Asked by help please

HOW CAN I SOLVE THIS? sin(350°+θ)cos(110°+θ)-cos(350°+θ)sin(110°+θ)
Answered by mathhelper
Use the sin (A - B) identity:
Sin(A - B) = sinAcosB - cosAsinB

here you A = 350+θ , and B = 110+θ

sin(350°+θ)cos(110°+θ)-cos(350°+θ)sin(110°+θ)
= sin(350+θ - (110+θ) )
= sin(240°)
= - sin 60°
= - √3 / 2
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