geometric proof of $2cosAcosB=cos(A+B)+cos(A-B)$

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I have seen geometric proof of identities
$$cos(A+B)=cosAcosB-sinAsinB$$
and
$$cos(A-B)=cosAcosB+sinAsinB$$



By adding two equation, $$2cosAcosB=cos(A+B)+cos(A-B)$$.



But how to prove this by geometry?



Thank you.




geometry trigonometry




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  • 1




    If $cos(A+B)=cosAcosB-sinAsinB$ and $cos(A-B)=cosAcosB+sinAsinB$ were proven geometrically, doesn't that mean you have already shown $2cosAcosB=cos(A+B)+cos(A-B)$ geometrically?
    – graydad
    Feb 25 '15 at 16:52














up vote
4
down vote

favorite
1












I have seen geometric proof of identities
$$cos(A+B)=cosAcosB-sinAsinB$$
and
$$cos(A-B)=cosAcosB+sinAsinB$$



By adding two equation, $$2cosAcosB=cos(A+B)+cos(A-B)$$.



But how to prove this by geometry?



Thank you.




geometry trigonometry




share|cite|improve this question















  • 1




    If $cos(A+B)=cosAcosB-sinAsinB$ and $cos(A-B)=cosAcosB+sinAsinB$ were proven geometrically, doesn't that mean you have already shown $2cosAcosB=cos(A+B)+cos(A-B)$ geometrically?
    – graydad
    Feb 25 '15 at 16:52












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





I have seen geometric proof of identities
$$cos(A+B)=cosAcosB-sinAsinB$$
and
$$cos(A-B)=cosAcosB+sinAsinB$$



By adding two equation, $$2cosAcosB=cos(A+B)+cos(A-B)$$.



But how to prove this by geometry?



Thank you.




geometry trigonometry




share|cite|improve this question











I have seen geometric proof of identities
$$cos(A+B)=cosAcosB-sinAsinB$$
and
$$cos(A-B)=cosAcosB+sinAsinB$$



By adding two equation, $$2cosAcosB=cos(A+B)+cos(A-B)$$.



But how to prove this by geometry?



Thank you.





geometry trigonometry





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Feb 25 '15 at 16:34









kong

27516




27516







  • 1




    If $cos(A+B)=cosAcosB-sinAsinB$ and $cos(A-B)=cosAcosB+sinAsinB$ were proven geometrically, doesn't that mean you have already shown $2cosAcosB=cos(A+B)+cos(A-B)$ geometrically?
    – graydad
    Feb 25 '15 at 16:52












  • 1




    If $cos(A+B)=cosAcosB-sinAsinB$ and $cos(A-B)=cosAcosB+sinAsinB$ were proven geometrically, doesn't that mean you have already shown $2cosAcosB=cos(A+B)+cos(A-B)$ geometrically?
    – graydad
    Feb 25 '15 at 16:52







1




1




If $cos(A+B)=cosAcosB-sinAsinB$ and $cos(A-B)=cosAcosB+sinAsinB$ were proven geometrically, doesn't that mean you have already shown $2cosAcosB=cos(A+B)+cos(A-B)$ geometrically?
– graydad
Feb 25 '15 at 16:52




If $cos(A+B)=cosAcosB-sinAsinB$ and $cos(A-B)=cosAcosB+sinAsinB$ were proven geometrically, doesn't that mean you have already shown $2cosAcosB=cos(A+B)+cos(A-B)$ geometrically?
– graydad
Feb 25 '15 at 16:52










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$$beginalign
2 cos A cos B &= cos(A-B)+cos(A+B) \[6pt]
2 sin A ,sin B &= cos(A-B)-cos(A+B)
endalign$$



Note. Although not labeled (yet), these identities are also evident:



$$beginalign
2 ,sin A cos B &= sin(A+B)+sin(A-B) \[6pt]
2 cos A ,sin B &= sin(A+B)-sin(A-B)
endalign$$






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    enter image description here



    $$beginalign
    2 cos A cos B &= cos(A-B)+cos(A+B) \[6pt]
    2 sin A ,sin B &= cos(A-B)-cos(A+B)
    endalign$$



    Note. Although not labeled (yet), these identities are also evident:



    $$beginalign
    2 ,sin A cos B &= sin(A+B)+sin(A-B) \[6pt]
    2 cos A ,sin B &= sin(A+B)-sin(A-B)
    endalign$$






    share|cite|improve this answer



























      up vote
      6
      down vote













      enter image description here



      $$beginalign
      2 cos A cos B &= cos(A-B)+cos(A+B) \[6pt]
      2 sin A ,sin B &= cos(A-B)-cos(A+B)
      endalign$$



      Note. Although not labeled (yet), these identities are also evident:



      $$beginalign
      2 ,sin A cos B &= sin(A+B)+sin(A-B) \[6pt]
      2 cos A ,sin B &= sin(A+B)-sin(A-B)
      endalign$$






      share|cite|improve this answer

























        up vote
        6
        down vote










        up vote
        6
        down vote









        enter image description here



        $$beginalign
        2 cos A cos B &= cos(A-B)+cos(A+B) \[6pt]
        2 sin A ,sin B &= cos(A-B)-cos(A+B)
        endalign$$



        Note. Although not labeled (yet), these identities are also evident:



        $$beginalign
        2 ,sin A cos B &= sin(A+B)+sin(A-B) \[6pt]
        2 cos A ,sin B &= sin(A+B)-sin(A-B)
        endalign$$






        share|cite|improve this answer















        enter image description here



        $$beginalign
        2 cos A cos B &= cos(A-B)+cos(A+B) \[6pt]
        2 sin A ,sin B &= cos(A-B)-cos(A+B)
        endalign$$



        Note. Although not labeled (yet), these identities are also evident:



        $$beginalign
        2 ,sin A cos B &= sin(A+B)+sin(A-B) \[6pt]
        2 cos A ,sin B &= sin(A+B)-sin(A-B)
        endalign$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 25 '15 at 22:21


























        answered Feb 25 '15 at 20:30









        Blue

        43.7k868141




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