Rewriting $sin(x+frac {pi} {6})cos(x)$ as $frac {1} {4}(2sin(2x+frac {pi} {6})+1)$ [closed]












0












$begingroup$


If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$






trigonometry







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closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:06










  • $begingroup$
    Sorry I made a mistake
    $endgroup$
    – Were
    Dec 3 '18 at 1:11










  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 3 '18 at 2:28
















0












$begingroup$


If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$






trigonometry







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:06










  • $begingroup$
    Sorry I made a mistake
    $endgroup$
    – Were
    Dec 3 '18 at 1:11










  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 3 '18 at 2:28














0












0








0





$begingroup$


If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$






trigonometry







share|cite|improve this question











$endgroup$




If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 1:41









Blue

47.8k870152




47.8k870152










asked Dec 3 '18 at 0:57









WereWere

33




33




closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:06










  • $begingroup$
    Sorry I made a mistake
    $endgroup$
    – Were
    Dec 3 '18 at 1:11










  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 3 '18 at 2:28


















  • $begingroup$
    Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:06










  • $begingroup$
    Sorry I made a mistake
    $endgroup$
    – Were
    Dec 3 '18 at 1:11










  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 3 '18 at 2:28
















$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06




$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06












$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11




$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11












$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28




$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28










1 Answer
1






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oldest

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From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get



$$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$



Hence



$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$



And since $sin(pi/6)=1/2$,



$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get



    $$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$



    Hence



    $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$



    And since $sin(pi/6)=1/2$,



    $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get



      $$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$



      Hence



      $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$



      And since $sin(pi/6)=1/2$,



      $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get



        $$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$



        Hence



        $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$



        And since $sin(pi/6)=1/2$,



        $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$






        share|cite|improve this answer









        $endgroup$



        From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get



        $$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$



        Hence



        $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$



        And since $sin(pi/6)=1/2$,



        $$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 1:18









        Jean-Claude ArbautJean-Claude Arbaut

        14.7k63464




        14.7k63464















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