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Mathematics 69 Online
OpenStudy (anonymous):

how did this step come about? from 4sin1/2xcos1/2x =2sinx

OpenStudy (anonymous):

This is a trigonometric identity. Best know as sin2x=2sinxcosx

OpenStudy (anonymous):

ok.. since sin2x=2sinxcosx, then isn't sin1/2x=2sin1/2xcos1/2x?

OpenStudy (anonymous):

no sin1/2x=2sin1/4xcos1/4x

OpenStudy (anonymous):

then why 4sin1/2xcos1/2x =2sinx

OpenStudy (anonymous):

because sinx=2sin1/2xcos1/2x

OpenStudy (anonymous):

hope it is clear

OpenStudy (anonymous):

no im very confused...=(

OpenStudy (anonymous):

sin2x=2sinxcosx

OpenStudy (anonymous):

why sin x becomes 2sin1/2xcos1/2x? did you replace x

OpenStudy (anonymous):

you have sin2x=2sinxcosx so what happens is that you multiply by two than for the sincos part you divide by two to get from sin2x=2sinxcosx so sinx=2* the sincos part you have to devide by two that is sin1/2xcos1/2x so sinx=2sin1/2xcos1/2x

OpenStudy (phi):

First, 2sin(x) cos(x)= sin(2x) comes from sin(A+B)= sin(A)cos(B)+sin(B)cos(A) for sin(A+B) proof see http://www.youtube.com/watch?v=zw0waJCEc-w now for your particular problem \[4sin(x/2)cos(x/2)=2⋅(2sin(x/2)cos(x/2))\] 2⋅(2sin(x/2)cos(x/2))=2⋅(sin(2⋅x/2)=2sin(x)

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