How to show work for composite inverse trig functions?

Question

a. arcsin(-√2/2) = -π/4 
b. arccos(-√2/2) = 3π/4 
c. sin(arcsin(-√2/2)) = -1/√2 
d. arcsin(sin(7π/6)) = -π/6 

Answer 1

  

• a).  sin^{- 1} ( - √2/2).

= sin^{- 1} [ - √2/(√2 * √2)]

= sin^{- 1} [- 1/√2].

= sin^{- 1} ( sin (- π/4))

sin(sin^-1(x)) = x   for every x in the interval [-1, 1].

sin^-1(sin(x)) = x   for every x in the interval [ - π/2, π/2].

= - π/4.

∴ sin^{- 1} ( - √2/2) = - π/4.

• b).  cos^{- 1} ( - √2/2).

= cos^{- 1} [ - √2/(√2 * √2)]

= cos^{- 1} [- 1/√2].

The function cos(θ) is negative in second and third quadrant.

= cos^{- 1} ( cos (- π/4))

= cos^{- 1} ( cos (π - π/4))

= cos^{- 1} ( cos (3π/4))

cos(cos^-1(x)) = x   for every x in the interval [-1, 1].

cos^-1(cos(x)) = x   for every x in the interval [ - π/2, π/2].

= 3π/4.

∴ cos^{- 1} ( - √2/2) = 3π/4.

• c).  sin(sin^{- 1} ( - √2/2)).

= sin(sin^{- 1} [ - √2/(√2 * √2)])

= sin(sin^{- 1} [- 1/√2]).

sin(sin^-1(x)) = x   for every x in the interval [-1, 1].

sin^-1(sin(x)) = x   for every x in the interval [ - π/2, π/2].

= - 1/√2.

∴ sin(sin^{- 1} ( - √2/2)) = - 1/√2.

• d).  sin^{- 1}(sin (7π/6)).

= sin^{- 1}(sin (π + π/6))

= sin^{- 1}(sin[- π/6])

sin(sin^-1(x)) = x   for every x in the interval [-1, 1].

sin^-1(sin(x)) = x   for every x in the interval [ - π/2, π/2].

= - π/6.

∴sin^{- 1}(sin (7π/6)) = - π/6.