The inverse trig functions θ = sin⁻¹x, tan⁻¹x, cot⁻¹x, sec⁻¹x always return values in QI or QIV (between −π/2 and π/2). This is just to give you a general idea of where θ is located for any of these functions. The range for each is slightly different because the trig functions sinx, tanx, cotx, secx are not all defined at all values in this interval.
The inverse trig functions θ = cos⁻¹x, sec⁻¹x always return values in QI or QII (between 0 and π). Again the range for cos⁻¹x and sec⁻¹x is slightly different, but you just need to have an idea where θ is located for these functions.
1.
csc⁻¹(−1) = some angle θ in QI or QIV where cscθ = −1 ----> θ = −π/2
2.
cos⁻¹(cos(7π6)) = some angle θ in QI or QII where cosθ = cos(7π6)
Since 7π6 is located in QIII where cos < 0, then θ is in QII where cos < 0
Also, θ must be same distance from negative x-axis as 7π/6 is
θ = 5π/6
3.
−3 sin⁻¹(2x) = π
sin⁻¹(2x) = −π/3 ---> valid since −π/3 is between −π/2 and π/2
2x = sin(−π/3)
2x = −√3/2
x = −√3/4
4.
sec(sin⁻¹(−8/9))
Let θ = sin⁻¹(−8/9)
Then sinθ = −8/9
Since sin⁻¹ returns value in QI or QIV, and since −8/9 < 0, then θ must be in QIV, where sinθ < 0. But cosθ > 0 in QIV
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The inverse trig functions θ = sin⁻¹x, tan⁻¹x, cot⁻¹x, sec⁻¹x always return values in QI or QIV (between −π/2 and π/2). This is just to give you a general idea of where θ is located for any of these functions. The range for each is slightly different because the trig functions sinx, tanx, cotx, secx are not all defined at all values in this interval.
The inverse trig functions θ = cos⁻¹x, sec⁻¹x always return values in QI or QII (between 0 and π). Again the range for cos⁻¹x and sec⁻¹x is slightly different, but you just need to have an idea where θ is located for these functions.
1.
csc⁻¹(−1) = some angle θ in QI or QIV where cscθ = −1 ----> θ = −π/2
2.
cos⁻¹(cos(7π6)) = some angle θ in QI or QII where cosθ = cos(7π6)
Since 7π6 is located in QIII where cos < 0, then θ is in QII where cos < 0
Also, θ must be same distance from negative x-axis as 7π/6 is
θ = 5π/6
3.
−3 sin⁻¹(2x) = π
sin⁻¹(2x) = −π/3 ---> valid since −π/3 is between −π/2 and π/2
2x = sin(−π/3)
2x = −√3/2
x = −√3/4
4.
sec(sin⁻¹(−8/9))
Let θ = sin⁻¹(−8/9)
Then sinθ = −8/9
Since sin⁻¹ returns value in QI or QIV, and since −8/9 < 0, then θ must be in QIV, where sinθ < 0. But cosθ > 0 in QIV
cos²θ = 1 − sin²θ = 1 − 64/81 = 17/81
Since cosθ > 0, take positive square root
cosθ = √17/9
sec(sin⁻¹(−8/9)) = secθ = 9/√17