What is $ sin(x)+sin(x−π)+sin(x+π) $?
$begingroup$
So I have this trig question:
$ sin(x)+sin(x−π)+sin(x+π) = $ _____
The answer is $- sin(x)$
I can't figure out how to solve it.
Any help?
trigonometry
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
asked Mar 19 at 13:16
ArkiloArkilo
555
$endgroup$
add a comment |
$begingroup$
So I have this trig question:
$ sin(x)+sin(x−π)+sin(x+π) = $ _____
The answer is $- sin(x)$
I can't figure out how to solve it.
Any help?
trigonometry
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
asked Mar 19 at 13:16
ArkiloArkilo
555
$endgroup$
$begingroup$
Do you know the formula of sine of a sum?
$endgroup$
– J. W. Tanner
Mar 19 at 13:18
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Yep I do.......
$endgroup$
– Arkilo
Mar 19 at 13:19
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@Afzal Then what does that formula tell you about $sin(x-pi)$? What about $sin(x + pi)$?
$endgroup$
– Arthur
Mar 19 at 13:20
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sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point
$endgroup$
– Arkilo
Mar 19 at 13:24
1
$begingroup$
Do you know $sin(pi)$ and $cos(pi)$?
$endgroup$
– J. W. Tanner
Mar 19 at 13:25
add a comment |
$begingroup$
So I have this trig question:
$ sin(x)+sin(x−π)+sin(x+π) = $ _____
The answer is $- sin(x)$
I can't figure out how to solve it.
Any help?
trigonometry
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
asked Mar 19 at 13:16
ArkiloArkilo
555
$endgroup$
So I have this trig question:
$ sin(x)+sin(x−π)+sin(x+π) = $ _____
The answer is $- sin(x)$
I can't figure out how to solve it.
Any help?
trigonometry
trigonometry
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
asked Mar 19 at 13:16
ArkiloArkilo
555
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
asked Mar 19 at 13:16
ArkiloArkilo
555
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
edited Mar 20 at 1:18
J. W. Tanner
3,6631320
3,6631320
asked Mar 19 at 13:16
ArkiloArkilo
555
asked Mar 19 at 13:16
ArkiloArkilo
555
asked Mar 19 at 13:16
ArkiloArkilo
555
555
$begingroup$
Do you know the formula of sine of a sum?
$endgroup$
– J. W. Tanner
Mar 19 at 13:18
$begingroup$
Yep I do.......
$endgroup$
– Arkilo
Mar 19 at 13:19
$begingroup$
@Afzal Then what does that formula tell you about $sin(x-pi)$? What about $sin(x + pi)$?
$endgroup$
– Arthur
Mar 19 at 13:20
$begingroup$
sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point
$endgroup$
– Arkilo
Mar 19 at 13:24
1
$begingroup$
Do you know $sin(pi)$ and $cos(pi)$?
$endgroup$
– J. W. Tanner
Mar 19 at 13:25
add a comment |
$begingroup$
Do you know the formula of sine of a sum?
$endgroup$
– J. W. Tanner
Mar 19 at 13:18
$begingroup$
Yep I do.......
$endgroup$
– Arkilo
Mar 19 at 13:19
$begingroup$
@Afzal Then what does that formula tell you about $sin(x-pi)$? What about $sin(x + pi)$?
$endgroup$
– Arthur
Mar 19 at 13:20
$begingroup$
sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point
$endgroup$
– Arkilo
Mar 19 at 13:24
1
$begingroup$
Do you know $sin(pi)$ and $cos(pi)$?
$endgroup$
– J. W. Tanner
Mar 19 at 13:25
$begingroup$
Do you know the formula of sine of a sum?
$endgroup$
– J. W. Tanner
Mar 19 at 13:18
$begingroup$
Do you know the formula of sine of a sum?
$endgroup$
– J. W. Tanner
Mar 19 at 13:18
$begingroup$
Yep I do.......
$endgroup$
– Arkilo
Mar 19 at 13:19
$begingroup$
Yep I do.......
$endgroup$
– Arkilo
Mar 19 at 13:19
$begingroup$
@Afzal Then what does that formula tell you about $sin(x-pi)$? What about $sin(x + pi)$?
$endgroup$
– Arthur
Mar 19 at 13:20
$begingroup$
@Afzal Then what does that formula tell you about $sin(x-pi)$? What about $sin(x + pi)$?
$endgroup$
– Arthur
Mar 19 at 13:20
$begingroup$
sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point
$endgroup$
– Arkilo
Mar 19 at 13:24
$begingroup$
sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point
$endgroup$
– Arkilo
Mar 19 at 13:24
1
1
$begingroup$
Do you know $sin(pi)$ and $cos(pi)$?
$endgroup$
– J. W. Tanner
Mar 19 at 13:25
$begingroup$
Do you know $sin(pi)$ and $cos(pi)$?
$endgroup$
– J. W. Tanner
Mar 19 at 13:25
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
$$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$
using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
$endgroup$
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
add a comment |
$begingroup$
As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmbox{and}quad sin(x+pi)=-sin x$$
If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.
The image below should help you understand why $sin(x+pi)=-sin x$.
Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
$$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
$endgroup$
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
add a comment |
$begingroup$
You probably know, that $$ sin(x−pi) = -sin(x).$$
Also $$sin(x+pi) = sin(x-pi + 2pi) = sin(x-pi)$$ so your given expression reduces to $$sin x - sin x - sin x$$
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
$endgroup$
add a comment |
$begingroup$
Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:
$$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$
$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
$endgroup$
$begingroup$
Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
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– Arkilo
Mar 19 at 13:26
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@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
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– Vasya
Mar 19 at 13:37
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Yea I wasn't evaluating the value of π in sine so that was messing it all up.
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– Arkilo
Mar 19 at 13:40
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@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
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– Vasya
Mar 19 at 13:44
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sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
$endgroup$
– Arkilo
Mar 19 at 14:05
|
show 2 more comments
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
$$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$
using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
$endgroup$
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
add a comment |
$begingroup$
$$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
$$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$
using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
$endgroup$
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
add a comment |
$begingroup$
$$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
$$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$
using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
$endgroup$
$$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
$$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$
using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
edited Mar 19 at 13:35
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
answered Mar 19 at 13:29
J. W. TannerJ. W. Tanner
3,6631320
3,6631320
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
add a comment |
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it.
$endgroup$
– Arkilo
Mar 19 at 13:39
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
You could say $sin(x-theta)=sin(x)cos(theta)-cos(x)sin(theta)$ or $sin(x+(-theta))=sin(x)cos(-theta)+cos(x)sin(-theta);$ they're the same because $cos(theta)=cos(-theta)$ and $-sin(theta)=sin(-theta)$
$endgroup$
– J. W. Tanner
Mar 19 at 13:45
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
$begingroup$
This is great, thanks man.
$endgroup$
– Arkilo
Mar 19 at 14:23
add a comment |
$begingroup$
As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmbox{and}quad sin(x+pi)=-sin x$$
If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.
The image below should help you understand why $sin(x+pi)=-sin x$.
Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
$$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
$endgroup$
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
add a comment |
$begingroup$
As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmbox{and}quad sin(x+pi)=-sin x$$
If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.
The image below should help you understand why $sin(x+pi)=-sin x$.
Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
$$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
$endgroup$
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
add a comment |
$begingroup$
As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmbox{and}quad sin(x+pi)=-sin x$$
If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.
The image below should help you understand why $sin(x+pi)=-sin x$.
Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
$$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
$endgroup$
As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmbox{and}quad sin(x+pi)=-sin x$$
If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.
The image below should help you understand why $sin(x+pi)=-sin x$.
Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
$$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
answered Mar 19 at 13:42
StackTDStackTD
23.9k2254
23.9k2254
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
add a comment |
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
Wow man, this was very neat
$endgroup$
– Arkilo
Mar 19 at 13:57
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
$begingroup$
You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions.
$endgroup$
– StackTD
Mar 19 at 14:37
add a comment |
$begingroup$
You probably know, that $$ sin(x−pi) = -sin(x).$$
Also $$sin(x+pi) = sin(x-pi + 2pi) = sin(x-pi)$$ so your given expression reduces to $$sin x - sin x - sin x$$
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
$endgroup$
add a comment |
$begingroup$
You probably know, that $$ sin(x−pi) = -sin(x).$$
Also $$sin(x+pi) = sin(x-pi + 2pi) = sin(x-pi)$$ so your given expression reduces to $$sin x - sin x - sin x$$
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
$endgroup$
add a comment |
$begingroup$
You probably know, that $$ sin(x−pi) = -sin(x).$$
Also $$sin(x+pi) = sin(x-pi + 2pi) = sin(x-pi)$$ so your given expression reduces to $$sin x - sin x - sin x$$
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
$endgroup$
You probably know, that $$ sin(x−pi) = -sin(x).$$
Also $$sin(x+pi) = sin(x-pi + 2pi) = sin(x-pi)$$ so your given expression reduces to $$sin x - sin x - sin x$$
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
answered Mar 19 at 13:31
CiaPanCiaPan
10.1k11247
10.1k11247
add a comment |
add a comment |
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Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:
$$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$
$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
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Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
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– Arkilo
Mar 19 at 13:26
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@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
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– Vasya
Mar 19 at 13:37
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Yea I wasn't evaluating the value of π in sine so that was messing it all up.
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– Arkilo
Mar 19 at 13:40
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@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
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– Vasya
Mar 19 at 13:44
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sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
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– Arkilo
Mar 19 at 14:05
|
show 2 more comments
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Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:
$$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$
$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
$endgroup$
$begingroup$
Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
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– Arkilo
Mar 19 at 13:26
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@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
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– Vasya
Mar 19 at 13:37
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Yea I wasn't evaluating the value of π in sine so that was messing it all up.
$endgroup$
– Arkilo
Mar 19 at 13:40
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@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
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– Vasya
Mar 19 at 13:44
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sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
$endgroup$
– Arkilo
Mar 19 at 14:05
|
show 2 more comments
$begingroup$
Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:
$$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$
$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
$endgroup$
Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:
$$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$
$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
edited Mar 19 at 13:33
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
answered Mar 19 at 13:24
Paras KhoslaParas Khosla
2,543322
2,543322
$begingroup$
Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
$endgroup$
– Arkilo
Mar 19 at 13:26
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@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
$endgroup$
– Vasya
Mar 19 at 13:37
$begingroup$
Yea I wasn't evaluating the value of π in sine so that was messing it all up.
$endgroup$
– Arkilo
Mar 19 at 13:40
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@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
$endgroup$
– Vasya
Mar 19 at 13:44
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sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
$endgroup$
– Arkilo
Mar 19 at 14:05
|
show 2 more comments
$begingroup$
Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
$endgroup$
– Arkilo
Mar 19 at 13:26
$begingroup$
@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
$endgroup$
– Vasya
Mar 19 at 13:37
$begingroup$
Yea I wasn't evaluating the value of π in sine so that was messing it all up.
$endgroup$
– Arkilo
Mar 19 at 13:40
$begingroup$
@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
$endgroup$
– Vasya
Mar 19 at 13:44
$begingroup$
sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
$endgroup$
– Arkilo
Mar 19 at 14:05
$begingroup$
Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
$endgroup$
– Arkilo
Mar 19 at 13:26
$begingroup$
Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it?
$endgroup$
– Arkilo
Mar 19 at 13:26
$begingroup$
@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
$endgroup$
– Vasya
Mar 19 at 13:37
$begingroup$
@Afzal: Did you try to apply $sin(alpha+beta)$ formula? Try and see that $sin(pi+x)=-sin x$
$endgroup$
– Vasya
Mar 19 at 13:37
$begingroup$
Yea I wasn't evaluating the value of π in sine so that was messing it all up.
$endgroup$
– Arkilo
Mar 19 at 13:40
$begingroup$
Yea I wasn't evaluating the value of π in sine so that was messing it all up.
$endgroup$
– Arkilo
Mar 19 at 13:40
$begingroup$
@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
$endgroup$
– Vasya
Mar 19 at 13:44
$begingroup$
@Afzal: Good, now you may try to obtain other useful formulas for $sin(pi/2-x)$, $sin(pi/2+x)$
$endgroup$
– Vasya
Mar 19 at 13:44
$begingroup$
sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
$endgroup$
– Arkilo
Mar 19 at 14:05
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sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
$endgroup$
– Arkilo
Mar 19 at 14:05
|
show 2 more comments
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Do you know the formula of sine of a sum?
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– J. W. Tanner
Mar 19 at 13:18
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Yep I do.......
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– Arkilo
Mar 19 at 13:19
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@Afzal Then what does that formula tell you about $sin(x-pi)$? What about $sin(x + pi)$?
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– Arthur
Mar 19 at 13:20
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sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point
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– Arkilo
Mar 19 at 13:24
1
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Do you know $sin(pi)$ and $cos(pi)$?
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– J. W. Tanner
Mar 19 at 13:25