Practice Exercise 2.1

The principal value branch of sin⁻¹ is [-π/2, π/2].

Let sin⁻¹(-√3/2) = y.
Then sin y = -√3/2.
We know sin(π/3) = √3/2.
Since sin(-x) = -sin(x), sin(-π/3) = -√3/2.
-π/3 lies in [-π/2, π/2].

Principal branch of cos⁻¹ is [0, π].

cos y = 1/2.
We know cos(Ï€/3) = 1/2.
π/3 lies in [0, π].

[-π/2, π/2] - {0}

cosec y = √2.
Since sin(π/4) = 1/√2, cosec(π/4) = √2.
Ï€/4 is in the valid range.

(-π/2, π/2)

tan y = -1/√3.
tan(π/6) = 1/√3.
Since tan(-x) = -tan(x), tan(-π/6) = -1/√3.

[0, π]

cos y = -1/√2.
cos(π/4) = 1/√2.
Since cos is negative in 2nd quadrant: cos(π - π/4) = cos(3π/4) = -1/√2.

(-π/2, π/2)

tan y = -√3.
tan(π/3) = √3.
tan(-π/3) = -√3.

[0, π] - {π/2}

sec y = 2.
cos y = 1/2.
y = π/3.

(0, π)

cot y = -1.
cot(Ï€/4) = 1.
For cot, negative values are in 2nd quadrant: π - π/4 = 3π/4.

[0, π]

cos y = -√3/2.
cos(π/6) = √3/2.
2nd quadrant: π - π/6 = 5π/6.

[-π/2, π/2] - {0}

cosec y = -2.
sin y = -1/2.
y = -Ï€/6.

= π/3

sec y = -2 (2nd quad for sec⁻¹). y = π - π/3 = 2π/3.

cosec y = -2/√3. sin y = -√3/2. y = -π/3.

π/3 + 2π/3 - π/3 = 2π/3.

cos⁻¹(1/2) = π/3.
sin⁻¹(1/2) = π/6.

π/3 + 2(π/6) = π/3 + π/3 = 2π/3.

The principal value branch (range) of sin⁻¹x is the closed interval [-π/2, π/2].

= π/4

= π - π/4 = 3π/4 (cot⁻¹ range is (0, π)).

Ï€/4 - 3Ï€/4 = -2Ï€/4 = -Ï€/2.