Inverse trigonometric branches and principal values
Введено| English | Значение |
|---|---|
| principal value/ˈprɪnsɪpl ˈvæljuː/ | главное значение |
How does a calculator choose just one of them?
- A height ratio gives many possible wheel angles.
- This lesson studies principal value 主值: The single inverse output chosen from a stated restricted interval.
The method
- Restrict sine to [−π/2,π/2] to define arcsin with domain [−1,1] and range [−π/2,π/2].
- Restrict cosine to [0,π] for arccos, with domain [−1,1] and range [0,π].
- Restrict tangent to (−π/2,π/2) for arctan, with every real input and range (−π/2,π/2).
- These inverse graphs reflect the restricted base graphs in y=x.
Why does arcsin(sin(5π/6)) equal π/6, not 5π/6?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Sine is restricted to [-π/2,π/2] to make it one-to-one; the inverse can only return values from that branch. The original angle is recovered only when it already lies in the restricted interval.
A worked case
arcsin(1/2)=π/6, arccos(−1/2)=2π/3 and arctan(−1)=−π/4. Although sin(5π/6)=1/2, arcsin(sin(5π/6))=π/6, not 5π/6. Similarly arccos(cos(4π/3))=2π/3 and arctan(tan(3π/4))=−π/4. For sin x=1/2 on [0,2π), the principal value α=π/6 generates both α and π−α=5π/6. For cos x=−1/2 the solutions are 2π/3 and 4π/3; tangent repeats its principal value after π.
Inverse trigonometric branches and principal values
Restrict sine to [−π/2,π/2] to define arcsin with domain [−1,1] and range [−π/2,π/2]
Explain which denominator or branch restriction makes each step valid.
Write arcsin(1/2)=kπ. Find k. Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The principal sine angle is π/6, so k=1/6.
Where it goes wrong
- sin(arcsin u)=u only for −1≤u≤1; arcsin(sin x)=x only on the restricted sine branch.
- Arccos is decreasing, so arccos(−1)=π and arccos(1)=0.
- Arctan approaches ±π/2 as input grows in magnitude but never attains those limits.
- A principal value alone is not a full interval solution set.
Not true: Applying inverse sine after sine returns the original angle for every real angle. arcsin returns an angle between −π/2 and π/2, so angles outside that range are not returned.
Write arccos(−1/2)=kπ. Find k. Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Cosine is −1/2 at 2π/3 on [0,π], so k=2/3.
Applying inverse sine after sine returns the original angle for every real angle.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
arcsin returns an angle between −π/2 and π/2, so angles outside that range are not returned.
Applying it
- Mark domain endpoints and output intervals before composing functions.
- On the inverse graphs arcsin runs from (−1,−π/2) to (1,π/2), arccos from (−1,π) to (1,0), and arctan passes through (0,0) between horizontal asymptotes.
- Keep answers in radians unless degrees are requested, and check all equation candidates in the original interval.
Write arctan(−1)=kπ. Find k.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Tangent is −1 at −π/4 on (−π/2,π/2), so k=−1/4.
Match each inverse function to its restricted range.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Each branch is chosen to make the original function one-to-one; a principal value then generates the full solution set via symmetry and period.
In your course
- AQA A-level Mathematics (7357), specification section E.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Principal value: The single inverse output chosen from a stated restricted interval.
- Restrict sine to [−π/2,π/2] to define arcsin with domain [−1,1] and range [−π/2,π/2].
- Watch for: sin(arcsin u)=u only for −1≤u≤1; arcsin(sin x)=x only on the restricted sine branch.