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










2












$begingroup$


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?










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
    5 hours ago










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















2












$begingroup$


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?










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
    5 hours ago










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













2












2








2





$begingroup$


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?










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






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 5 hours ago









MarianD

1,9521617




1,9521617






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 5 hours ago









zaz9999zaz9999

112




112




New contributor




zaz9999 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    5 hours ago










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
















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










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















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




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












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




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










3 Answers
3






active

oldest

votes


















4












$begingroup$

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






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    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











    $endgroup$




















      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









        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 5 hours ago









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            78.4k42867




            78.4k42867





















                4












                $begingroup$

                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












                  $begingroup$

                  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












                    4








                    4





                    $begingroup$

                    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 5 hours ago

























                    answered 5 hours ago









                    MarianDMarianD

                    1,9521617




                    1,9521617





















                        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 5 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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