Pre calc inverse help?

sin⁻¹(1) → here, the goal is to find an angle where the sine is 1 → the angle is (π/2)

cos⁻¹(1/2) → here, the goal is to find an angle where the cosine is 1/2 → the angle is (π/3)


= sin[sin⁻¹(1) + cos⁻¹(1/2)]

= sin[(π/2) + (π/3)]

= sin[(3π/6) + (2π/6)]

= sin(5π/6)

= sin[(6π - π)/6]

= sin[(6π/6) - (π/6)]

= sin[π - (π/6)] → recall: sin(π - x) = sin(x)

= sin(π/6)

= 1/2

Let sin⁻¹(1) = θ
sin(θ) = 1
θ = 90°

Let cos⁻¹(1/2) = ϕ
cos(ϕ) = 1/2
ϕ = 60° or ϕ = 300°

Let y = sin[sin⁻¹(1) + cos⁻¹(1/2)]
y = sin(θ + ϕ)
y = sin(90° + 60°)   or  y = sin(90° + 300°)
y = sin(150°)           or  y = sin(390°)
y = sin(180° - 30°)  or  y = sin(360° - 30)
y = sin(30°)
y = 1/2

Hence, sin[sin⁻¹(1) + cos⁻¹(1/2)] = 1/2

sin[sin^-1(1)+cos^-1(1/2)]
=
sin[sin^-1(1)cos[cos^-1(1/2)]+cos[sin^-1(1)]sin[cos^-1(1/2)]
=
1(1/2)+cos[pi/2]sin[pi/3]
{for sin(pi/2)=1 & cos(pi/3)=1/2}
=
1/2+0*sin[pi/3]
=
1/2 (answer)

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)
sin(x)^2 + cos(x)^2 = 1
sin(arcsin(t)) = t
cos(arccos(t)) = t

Using those identities, we can move forward

sin(arcsin(1) + arccos(1/2)) =>
sin(arcsin(1)) * cos(arccos(1/2)) + sin(arccos(1/2)) * cos(arcsin(1)) =>
1 * (1/2) + sqrt(1 - cos(arccos(1/2))^2) * sqrt(1 - sin(arcsin(1))^2) =>
(1/2) + sqrt(1 - (1/2)^2) * sqrt(1 - 1^2) =>
(1/2) + sqrt(1 - 1/4) * sqrt(1 - 1) =>
(1/2) + sqrt(3/4) * sqrt(0) =>
(1/2) + (sqrt(3)/2) * 0 =>
(1/2) + 0 =>
(1/2)

Or, we can find the angles for arcsin(1) and arccos(1/2)

arcsin(1) = pi/2
arccos(1/2) = pi/3

pi/2 + pi/3 = 3pi/6 + 2pi/6 = 5pi/6

sin(5pi/6) = 1/2

sin[ pi/2 + pi/3] = 1/2, by common triangle facts.

Sin[ sin^-1 1+ cos^-1 1/2)
expand it sin (sin^-1 1) cos(cos^-1 1/2) + sin (cos^-1 1/2) cos (sin^-1 1)=1/2+ sin 60 cos 90
=1/2,  cos 60 =1/2, 60=cos^-1 (1/2), sin 90=1
90= sin^-1 1