How do you find the exact value of sin[arcsin(1/2) + arccos (0) ] sin[arcsin(12)+arccos(0)]?

1 Answer
Apr 29, 2018

sqrt3/232.

Explanation:

Recall the following definitions of arc sin and arc cosarcsinandarccos functions :

arc sinx=theta, |x|le1 iff sintheta=x, |theta|lepi/2arcsinx=θ,|x|1sinθ=x,|θ|π2.

arc cosx=theta, |x|le1 iff costheta=x, theta in [0,pi]arccosx=θ,|x|1cosθ=x,θ[0,π].

Now, sin(pi/6)=1/2, and, |pi/6| lepi/2 :. arcsin(1/2)=pi/6.

Similarly, arc cos0=pi/2.

:. sin[arcsin(1/2)+arc cos0]=sin(pi/6+pi/2),

=cos(pi/6),

=sqrt3/2.