prove the following trig identity:

Question

a) sin(pi/6 + x) + sin (pi/3 + x) + sin (pi/2 + x) = ((sqrt3) +1)/2 (sinx +(sqrt3)cosx)

b) sin(pi/4 + x) + sin(pi/4 - 4)= (sqrt2)cosx

Answer 1

remember
sin(A+B) = sinAcosB + cosAsinB
and
sin pi/6 = 1/2, cos pi/6 = √3/2
sin pi/3 = √3/2 , cos pi/3 = 1/2
sin pi/2 = 1 , cos pi/2 = 0

simplify your expanded left side, by factoring and collection like terms.

both questions should come out ,
let me know where you get stuck.

Answer 2

I get stuck on the last step of both questions:

ie:

I have:
0.5cosx+sqrt3/2cosx+cosx+sqrt3/2sinx+0.5sinx

then if I factor out
(sqrt3+1/2) I get sinx but i don't get sqrt3cosx

for the second q
I don't know how to get the sqrt2cosx, but I end up with 2cosx instead, where is the sqrt?

Answer 3

for the second, I believe you have a typo in
sin(pi/4 + x) + sin(pi/4 - 4)= (sqrt2)cosx
and you meant
sin(pi/4 + x) + sin(pi/4 - x)= (sqrt2)cosx

LS =
sinpi/4 cosx + cospi/4 sinx + sinpi/4 cosx - cospi/4 sinx
= (√2/2)cosx + (√2/2)cosx
= √2cosx
= RS

YOu might be using sinpi/4 = 1/√2
which of course is √2/2 after rationalizing.