Inverse sine and inverse cosine
A calculator's "sin⁻¹" key works backwards: you tell it the ratio and it tells you the angle. But many angles share one ratio, so it must agree to give only one of them.
Since sine repeats, we restrict it to [−π/2, π/2] to make it one-one and onto [−1, 1]. Then sin⁻¹ : [−1, 1] → [−π/2, π/2]. In the same way cos⁻¹ : [−1, 1] → [0, π]. The value in this range is the principal value.
Formula: sin⁻¹ x lies in [−π/2, π/2]. cos⁻¹ x lies in [0, π].
In the figure: The graph y = sin⁻¹ x lives only for x from −1 to 1 and its height never leaves −π/2 to π/2 (about −1.57 to 1.57). The marked points are (1/2, π/6) and (−1, −π/2).
Rule 1: Read the table — Use the standard angles 0, π/6, π/4, π/3, π/2.
- Easy: sin⁻¹(1/2) = π/6 and cos⁻¹(1/2) = π/3.
- Medium: sin⁻¹(−1/2) = −π/6 (sine may give negative angles), but cos⁻¹(−1/2) = 2π/3 (cosine stays in 0 to π).
- Exam-type: sin⁻¹(−√3/2) = −π/3 and cos⁻¹(−√3/2) = 5π/6.
Rule 2: Check that the value is allowed — The number inside must lie between −1 and 1.
- Easy: sin⁻¹(2) does not exist because the sine of an angle is never 2.
- Medium: cos⁻¹(−1) = π and sin⁻¹(−1) = −π/2.
- Exam-type: sin⁻¹(sin(5π/6)) is not 5π/6: sin(5π/6) = 1/2 and sin⁻¹(1/2) = π/6.
The answer of sin⁻¹ is an angle in [−π/2, π/2]; the answer of cos⁻¹ is an angle in [0, π].
Giving an answer outside the principal range, such as sin⁻¹(1/2) = 5π/6.
Sine lives in the right half (−π/2 to π/2); cosine lives on top (0 to π).
Practice questions
- 1Find sin⁻¹(√3/2).
- 2Find cos⁻¹(1/2).
- 3Find sin⁻¹(−1).
- 4Find cos⁻¹(−1/√2).
- 5State the domain and the range of cos⁻¹ x.
- 6Find sin⁻¹(−√3/2).
- 7Is sin⁻¹(3/2) defined? Give a reason.
- 8Find sin⁻¹(sin(2π/3)).
Show answers (try them yourself first)
- 1π/3
- 2π/3
- 3−π/2
- 43π/4
- 5Domain [−1, 1]; range [0, π]
- 6−π/3
- 7No, because 3/2 is bigger than 1
- 8sin(2π/3) = √3/2, so the answer is π/3 (not 2π/3)
Inverse tangent and inverse cotangent
A ramp rises 1 metre over 1 metre of ground. Its slope is 1, and the angle of the ramp is the "inverse tangent" of 1.
Tangent is restricted to (−π/2, π/2), giving tan⁻¹ : R → (−π/2, π/2). Cotangent is restricted to (0, π), giving cot⁻¹ : R → (0, π). Both accept every real number.
Formula: tan⁻¹ x lies in (−π/2, π/2). cot⁻¹ x lies in (0, π).
In the figure: The graph y = tan⁻¹ x flattens out towards π/2 on the right and −π/2 on the left but never touches them. The marked point is (1, π/4).
Rule 1: Read the table — Remember tan 45° = 1 and tan 60° = √3.
- Easy: tan⁻¹(1) = π/4 and tan⁻¹(−1) = −π/4.
- Medium: tan⁻¹(√3) = π/3 and tan⁻¹(1/√3) = π/6.
- Exam-type: cot⁻¹(1) = π/4, cot⁻¹(−1) = 3π/4 and cot⁻¹(−√3) = 5π/6.
Rule 2: Why the ends are missing — The end values are not reached.
- Easy: tan⁻¹ x is never π/2 because tan(π/2) does not exist.
- Medium: cot⁻¹ x is never 0 or π.
- Exam-type: tan⁻¹(tan(3π/4)) = tan⁻¹(−1) = −π/4.
tan⁻¹ gives angles in (−π/2, π/2), cot⁻¹ gives angles in (0, π); both ranges are open at the ends.
Writing cot⁻¹(−1) = −π/4; cot⁻¹ never gives a negative angle.
tan⁻¹ is "odd" (negative in, negative out); cot⁻¹ stays between 0 and π.
Practice questions
- 1Find tan⁻¹(√3).
- 2Find tan⁻¹(−1).
- 3Find cot⁻¹(√3).
- 4Find cot⁻¹(−1).
- 5Find tan⁻¹(1/√3).
- 6Find cot⁻¹(−√3).
- 7Find tan⁻¹(1) + cot⁻¹(1).
- 8Find tan⁻¹(tan(3π/4)).
Show answers (try them yourself first)
- 1π/3
- 2−π/4
- 3π/6
- 43π/4
- 5π/6
- 65π/6
- 7π/4 + π/4 = π/2
- 8tan(3π/4) = −1, so the answer is −π/4
Inverse secant and inverse cosecant
Secant and cosecant are the reciprocals 1 / cos and 1 / sin. A ratio of 2 for secant means the cosine is one half.
sec⁻¹ accepts numbers with |x| ≥ 1 and gives angles in [0, π] except π/2. cosec⁻¹ accepts |x| ≥ 1 and gives angles in [−π/2, π/2] except 0. Use the shortcuts sec⁻¹ x = cos⁻¹(1/x) and cosec⁻¹ x = sin⁻¹(1/x).
Formula: sec⁻¹ x = cos⁻¹(1/x), cosec⁻¹ x = sin⁻¹(1/x), |x| ≥ 1
Rule 1: Use the reciprocal shortcut — Flip the number and use the cos⁻¹ or sin⁻¹ table.
- Easy: sec⁻¹(2) = cos⁻¹(1/2) = π/3 and cosec⁻¹(2) = sin⁻¹(1/2) = π/6.
- Medium: sec⁻¹(√2) = π/4 and sec⁻¹(−2) = cos⁻¹(−1/2) = 2π/3.
- Exam-type: cosec⁻¹(−2) = sin⁻¹(−1/2) = −π/6 and cosec⁻¹(−√2) = −π/4.
Rule 2: Check the domain — Numbers between −1 and 1 are not allowed.
- Easy: sec⁻¹(1/2) is not defined.
- Medium: sec⁻¹(1) = 0 and cosec⁻¹(1) = π/2.
- Exam-type: sec⁻¹(2/√3) = cos⁻¹(√3/2) = π/6.
For sec⁻¹ and cosec⁻¹, the inside number must satisfy |x| ≥ 1.
Using sec⁻¹ x = 1 / cos⁻¹ x; the reciprocal goes inside the bracket, not outside.
Flip inside the bracket: sec⁻¹ x = cos⁻¹(1/x).
Practice questions
- 1Find sec⁻¹(2).
- 2Find cosec⁻¹(2).
- 3Find sec⁻¹(√2).
- 4Find sec⁻¹(−2).
- 5Find cosec⁻¹(−2).
- 6Find cosec⁻¹(−√2).
- 7Is sec⁻¹(1/2) defined? Why?
- 8Find sec⁻¹(2/√3).
Show answers (try them yourself first)
- 1π/3
- 2π/6
- 3π/4
- 42π/3
- 5−π/6
- 6−π/4
- 7No, since |1/2| is less than 1
- 8cos⁻¹(√3/2) = π/6