Multiple-angle equations and transformed intervals
導入済み| English | 意味 |
|---|---|
| transformed interval/trænsˈfɔːmd ˈɪntəvl/ | 変換区間 |
Doubling an angle doubles the number of cycles searched. How can we avoid losing half the solutions?
- Doubling an angle doubles the number of cycles searched. How can we avoid losing half the solutions?
- This lesson studies transformed interval 变换后的区间: The interval for a substituted angle after applying the same input change to its endpoints.
Choose the mathematical structure
- For an argument u=kx+c, transform the whole interval for x into the corresponding interval for u. If k is negative, reverse endpoint order and preserve whether each endpoint is included. Solve in u using all repeated cycles, then recover x=(u−c)/k and check the original interval. Sine/cosine repeat after 2π; tangent repeats after π.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Why does solving sin 2x=1/2 on [0,2π) need the interval [0,4π) for u=2x?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Transforming the whole interval gives 0≤u<4π, where sine hits 1/2 four times: π/6, 5π/6, 13π/6, 17π/6. Halving recovers four x values; a single u-cycle finds only two.
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.
Solve sin2x=1/2 on 0≤x<2π. Then 0≤u=2x<4π, giving u=π/6,5π/6,13π/6,17π/6. Divide by 2: x=π/12,5π/12,13π/12,17π/12. For cos(2x−π/3)=1/2 on [0,2π), the new interval is [−π/3,11π/3). Its allowed u values are −π/3,π/3,5π/3,7π/3, hence x=0,π/3,π,4π/3. For tan3x=1 on [0,π), u ranges over [0,3π), so x=π/12,5π/12,3π/4. For sin(−2x)=0 on [0,π), the transformed interval is (−2π,0]; its roots u=−π,0 give x=π/2,0.
Multiple-angle equations and transformed intervals
For an argument u=kx+c, transform the whole interval for x into the corresponding interval for u
Explain the original interval, denominator or physical reference before using a trig equation.
How many solutions does sin2x=1/2 have on [0,2π)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
2x covers two full turns, producing four sine roots.
Test a tempting shortcut
- Solving only one cycle of u misses roots. An excluded upper endpoint for x need not be the upper endpoint for u when k<0. Dividing by a trig factor can remove zero roots; squaring can introduce extra roots. Substitute into the original equation, whose reciprocal/tangent denominators may still exclude a candidate.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Solving sin2x on one full turn of 2x always finds every x solution on [0,2π). This claim is false. Explain which definition or assumption it violates.
How many solutions does tan3x=1 have on [0,π)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
3x covers three tangent periods, giving three roots.
Solving sin2x on one full turn of 2x always finds every x solution on [0,2π).
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Solving only one cycle of u misses roots. An excluded upper endpoint for x need not be the upper endpoint for u when k<0. Dividing by a trig factor can remove zero roots; squaring can introduce extra roots. Substitute into the original equation, whose reciprocal/tangent denominators may still exclude a candidate.
Interpret a new situation
- Use general families such as u=α+2jπ or π−α+2jπ for sine, u=±arccos a+2jπ for cosine, and u=arctan a+jπ for tangent, with integer j. Keep only values in the transformed interval. At sine/cosine extrema, deduplicate the families. Return the answer in the requested angle unit and original variable.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
How many solutions does sin(−2x)=0 have on [0,π)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The transformed interval is (−2π,0], giving u=−π,0 and hence two x values.
Match each multiple-angle step to its rule.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Tangent repeats after π, sine and cosine after 2π; the number of cycles scales with |k|.
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 interval for a substituted angle after applying the same input change to its endpoints. Choose the relationship, show the method, check its assumptions and interpret the result.