Question 579604
What is the exact value of sin 19π/12?
(19π/12)=(10π/12+9π/12)=(5π/6+3π/4)
sin(19π/12)=sin(5π/6+3π/4)
sin addition formula:
sin(s+t)=sin s cos t + cos s sin t
s=5π/6
t=3π/4
Both angles in quadrant II where sin>0, and cos<0
sin(19&#960;/12)=sin(5&#960;/6+3&#960;/4)
sin(19&#960;/12)=sin(5&#960;/6)cos(3&#960;/4)+cos(5&#960;/6)sin(3&#960;/4)
sin(19&#960;/12)=1/2*-&#8730;2/2+-&#8730;3/2*&#8730;2/2
sin(19&#960;/12)=-&#8730;2/4+-&#8730;6/4=-(&#8730;2+&#8730;6)/4