Reciprocal identities and one-sided proof chains
Введено| English | Значение |
|---|---|
| identity/aɪˈdentɪti/ | тождество |
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?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
sin²θ=1−cos²θ=3/4; its reciprocal is 4/3.
Checking a proposed identity at one angle proves it for all allowed angles.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
cot²θ=cosec²θ−1=4/3−1=1/3.
Match each derived identity to its source.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
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.