Prove sin(x+2y) + sin(x+2z) = 2cos(y-z) if x+y+z=90?

Question

don't forget that x+y+z=90
Thanks

Answer 1

solve the left side...
recall the rule for addition of sines
i-e.
Sin (A) + Sin(B) = 2Sin{(A+B)/2}.Cos{(A-B)/2}

Apply to the LHS

2Sin{(2X+2Y+2Z)/2}.Cos{(X + 2Y - X - 2Z)/2}
2Sin(90)Cos(Y-Z)
2Cos(Y-Z)

= RHS

Answer 2

x+z =90-y
x+2z =90-y+z= 90-(y-z)
So sin(x+2z)= sin(90-(y-z)) = cos(y-z)
So sin(x+2y) = cos(y-z)
x+y=90-z
x+2y=90-z+y= 90-(z-y))
So sin(x+2y) = sin(90-(z-y) =sin(90+(y-z)= cos(y-z )
cos(y-z)=cos(y-z)

Answer 3

Use sin a + sin b = 2sin[ 0.5 (a + b)] cos[0.5 (a - b)].

Trivial.