Simplify trigonometric expression using trigonometric identities The Next CEO of Stack OverflowSimplify a trigonometric expressionbest way to detect the trigonometric identites that shall work on a given expression so as to simplify it accordingly?Trigonometric Identities HWsimplyfing trigonometric expression using phase shift identitiesSimplify the expression using trigonometric identitiesHow to simplify trigonometric expressionSimplify Trigonometric ExpressionHow to simplify a trigonometric expression with the identitiesTrigonometric IdentitiesHow to evaluate this trigonometric expression

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Simplify trigonometric expression using trigonometric identities



The Next CEO of Stack OverflowSimplify a trigonometric expressionbest way to detect the trigonometric identites that shall work on a given expression so as to simplify it accordingly?Trigonometric Identities HWsimplyfing trigonometric expression using phase shift identitiesSimplify the expression using trigonometric identitiesHow to simplify trigonometric expressionSimplify Trigonometric ExpressionHow to simplify a trigonometric expression with the identitiesTrigonometric IdentitiesHow to evaluate this trigonometric expression










3












$begingroup$


I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?






trigonometry







share|cite|improve this question









New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    7 hours ago










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    7 hours ago










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    32 mins ago















3












$begingroup$


I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?






trigonometry







share|cite|improve this question









New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    7 hours ago










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    7 hours ago










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    32 mins ago













3












3








3





$begingroup$


I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?






trigonometry







share|cite|improve this question









New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?






trigonometry




trigonometry






share|cite|improve this question









New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









MarianD

1,9621617




1,9621617






New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









zaz9999zaz9999

162




162




New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    7 hours ago










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    7 hours ago










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    32 mins ago
















  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    7 hours ago










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    7 hours ago










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    32 mins ago















$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
7 hours ago




$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
7 hours ago












$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
7 hours ago




$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
7 hours ago












$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
32 mins ago




$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
32 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

Using the formula for $sin (alpha - beta)$ you obtain



beginalign
2&sin x +2sin left(fracpi 3 -xright)\
= 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
= 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
endalign






share|cite|improve this answer











$endgroup$




















    4












    $begingroup$

    Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

      Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        Using the formula for $sin (alpha - beta)$ you obtain



        beginalign
        2&sin x +2sin left(fracpi 3 -xright)\
        = 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
        = 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
        = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
        endalign






        share|cite|improve this answer











        $endgroup$

















          5












          $begingroup$

          Using the formula for $sin (alpha - beta)$ you obtain



          beginalign
          2&sin x +2sin left(fracpi 3 -xright)\
          = 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
          = 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
          = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
          endalign






          share|cite|improve this answer











          $endgroup$















            5












            5








            5





            $begingroup$

            Using the formula for $sin (alpha - beta)$ you obtain



            beginalign
            2&sin x +2sin left(fracpi 3 -xright)\
            = 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
            = 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
            = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
            endalign






            share|cite|improve this answer











            $endgroup$



            Using the formula for $sin (alpha - beta)$ you obtain



            beginalign
            2&sin x +2sin left(fracpi 3 -xright)\
            = 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
            = 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
            = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
            endalign







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 6 hours ago

























            answered 7 hours ago









            MarianDMarianD

            1,9621617




            1,9621617





















                4












                $begingroup$

                Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






                share|cite|improve this answer









                $endgroup$

















                  4












                  $begingroup$

                  Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






                  share|cite|improve this answer









                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






                    share|cite|improve this answer









                    $endgroup$



                    Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 7 hours ago









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    78.4k42867




                    78.4k42867





















                        1












                        $begingroup$

                        Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$






                            share|cite|improve this answer









                            $endgroup$



                            Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 7 hours ago









                            VasyaVasya

                            4,1571618




                            4,1571618




















                                zaz9999 is a new contributor. Be nice, and check out our Code of Conduct.









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                                zaz9999 is a new contributor. Be nice, and check out our Code of Conduct.












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