Mixing
Determining $sin(2x)$
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up vote
2
down vote
favorite
Given that
$$sin (y-x)cos(x+y)=dfrac 1 2$$
$$sin (x+y)cos (x-y) = dfrac 1 3 $$
Determine $sin (2x)$.
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $sin(2x)$ is given by
$$sin(2x) = 2sincos$$
Let us try simpiflying the second equation
$$sin(x+y)-cos(x-y)=sin x cos y+cos x sin y-cos x cos y-sin xsin y=
cos x(sin y-cos y)+sin x(cos y-sin y)=colorblue(cos x-sin x)(sin y-cos y)$$
However, there seems to be nothing useful.
trigonometry
asked Jul 20 at 14:17
Enzo
425
add a comment |Â
up vote
2
down vote
favorite
Given that
$$sin (y-x)cos(x+y)=dfrac 1 2$$
$$sin (x+y)cos (x-y) = dfrac 1 3 $$
Determine $sin (2x)$.
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $sin(2x)$ is given by
$$sin(2x) = 2sincos$$
Let us try simpiflying the second equation
$$sin(x+y)-cos(x-y)=sin x cos y+cos x sin y-cos x cos y-sin xsin y=
cos x(sin y-cos y)+sin x(cos y-sin y)=colorblue(cos x-sin x)(sin y-cos y)$$
However, there seems to be nothing useful.
trigonometry
asked Jul 20 at 14:17
Enzo
425
2
$2sin(x+y)cos(x-y)=sin2x+sin 2y$ surely?
– Lord Shark the Unknown
Jul 20 at 14:22
The second equation doesn't subtract $sin(x+y)$ and $cos(x-y)$, it multiplies them.
– Barry Cipra
Jul 20 at 14:39
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given that
$$sin (y-x)cos(x+y)=dfrac 1 2$$
$$sin (x+y)cos (x-y) = dfrac 1 3 $$
Determine $sin (2x)$.
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $sin(2x)$ is given by
$$sin(2x) = 2sincos$$
Let us try simpiflying the second equation
$$sin(x+y)-cos(x-y)=sin x cos y+cos x sin y-cos x cos y-sin xsin y=
cos x(sin y-cos y)+sin x(cos y-sin y)=colorblue(cos x-sin x)(sin y-cos y)$$
However, there seems to be nothing useful.
trigonometry
asked Jul 20 at 14:17
Enzo
425
Given that
$$sin (y-x)cos(x+y)=dfrac 1 2$$
$$sin (x+y)cos (x-y) = dfrac 1 3 $$
Determine $sin (2x)$.
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $sin(2x)$ is given by
$$sin(2x) = 2sincos$$
Let us try simpiflying the second equation
$$sin(x+y)-cos(x-y)=sin x cos y+cos x sin y-cos x cos y-sin xsin y=
cos x(sin y-cos y)+sin x(cos y-sin y)=colorblue(cos x-sin x)(sin y-cos y)$$
However, there seems to be nothing useful.
trigonometry
asked Jul 20 at 14:17
Enzo
425
asked Jul 20 at 14:17
Enzo
425
asked Jul 20 at 14:17
Enzo
425
asked Jul 20 at 14:17
Enzo
425
425
2
$2sin(x+y)cos(x-y)=sin2x+sin 2y$ surely?
– Lord Shark the Unknown
Jul 20 at 14:22
The second equation doesn't subtract $sin(x+y)$ and $cos(x-y)$, it multiplies them.
– Barry Cipra
Jul 20 at 14:39
add a comment |Â
2
$2sin(x+y)cos(x-y)=sin2x+sin 2y$ surely?
– Lord Shark the Unknown
Jul 20 at 14:22
The second equation doesn't subtract $sin(x+y)$ and $cos(x-y)$, it multiplies them.
– Barry Cipra
Jul 20 at 14:39
2
2
$2sin(x+y)cos(x-y)=sin2x+sin 2y$ surely?
– Lord Shark the Unknown
Jul 20 at 14:22
$2sin(x+y)cos(x-y)=sin2x+sin 2y$ surely?
– Lord Shark the Unknown
Jul 20 at 14:22
The second equation doesn't subtract $sin(x+y)$ and $cos(x-y)$, it multiplies them.
– Barry Cipra
Jul 20 at 14:39
The second equation doesn't subtract $sin(x+y)$ and $cos(x-y)$, it multiplies them.
– Barry Cipra
Jul 20 at 14:39
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Recall that by Product to sum identity
$$2sin theta cos varphi = sin(theta + varphi) + sin(theta - varphi) $$
that is
$$begincasessin (y-x)cos(x+y)=frac12sin (2y)+frac12 sin (-2x)=dfrac 1 2\sin (x+y)cos (x-y) = frac12sin (2x)+frac12 sin (2y)=dfrac 1 3endcases $$
$$begincases
sin (2y)-sin (2x)=1\
sin (2x)+sin (2y)=dfrac 2 3
endcases $$
and subtracting the first equation from the second we obtain
$$2sin (2x)=-frac13 implies sin (2x)=-frac16$$
answered Jul 20 at 14:26
gimusi
65.4k73584
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
add a comment |Â
up vote
2
down vote
Hint
$$
sin 2x=sin((x+y)+(x-y))=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)
$$
We know the value of the first summand. For the second use the fact that sine is odd.
answered Jul 20 at 14:28
Foobaz John
18.1k41245
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
add a comment |Â
up vote
1
down vote
$$ 2x = ( x+y) + (x-y)$$
$$ sin ( alpha + beta ) = sin (alpha) cos (beta) + cos (alpha) sin (beta) $$
$$ sin ( 2x ) = sin (x+y) cos (x-y) + cos (x+y) sin (x-y)=1/3-1/2 =-1/6$$
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
You were on the right track but the equations contain the product, not the difference.
beginalign
frac12 &= sin (y-x)cos(x+y)\
&=(sin ycos x - sin x cos y)(cos xcos y - sin xsin y) \
&= sin ycos y - cos x sin x
endalign
beginalign
frac13 &= sin (x+y)cos(x-y)\
&=(sin xcos y + sin y cos x)(cos xcos y + sin xsin y) \
&= sin ycos y + cos x sin x
endalign
So subtracting them gives
$$sin 2x = 2cos xsin x = ( sin ycos y + cos x sin x) - ( sin ycos y - cos x sin x) = frac13 - frac12 = -frac16$$
answered Jul 20 at 14:49
mechanodroid
22.2k52041
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Recall that by Product to sum identity
$$2sin theta cos varphi = sin(theta + varphi) + sin(theta - varphi) $$
that is
$$begincasessin (y-x)cos(x+y)=frac12sin (2y)+frac12 sin (-2x)=dfrac 1 2\sin (x+y)cos (x-y) = frac12sin (2x)+frac12 sin (2y)=dfrac 1 3endcases $$
$$begincases
sin (2y)-sin (2x)=1\
sin (2x)+sin (2y)=dfrac 2 3
endcases $$
and subtracting the first equation from the second we obtain
$$2sin (2x)=-frac13 implies sin (2x)=-frac16$$
answered Jul 20 at 14:26
gimusi
65.4k73584
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
add a comment |Â
up vote
1
down vote
accepted
Recall that by Product to sum identity
$$2sin theta cos varphi = sin(theta + varphi) + sin(theta - varphi) $$
that is
$$begincasessin (y-x)cos(x+y)=frac12sin (2y)+frac12 sin (-2x)=dfrac 1 2\sin (x+y)cos (x-y) = frac12sin (2x)+frac12 sin (2y)=dfrac 1 3endcases $$
$$begincases
sin (2y)-sin (2x)=1\
sin (2x)+sin (2y)=dfrac 2 3
endcases $$
and subtracting the first equation from the second we obtain
$$2sin (2x)=-frac13 implies sin (2x)=-frac16$$
answered Jul 20 at 14:26
gimusi
65.4k73584
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Recall that by Product to sum identity
$$2sin theta cos varphi = sin(theta + varphi) + sin(theta - varphi) $$
that is
$$begincasessin (y-x)cos(x+y)=frac12sin (2y)+frac12 sin (-2x)=dfrac 1 2\sin (x+y)cos (x-y) = frac12sin (2x)+frac12 sin (2y)=dfrac 1 3endcases $$
$$begincases
sin (2y)-sin (2x)=1\
sin (2x)+sin (2y)=dfrac 2 3
endcases $$
and subtracting the first equation from the second we obtain
$$2sin (2x)=-frac13 implies sin (2x)=-frac16$$
answered Jul 20 at 14:26
gimusi
65.4k73584
Recall that by Product to sum identity
$$2sin theta cos varphi = sin(theta + varphi) + sin(theta - varphi) $$
that is
$$begincasessin (y-x)cos(x+y)=frac12sin (2y)+frac12 sin (-2x)=dfrac 1 2\sin (x+y)cos (x-y) = frac12sin (2x)+frac12 sin (2y)=dfrac 1 3endcases $$
$$begincases
sin (2y)-sin (2x)=1\
sin (2x)+sin (2y)=dfrac 2 3
endcases $$
and subtracting the first equation from the second we obtain
$$2sin (2x)=-frac13 implies sin (2x)=-frac16$$
answered Jul 20 at 14:26
gimusi
65.4k73584
edited Jul 20 at 14:59
answered Jul 20 at 14:26
gimusi
65.4k73584
answered Jul 20 at 14:26
gimusi
65.4k73584
answered Jul 20 at 14:26
gimusi
65.4k73584
65.4k73584
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
add a comment |Â
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
How did you obtain those values? My apologies, I couldn't get it well.
– Enzo
Jul 20 at 14:29
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
@Enzo Refer to en.wikipedia.org/wiki/…
– gimusi
Jul 20 at 14:30
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
I'm stuck at this step $$begincases sin (2y)-sin (2x)=1\ sin (2x)+sin (2y)=dfrac 2 3 endcases$$
– Enzo
Jul 20 at 14:55
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
@Enzo It is a linear system in $sin(2x)$ and $sin (2y)$ therefore we can easily obtain the value. HINT: subtract the first equation from the second.
– gimusi
Jul 20 at 14:57
add a comment |Â
up vote
2
down vote
Hint
$$
sin 2x=sin((x+y)+(x-y))=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)
$$
We know the value of the first summand. For the second use the fact that sine is odd.
answered Jul 20 at 14:28
Foobaz John
18.1k41245
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
add a comment |Â
up vote
2
down vote
Hint
$$
sin 2x=sin((x+y)+(x-y))=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)
$$
We know the value of the first summand. For the second use the fact that sine is odd.
answered Jul 20 at 14:28
Foobaz John
18.1k41245
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint
$$
sin 2x=sin((x+y)+(x-y))=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)
$$
We know the value of the first summand. For the second use the fact that sine is odd.
answered Jul 20 at 14:28
Foobaz John
18.1k41245
Hint
$$
sin 2x=sin((x+y)+(x-y))=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)
$$
We know the value of the first summand. For the second use the fact that sine is odd.
answered Jul 20 at 14:28
Foobaz John
18.1k41245
answered Jul 20 at 14:28
Foobaz John
18.1k41245
answered Jul 20 at 14:28
Foobaz John
18.1k41245
answered Jul 20 at 14:28
Foobaz John
18.1k41245
18.1k41245
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
add a comment |Â
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
What did you mean by sine is odd?
– Enzo
Jul 20 at 14:36
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
@Enzo $$sin(-x)=-sin(x)$$ so that $$sin(y-x)=-sin(x-y)$$
– Isham
Jul 20 at 14:37
add a comment |Â
up vote
1
down vote
$$ 2x = ( x+y) + (x-y)$$
$$ sin ( alpha + beta ) = sin (alpha) cos (beta) + cos (alpha) sin (beta) $$
$$ sin ( 2x ) = sin (x+y) cos (x-y) + cos (x+y) sin (x-y)=1/3-1/2 =-1/6$$
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
1
down vote
$$ 2x = ( x+y) + (x-y)$$
$$ sin ( alpha + beta ) = sin (alpha) cos (beta) + cos (alpha) sin (beta) $$
$$ sin ( 2x ) = sin (x+y) cos (x-y) + cos (x+y) sin (x-y)=1/3-1/2 =-1/6$$
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$ 2x = ( x+y) + (x-y)$$
$$ sin ( alpha + beta ) = sin (alpha) cos (beta) + cos (alpha) sin (beta) $$
$$ sin ( 2x ) = sin (x+y) cos (x-y) + cos (x+y) sin (x-y)=1/3-1/2 =-1/6$$
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
$$ 2x = ( x+y) + (x-y)$$
$$ sin ( alpha + beta ) = sin (alpha) cos (beta) + cos (alpha) sin (beta) $$
$$ sin ( 2x ) = sin (x+y) cos (x-y) + cos (x+y) sin (x-y)=1/3-1/2 =-1/6$$
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
edited Jul 20 at 16:17
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 14:44
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
up vote
0
down vote
You were on the right track but the equations contain the product, not the difference.
beginalign
frac12 &= sin (y-x)cos(x+y)\
&=(sin ycos x - sin x cos y)(cos xcos y - sin xsin y) \
&= sin ycos y - cos x sin x
endalign
beginalign
frac13 &= sin (x+y)cos(x-y)\
&=(sin xcos y + sin y cos x)(cos xcos y + sin xsin y) \
&= sin ycos y + cos x sin x
endalign
So subtracting them gives
$$sin 2x = 2cos xsin x = ( sin ycos y + cos x sin x) - ( sin ycos y - cos x sin x) = frac13 - frac12 = -frac16$$
answered Jul 20 at 14:49
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
You were on the right track but the equations contain the product, not the difference.
beginalign
frac12 &= sin (y-x)cos(x+y)\
&=(sin ycos x - sin x cos y)(cos xcos y - sin xsin y) \
&= sin ycos y - cos x sin x
endalign
beginalign
frac13 &= sin (x+y)cos(x-y)\
&=(sin xcos y + sin y cos x)(cos xcos y + sin xsin y) \
&= sin ycos y + cos x sin x
endalign
So subtracting them gives
$$sin 2x = 2cos xsin x = ( sin ycos y + cos x sin x) - ( sin ycos y - cos x sin x) = frac13 - frac12 = -frac16$$
answered Jul 20 at 14:49
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You were on the right track but the equations contain the product, not the difference.
beginalign
frac12 &= sin (y-x)cos(x+y)\
&=(sin ycos x - sin x cos y)(cos xcos y - sin xsin y) \
&= sin ycos y - cos x sin x
endalign
beginalign
frac13 &= sin (x+y)cos(x-y)\
&=(sin xcos y + sin y cos x)(cos xcos y + sin xsin y) \
&= sin ycos y + cos x sin x
endalign
So subtracting them gives
$$sin 2x = 2cos xsin x = ( sin ycos y + cos x sin x) - ( sin ycos y - cos x sin x) = frac13 - frac12 = -frac16$$
answered Jul 20 at 14:49
mechanodroid
22.2k52041
You were on the right track but the equations contain the product, not the difference.
beginalign
frac12 &= sin (y-x)cos(x+y)\
&=(sin ycos x - sin x cos y)(cos xcos y - sin xsin y) \
&= sin ycos y - cos x sin x
endalign
beginalign
frac13 &= sin (x+y)cos(x-y)\
&=(sin xcos y + sin y cos x)(cos xcos y + sin xsin y) \
&= sin ycos y + cos x sin x
endalign
So subtracting them gives
$$sin 2x = 2cos xsin x = ( sin ycos y + cos x sin x) - ( sin ycos y - cos x sin x) = frac13 - frac12 = -frac16$$
answered Jul 20 at 14:49
mechanodroid
22.2k52041
answered Jul 20 at 14:49
mechanodroid
22.2k52041
answered Jul 20 at 14:49
mechanodroid
22.2k52041
answered Jul 20 at 14:49
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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2
$2sin(x+y)cos(x-y)=sin2x+sin 2y$ surely?
– Lord Shark the Unknown
Jul 20 at 14:22
The second equation doesn't subtract $sin(x+y)$ and $cos(x-y)$, it multiplies them.
– Barry Cipra
Jul 20 at 14:39