Multiple-angle equations and transformed intervals
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| transformed interval/trænsˈfɔːmd ˈɪntəvl/ | khoảng biến đổi |
How can we avoid losing half the solutions?
- Doubling an angle doubles the number of cycles searched.
- 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.
The method
- 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.
Why does solving sin 2x=1/2 on [0,2π) need the interval [0,4π) for u=2x?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.
A worked case
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)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
2x covers two full turns, producing four sine roots.
Where it goes wrong
- 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.
Not true: Solving sin2x on one full turn of 2x finds every x solution on [0,2π). If x runs over [0,2π), then 2x runs over [0,4π), so solve for 2x over two full turns.
How many solutions does tan3x=1 have on [0,π)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
3x covers three tangent periods, giving three roots.
Solving sin2x on one full turn of 2x finds every x solution on [0,2π).
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
If x runs over [0,2π), then 2x runs over [0,4π), so solve for 2x over two full turns.
Applying it
- 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.
How many solutions does sin(−2x)=0 have on [0,π)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The transformed interval is (−2π,0], giving u=−π,0 and hence two x values.
Match each multiple-angle step to its rule.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Tangent repeats after π, sine and cosine after 2π; the number of cycles scales with |k|.
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.
- Transformed interval: The interval for a substituted angle after applying the same input change to its endpoints.
- For an argument u=kx+c, transform the whole interval for x into the corresponding interval for u.
- Watch for: solving only one cycle of u misses roots.