Question

How do I prove that `2 sin ((C+D)/2) cos ((C-D)/2) = sin C+sin D`?

Answer 1

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)sin(A+B)=sinAcosB+cosAsinB`

`•color(white)(x)sin(A-B)=sinAcosB-cosAsinB`

`"Adding the 2 equations gives"`

`sin(A+B)+sin(A-B)=2sinAcosB`

`"Subtracting the 2 equations gives"`

`sin(A+B)-sin(A-B)=2cosAsinB`

`"let "C=A+B" and "D=A-B`

`rArrA=(C+D)/2" and "B=(C-D)/2`

`rArrsinC+sinD=2sin((C+D)/2)cos((C-D)/2)`

Answer 2

See the proof below

Explanation:

We need

`sin(a+b)=sinacosb+sinbcosa`

`cos(a-b)=cosacosb+sinasinb`

`sin^2a+cos^2a=1`

`sin2a=2sinacosa`

Therefore,

`LHS=2sin((C+D)/2)cos((C-D)/2)`

`=2(sin(C/2)cos(D/2)+sin(D/2)cos(C/2))(cos(C/2)cos(D/2)+sin(C/2)sin(D/2))`

`=2(sin(C/2)cos(C/2)cos^2(D/2)+sin(D/2)cos(D/2)sin^2(C/2)+sin(D/2)cos(D/2)cos^2(C/2)+sin(C/2)cos(C/2)sin^2(D/2))`

`=2*1/2(sinCcos^2(D/2)+sinDsin^2(C/2)+sinDcos^2(C/2)+sinCsin^2(D/2))`

`=sinC(cos^2(D/2)+sin^2(D/2))+sinD(sin^2(C/2)+cos^2(C/2))`

`=sinC+sinD`

`=RHS`

`QED`