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What is $ sin(x)+sin(x−π)+sin(x+π) $?

How to solve $sin x +cos x = 1$?solve $3sintheta+4costheta=0$Trig identity $fraccos xsec x + fracsin xcsc x = csc^2x - cot^2x$solving the integral $sqrtsin(y)(4-y)$What identity was used in this Trigonometry problem?Use Trig Identities to solve $sin x cos^4 x+cos^6 x$Solve $cos 2x - sin 2x = sqrt 3cos 4x$Trigonometric simple equationThe lengths of a triangle's sides are consecutive integers. If the largest angle is twice the smallest angle, find the cosine of the smallest angle?Trouble proving the trigonometric identity $frac1-2sin(x)sec(x)=fraccos(3x)1+2sin(x)$

4

2

$begingroup$

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

share|cite|improve this question

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

4

2

$begingroup$

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

share|cite|improve this question

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

$begingroup$

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

share|cite|improve this question

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

asked Mar 19 at 13:16

ArkiloArkilo

555

$endgroup$

share|cite|improve this question

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

asked Mar 19 at 13:16

ArkiloArkilo

555

share|cite|improve this question

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

asked Mar 19 at 13:16

ArkiloArkilo

555

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

edited Mar 20 at 1:18

J. W. Tanner

4,4691320

asked Mar 19 at 13:16

ArkiloArkilo

555

asked Mar 19 at 13:16

ArkiloArkilo

555

4 Answers
4

active

oldest

votes

3

$begingroup$

$$sin(x)+colorgreensin(x-pi)+colorredsin(x+pi)$$ $$=sin(x)+colorgreensin(x)cos(-pi)+cos(x)sin(-pi)+colorredsin(x)cos(pi)+cos(x)sin(pi)$$
$$=sin(x)colorgreen-sin(x)colorred-sin(x)=-sin(x)$$

using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$

share|cite|improve this answer

edited Mar 19 at 13:35

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

4

$begingroup$

As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmboxandquad sin(x+pi)=-sin x$$
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 $x-pi$:
$$sin(x-pi)=sin(x-picolorblue+2pi)=sin(x+pi)=-sin x$$

share|cite|improve this answer

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

1

$begingroup$

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 $$sin x - sin x - sin x$$

share|cite|improve this answer

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

$endgroup$

add a comment | 

0

$begingroup$

Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:

$$beginalignedlambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xendaligned$$

$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$

share|cite|improve this answer

edited Mar 19 at 13:33

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – 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
  

  
  
  
  

 | 
show 2 more comments

Your Answer

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3

$begingroup$

$$sin(x)+colorgreensin(x-pi)+colorredsin(x+pi)$$ $$=sin(x)+colorgreensin(x)cos(-pi)+cos(x)sin(-pi)+colorredsin(x)cos(pi)+cos(x)sin(pi)$$
$$=sin(x)colorgreen-sin(x)colorred-sin(x)=-sin(x)$$

using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$

share|cite|improve this answer

edited Mar 19 at 13:35

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

3

$begingroup$

$$sin(x)+colorgreensin(x-pi)+colorredsin(x+pi)$$ $$=sin(x)+colorgreensin(x)cos(-pi)+cos(x)sin(-pi)+colorredsin(x)cos(pi)+cos(x)sin(pi)$$
$$=sin(x)colorgreen-sin(x)colorred-sin(x)=-sin(x)$$

using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$

share|cite|improve this answer

edited Mar 19 at 13:35

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

$begingroup$

$$sin(x)+colorgreensin(x-pi)+colorredsin(x+pi)$$ $$=sin(x)+colorgreensin(x)cos(-pi)+cos(x)sin(-pi)+colorredsin(x)cos(pi)+cos(x)sin(pi)$$
$$=sin(x)colorgreen-sin(x)colorred-sin(x)=-sin(x)$$

using the formula for $sin(x+theta)$ and the facts that $cos(pmpi)=-1$ and $sin(pmpi)=0$

share|cite|improve this answer

edited Mar 19 at 13:35

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

share|cite|improve this answer

edited Mar 19 at 13:35

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

answered Mar 19 at 13:29

J. W. TannerJ. W. Tanner

4,4691320

4

$begingroup$

As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmboxandquad sin(x+pi)=-sin x$$
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 $x-pi$:
$$sin(x-pi)=sin(x-picolorblue+2pi)=sin(x+pi)=-sin x$$

share|cite|improve this answer

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

4

$begingroup$

As shown in some other answers, this is very simple if you know that:
$$sin(x-pi)=-sin x quadmboxandquad sin(x+pi)=-sin x$$
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 $x-pi$:
$$sin(x-pi)=sin(x-picolorblue+2pi)=sin(x+pi)=-sin x$$

share|cite|improve this answer

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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:
$$sin(x-pi)=-sin x quadmboxandquad sin(x+pi)=-sin x$$
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 $x-pi$:
$$sin(x-pi)=sin(x-picolorblue+2pi)=sin(x+pi)=-sin x$$

share|cite|improve this answer

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

$endgroup$

share|cite|improve this answer

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

answered Mar 19 at 13:42

StackTDStackTD

24.5k2254

1

$begingroup$

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 $$sin x - sin x - sin x$$

share|cite|improve this answer

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

$endgroup$

add a comment | 

1

$begingroup$

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 $$sin x - sin x - sin x$$

share|cite|improve this answer

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

$endgroup$

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 $$sin x - sin x - sin x$$

share|cite|improve this answer

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

$endgroup$

share|cite|improve this answer

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

answered Mar 19 at 13:31

CiaPanCiaPan

10.3k11248

0

$begingroup$

Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:

$$beginalignedlambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xendaligned$$

$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$

share|cite|improve this answer

edited Mar 19 at 13:33

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – 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
  
  

  
  
  
  


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  $begingroup$
  @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
  

  
  
  
  

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Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:

$$beginalignedlambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xendaligned$$

$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$

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edited Mar 19 at 13:33

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523

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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 $sin(pi/2-x)$, $sin(pi/2+x)$
  $endgroup$
  – 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

$begingroup$

Note that $sin(pi -x)=sin x$ and $sin(pi+x)=-sin x$, using which we get:

$$beginalignedlambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xendaligned$$

$$sin(pi -x)=sin pi cos x-sin xcospi=+sin x \ sin(pi+x)=sinpicos x+sin xcos pi =-sin x$$

share|cite|improve this answer

edited Mar 19 at 13:33

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523

$endgroup$

share|cite|improve this answer

edited Mar 19 at 13:33

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523

answered Mar 19 at 13:24

Paras KhoslaParas Khosla

2,857523