Question

Question #1e1b5

Answer 1

Please see below.

Explanation:

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

`sin(x+pi/3)=sinxcos(pi/3)+cosxsin(pi/3)=sinx*(1/2)+cosx*(sqrt3/2)=sinx/2+(sqrt3cosx)/2`

`sin(x+(2pi)/3)=sinxcos((2pi)/3)+cosxsin((2pi)/3)=six*(-1/2)+cosx*(sqrt3/2)=-sinx/2+(sqrt3cosx)/2`

Now we can plug these into the problem function:

`sin(x+pi/3)+sin(x+(2pi)/3)=sinx/2+(sqrt3cosx)/2-sinx/2+(sqrt3cosx)/2=cancelcolor(red)(sinx/2)+(sqrt3cosx)/2cancelcolor(red)(-sinx/2)+(sqrt3cosx)/2=(2sqrt3cosx)/2=sqrt3cosx`

Answer 2

Kindly refer to the Explanation.

Explanation:

Recall that, `sinC+sinD=2sin((C+D)/2)sin((C-D)/2).`

Hence, `sin(x+pi/3)+sin(x+2pi/3),`

`=2sin{1/2(x+pi/3)+(x+2pi/3)}cos{1/2(x+pi/3)-(x+2pi/3)},`

`=2sin(x+pi/2)cos(-pi/6),`

`=2(cosx)(sqrt3/2),`

`=sqrt3cosx.`