How do you find the exact value of sin[arccos(-2/3)+arcsin(1/4)]sin[arccos(−23)+arcsin(14)]?
1 Answer
Jul 26, 2016
Explanation:
Let
sin(alpha) = sqrt(1-cos^2(alpha)) = sqrt(1-(-2/3)^2) = sqrt(5/9) = sqrt(5)/3sin(α)=√1−cos2(α)=√1−(−23)2=√59=√53
cos(beta) = sqrt(1-sin^2(beta))= sqrt(1-(1/4)^2) = sqrt(15/16) = sqrt(15)/4cos(β)=√1−sin2(β)=√1−(14)2=√1516=√154
Then:
sin(alpha+beta)sin(α+β)
= sin alpha cos beta + sin beta cos alpha=sinαcosβ+sinβcosα
=(sqrt(5)/3)(sqrt(15)/4) + (1/4)(-2/3)=(√53)(√154)+(14)(−23)
=(5sqrt(3))/12-1/6=5√312−16
