Finding principal values
To find a principal value, ask which angle in the principal branch has the given ratio. Watch negative inputs: sin⁻¹(−1/2) = −π/6 but cos⁻¹(−1/2) = 2π/3.
The method
These steps work for all six functions.
- 1Let y equal the expression, e.g. y = cos⁻¹ x, and write the branch y must lie in, e.g. [0, π].
- 2Rewrite it without the inverse: cos y = x.
- 3Ignore the sign of x and find the acute angle α with cos α = |x| from the table of standard values.
- 4★If x ≥ 0, y = α. If x < 0: y = −α for sin⁻¹, tan⁻¹, cosec⁻¹; y = π − α for cos⁻¹, cot⁻¹, sec⁻¹.because sin(−α) = −sin α, tan(−α) = −tan α and cos(π − α) = −cos α
- 5Check that y lies in the branch and gives back x.
★ key step
Standard values you need
| θ | 0 | π/6 | π/4 | π/3 | π/2 |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | not defined |
Example 1: direct principal values
Find the principal values of (i) cos⁻¹(−√3/2), (ii) tan⁻¹(−1/√3), (iii) sec⁻¹(−2), (iv) cosec⁻¹(−2/√3).
- (i) cos α = √3/2 gives α = π/6. The input is negative and cos⁻¹ lies in [0, π], so y = π − π/6 = 5π/6.
- (ii) tan α = 1/√3 gives α = π/6. tan⁻¹ lies in (−π/2, π/2), so y = −π/6.
- (iii) sec y = −2 means cos y = −1/2. cos α = 1/2 gives α = π/3, so y = π − π/3 = 2π/3.
- (iv) cosec y = −2/√3 means sin y = −√3/2. sin α = √3/2 gives α = π/3, so y = −π/3.
- Check (iii): cos(2π/3) = −1/2, so sec(2π/3) = −2, and 2π/3 lies in [0, π] − {π/2}.
(i) 5π/6 (ii) −π/6 (iii) 2π/3 (iv) −π/3
Example 2: an expression (board style)
Evaluate tan⁻¹(√3) + cos⁻¹(−1/2) + sin⁻¹(−1/√2).
- Rule: find each principal value, then add.
- tan⁻¹(√3) = π/3, since tan(π/3) = √3 and π/3 lies in (−π/2, π/2).
- cos⁻¹(−1/2) = π − π/3 = 2π/3.
- sin⁻¹(−1/√2) = −π/4, since sin(π/4) = 1/√2.
- Sum = π/3 + 2π/3 − π/4.
- π/3 + 2π/3 = π, so the sum is π − π/4 = 3π/4.
3π/4
Example 3: the angle is outside the branch
Find (i) sin⁻¹(sin 5π/4), (ii) cos⁻¹(cos 4π/3), (iii) tan⁻¹(tan 2π/3).
- (i) 5π/4 is not in [−π/2, π/2]. sin(5π/4) = sin(π + π/4) = −sin(π/4) = −1/√2.
- sin⁻¹(−1/√2) = −π/4, which is in the branch.
- (ii) 4π/3 is not in [0, π]. cos(4π/3) = cos(π + π/3) = −cos(π/3) = −1/2.
- cos⁻¹(−1/2) = π − π/3 = 2π/3.
- (iii) 2π/3 is not in (−π/2, π/2). tan(2π/3) = tan(π − π/3) = −tan(π/3) = −√3.
- tan⁻¹(−√3) = −π/3.
(i) −π/4 (ii) 2π/3 (iii) −π/3
Common traps
- sin⁻¹(sin θ) = θ only when θ is in [−π/2, π/2]. Always check the branch first.
- A value of cos⁻¹, cot⁻¹ or sec⁻¹ is never negative.
- A value of sin⁻¹, tan⁻¹ or cosec⁻¹ is never more than π/2.
- Give answers in radians (π/3, not 60°) unless the question asks for degrees.
- For sec⁻¹ and cosec⁻¹, rewrite the equation first: sec y = 2 means cos y = 1/2, so y = π/3.
Surveyors at hill stations like Ooty use inverse tangent to turn a height and a distance into an angle of elevation. A tower 50 m tall seen from 50 m away gives tan⁻¹(1) = π/4, that is 45°.
1 marks: “Find the principal value of cot⁻¹(−1/√3).” cot α = 1/√3 gives α = π/3. The input is negative and cot⁻¹ lies in (0, π), so the value is π − π/3 = 2π/3.
2 marks: “Evaluate tan⁻¹(1) − sec⁻¹(−√2).” tan⁻¹(1) = π/4. For sec⁻¹(−√2): cos y = −1/√2, so y = π − π/4 = 3π/4. Answer: π/4 − 3π/4 = −π/2.
2 marks: “Find the value of sin⁻¹(sin 4π/5).” 4π/5 is not in [−π/2, π/2]. sin(4π/5) = sin(π − 4π/5) = sin(π/5), and π/5 is in the branch. So the value is π/5.
Negative input? Sine family: put a minus sign. Cosine family: subtract from π.