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

Question

4157 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-29T18:41:50

# sqrt3/2#.

Recall the following definitions of #arc sin and arc cos# functions :

#arc sinx=theta, |x|le1 iff sintheta=x, |theta|lepi/2#.

#arc cosx=theta, |x|le1 iff costheta=x, theta in [0,pi]#.

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#.