Reciprocal identities and one-sided proof chains
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| identity/aɪˈdentɪti/ | đẳng thức |
Does that prove they agree for every angle where they are defined?
- Two formulas agree at one angle.
- This lesson studies identity 恒等式: An equality true for every input in its stated common domain.
The method
- The unit-circle equation gives sin²θ+cos²θ=1.
- Divide by cos²θ when cosθ≠0 to obtain tan²θ+1=sec²θ.
- Divide by sin²θ when sinθ≠0 to obtain 1+cot²θ=cosec²θ.
- Record exclusions before cancelling or dividing.
- A proof starts from one side and reaches the other through valid equalities; numerical checks alone are not a proof.
Why is checking tan²θ-sin²θ=tan²θsin²θ at θ=π/4 not a proof?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
An identity claims equality at EVERY allowed input. The proof chains valid equalities from one side to the other: sin²θ(1/cos²θ-1)=sin⁴θ/cos²θ=tan²θsin²θ, with exclusions recorded.
A worked case
Prove tan²θ−sin²θ=tan²θ sin²θ for cosθ≠0. The left side is sin²θ(1/cos²θ−1)=sin²θ(1−cos²θ)/cos²θ=sin⁴θ/cos²θ=tan²θ sin²θ. No division by sinθ is needed, so sinθ=0 remains allowed. For sinθ≠0, (1−cosθ)/sinθ becomes (1−cosθ)(1+cosθ)/[sinθ(1+cosθ)]=sin²θ/[sinθ(1+cosθ)]=sinθ/(1+cosθ). The stated exclusion also makes 1+cosθ nonzero. If secθ=2 with θ acute, tan²θ=4−1=3, cosθ=1/2, sin²θ=3/4, cosec²θ=4/3 and cot²θ=1/3.
Reciprocal identities and one-sided proof chains
The unit-circle equation gives sin²θ+cos²θ=1
Explain which denominator or branch restriction makes each step valid.
If secθ=2 and θ is acute, find tan²θ.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
tan²θ=sec²θ−1=4−1=3.
Where it goes wrong
- Proving an identity does not mean assuming both sides equal and manipulating that assumption without justification.
- A squared identity does not fix a function’s sign: the quadrant does.
- The last expression sinθ/(1+cosθ) is defined at θ=0, but the original (1−cosθ)/sinθ is not; simplification does not repair the original domain.
Not true: Checking a proposed identity at one angle proves it for all allowed angles. An identity must hold for every allowed angle; one test can agree by chance.
Find cosec²θ for that angle. Give it to 3 significant figures.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
sin²θ=1−cos²θ=3/4; its reciprocal is 4/3.
Checking a proposed identity at one angle proves it for all allowed angles.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
An identity must hold for every allowed angle; one test can agree by chance.
Applying it
- Choose the more complicated side, replace reciprocal functions by sine/cosine, use a common denominator and apply sin²+cos²=1.
- Annotate every cancelled factor.
- To disprove a claimed identity, one valid counterexample suffices; to prove it, explain why the chain works for all angles in the common domain.
Find cot²θ for that angle. Give it to 3 significant figures.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
cot²θ=cosec²θ−1=4/3−1=1/3.
Match each derived identity to its source.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Simplification never repairs the original domain: a step valid only where sinθ≠0 leaves θ=0 excluded from the original statement.
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.
- Identity: An equality true for every input in its stated common domain.
- The unit-circle equation gives sin²θ+cos²θ=1.
- Watch for: proving an identity does not mean assuming both sides equal and manipulating that assumption without justification.