + cos(omega t) sin(omega t) sin(theta)
$$
which is the integrand on your second line.
answered Jan 22 at 1:12
themathochistthemathochist
13
$endgroup$
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
add a comment |
$begingroup$
Assuming that $F$ and $left| z_m right|$ are constants, they can come all the way outside the integral.
This leaves you with:
$$
cos (omega t) cos (omega t - theta)
$$
as the integrand.
Note the trig identity:
$$
cos(a - b) = cos(a) cos(b) + sin(a) * sin(b)
$$
substituting
$$
a = omega t qquad and qquad b = theta,
$$
it follows
$$
cos(omega t - theta) = cos(omega t) cos(theta) + sin(omega t) sin(theta)
$$
From this, it follows:
$$
cos(omega t)cos(omega t - theta) = cos^2(omega t) cos(theta) + cos(omega t) sin(omega t) sin(theta)
$$
which is the integrand on your second line.
answered Jan 22 at 1:12
themathochistthemathochist
13
$endgroup$
Assuming that $F$ and $left| z_m right|$ are constants, they can come all the way outside the integral.
This leaves you with:
$$
cos (omega t) cos (omega t - theta)
$$
as the integrand.
Note the trig identity:
$$
cos(a - b) = cos(a) cos(b) + sin(a) * sin(b)
$$
substituting
$$
a = omega t qquad and qquad b = theta,
$$
it follows
$$
cos(omega t - theta) = cos(omega t) cos(theta) + sin(omega t) sin(theta)
$$
From this, it follows:
$$
cos(omega t)cos(omega t - theta) = cos^2(omega t) cos(theta) + cos(omega t) sin(omega t) sin(theta)
$$
which is the integrand on your second line.
answered Jan 22 at 1:12
themathochistthemathochist
13
edited Jan 28 at 3:01
answered Jan 22 at 1:12
themathochistthemathochist
13
answered Jan 22 at 1:12
themathochistthemathochist
13
answered Jan 22 at 1:12
themathochistthemathochist
13
13
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
add a comment |
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
This helps very much thank you!
$endgroup$
– articat
Jan 22 at 3:43
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
$begingroup$
Welcome to the Math SE Community! Be sure to use typesetting whenever typing an answer or asking a query. Follow this link : math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Jan 23 at 18:28
add a comment |
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1
$begingroup$
You can move the constant factors from inside the integral to outside. Then expand $cos(a+b)$ using the standard trig identity. There's a $cos^2$ because you have $cos(a)cos(a+b)$.
$endgroup$
– Ethan Bolker
Jan 22 at 0:41
$begingroup$
It will help to note that $$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$$
$endgroup$
– clathratus
Jan 22 at 1:33