What is $ sin(x)+sin(x−π)+sin(x+π) $?












4












$begingroup$


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











$endgroup$












  • $begingroup$
    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
















4












$begingroup$


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











$endgroup$












  • $begingroup$
    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














4












4








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











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 1:18









J. W. Tanner

4,0461320




4,0461320










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


















  • $begingroup$
    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










4 Answers
4






active

oldest

votes


















3












$begingroup$

$$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
$$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$



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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















4












$begingroup$

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 $sin(x+pi)=-sin x$.



enter image description here



Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
$$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















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









$endgroup$





















    0












    $begingroup$

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



    $$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$



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






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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












    Your Answer





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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
    $$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$



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






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    3












    $begingroup$

    $$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
    $$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$



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






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    3












    3








    3





    $begingroup$

    $$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
    $$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$



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






    share|cite|improve this answer











    $endgroup$



    $$sin(x)+color{green}{sin(x-pi)}+color{red}{sin(x+pi)}$$ $$=sin(x)+color{green}{sin(x)cos(-pi)+cos(x)sin(-pi)}+color{red}{sin(x)cos(pi)+cos(x)sin(pi)}$$
    $$=sin(x)color{green}{-sin(x)}color{red}{-sin(x)}=-sin(x)$$



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







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 19 at 13:35

























    answered Mar 19 at 13:29









    J. W. TannerJ. W. Tanner

    4,0461320




    4,0461320












    • $begingroup$
      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










    • $begingroup$
      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











    4












    $begingroup$

    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 $sin(x+pi)=-sin x$.



    enter image description here



    Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
    $$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    4












    $begingroup$

    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 $sin(x+pi)=-sin x$.



    enter image description here



    Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
    $$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    4












    4








    4





    $begingroup$

    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 $sin(x+pi)=-sin x$.



    enter image description here



    Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
    $$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$






    share|cite|improve this answer









    $endgroup$



    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 $sin(x+pi)=-sin x$.



    enter image description here



    Then note that by "adding a full cirle", the same holds for the angle $x-pi$:
    $$sin(x-pi)=sin(x-picolor{blue}{+2pi})=sin(x+pi)=-sin x$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 19 at 13:42









    StackTDStackTD

    24.3k2254




    24.3k2254












    • $begingroup$
      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










    • $begingroup$
      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











    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









    $endgroup$


















      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









      $endgroup$
















        1












        1








        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









        $endgroup$



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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 19 at 13:31









        CiaPanCiaPan

        10.1k11247




        10.1k11247























            0












            $begingroup$

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



            $$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$



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






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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
















            0












            $begingroup$

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



            $$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$



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






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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














            0












            0








            0





            $begingroup$

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



            $$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$



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






            share|cite|improve this answer











            $endgroup$



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



            $$begin{aligned}lambda&=sin x+sin(x-pi)+sin(x+pi)\&= sin x-sin(pi -x)+sin(pi+x)\&=sin x-sin x-sin x=-sin xend{aligned}$$



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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 19 at 13:33

























            answered Mar 19 at 13:24









            Paras KhoslaParas Khosla

            2,681423




            2,681423












            • $begingroup$
              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










            • $begingroup$
              @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




            $begingroup$
            sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)?
            $endgroup$
            – Arkilo
            Mar 19 at 14:05


















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