answer : 5pi/4 because arcsin is the inverse operation of sin so doing sin then arcsin is the same as doing nothing!
2. cos (5pi/12)
we can use the addition formulae. eg 5/12 = 2/12 + 3/12 cos(5pi/12) = cos(2pi/12 + 3pi/12) = cos(pi/6 + pi/4) = cos(pi/6)cos(pi/4) - sin(pi/6)sin(pi/4) = [1/2][1/sqrt(2)] - [1/2][(1/2)sqrt(3)] = [1/4][sqrt(2) - sqrt(3)] = - [1/4][sqrt(3) - sqrt(2)].
3. sin(5pi/8) = cos(pi/8) because sin(x) = cos(x-pi/2) = cos(x-4pi/8).
Using cos(2x) = 2cos^2(x)-1 we get cos^2(x) = (1/2)[cos(2x)+1] cos^2(pi/8) = (1/2)[cos(pi/4)+1] = (1/2)[1/sqrt(2)+1] = (1/4)[sqrt(2)+2].
answer : cos(pi/8) = (1/2)sqrt((2+sqrt(2)).
4. sin(arcsin(3/5) - arccos(3/5)).
Imagine a 3-4-5 right triangle with base 4 units, opposite 3 units and hypotenuse 5 units. If base angle is A and the vertex angle is B then sinA = cosB = 3/5 A = arcsin(3/5) and B = arccos(3/5) sinB = cosA = 4/5.
