cos[sin inverse(-4/5)-cos inverse(4/5)]=

Question

options....

1)Π/2

2)-Π/2

3)0

4)Π

Answer 1

inverse sin is arcsin or sin^-1, inverse of cos arccos or cos^-1

cos[arcsin(-4/5)-arcscos(4/5)]=

=cos[arcsin(-4/5)]cos[arcscos(4/5)]+sin[arcsin(-4/5)]sin[arcscos(4/5)]=

     |cos[arccos(x)]= x, cos[arcsin(x)=√(1-x²)|

     |sin[arcsin(x)] = x, sin[arccos(x)=√(1-x²)|

 

=√(1-(-4/5)²*(4/5)+(-4/5)*(√(1-(4/5)²)=

=(3/5)*(4/5)-(4/5)*(3/5)=0.

Note: Don't forget about range and domain for inverse functions.