Inverse trigonometric branches and principal values
| English | 中文 | Pinyin |
|---|---|---|
| principal value/ˈprɪnsɪpl ˈvæljuː/ | 主值 | zhǔ zhí |
A height ratio gives many possible wheel angles. How does a calculator choose just one of them?
- A height ratio gives many possible wheel angles. How does a calculator choose just one of them?
- This lesson studies principal value 主值: The single inverse output chosen from a stated restricted interval.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines principal value?
The single inverse output chosen from a stated restricted interval.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
The principal sine angle is π/6, so k=1/6.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Applying inverse sine after sine returns the original angle for every real angle. This claim is false. Explain which definition or assumption it violates.
Write arccos(−1/2)=kπ. Find k.
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.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Write arctan(−1)=kπ. Find k.
Tangent is −1 at −π/4 on (−π/2,π/2), so k=−1/4.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · E. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The single inverse output chosen from a stated restricted interval. Choose the relationship, show the method, check its assumptions and interpret the result.