3.1.12 · D5Advanced Trigonometry
Question bank — Sum and difference formulas — sin(A±B), cos(A±B), tan(A±B) — proofs
Before you start, remember the three anchors:
- Sine keeps company (same signs): .
- Cosine is contrary (flipped signs): .
- Tangent's below is opposite: .
True or false — justify
Is ever true?
Only in special cases (e.g. , or ), never as an identity — because is a rotation that stirs cross-terms, so the general answer is .
True or false: .
False — cosine of a difference is ; the components multiply and add, they do not subtract term-by-term.
True or false: can be larger than and both.
False for the plus case in the first quadrant — adding angles pushes you toward where cosine shrinks; but generally cosine of a sum is just some other value in , not bounded by the parts.
True or false: .
True — the formula is symmetric ( reads the same after swapping and ), matching the fact that rotating by then equals then .
True or false: .
True — because cosine is even, , and geometrically the chord between the two points doesn't care about order.
True or false: .
False — sine is odd, so ; swapping the order flips the sign.
True or false: the tangent formula holds for all real .
False — it fails whenever any is undefined ( or a multiple of ) or when makes the denominator zero.
True or false: depends on terms.
False — the terms cancel, leaving ; this cancellation is exactly why sums of these formulas produce product-to-sum identities. See Product-to-Sum and Sum-to-Product.
True or false: setting in gives a valid double-angle formula.
True — , which is exactly the double angle result.
Spot the error
Find the error: "."
Wrong sign — the plus-form of cosine flips to a minus: . The written formula is actually .
Find the error: "."
The denominator sign is wrong — for a difference it must be : . The below mirrors the above.
Find the error: "."
The middle sign should be minus: . Sine keeps company, so the sign inside matches the sign of the whole expression.
Find the error: "."
Sign is wrong — (positive). The minus would appear for .
Find the error: a student proves then says "the same distance argument directly gives ."
The chord/distance method only yields ; sine must be reached separately via the cofunction bridge , not from the same equation.
Find the error: "."
Since we use , which has a plus: . The minus belongs to .
Why questions
Why do we prove first instead of ?
Because the chord/distance argument naturally produces — the chord depends only on the angle between the two points, which is — and every other identity falls out of it algebraically.
Why divide the tangent proof by specifically?
That divisor turns each ratio into a and cancels cleanly, converting the whole fraction into an expression purely in and .
Why does replacing with turn into ?
Because , and since cosine is even () and sine is odd (), the term flips to . See Cofunction Identities and Unit Circle and Trig Definitions.
Why is the cofunction identity needed at all?
The distance method only gives cosine formulas; the cofunction identity is the bridge that translates a cosine-of-difference into a sine-of-sum, letting us derive sine without a new geometric proof.
Why does prove the phase-shift fact rather than assume it?
Expanding gives , so the "shift sine left by " rule emerges from the formula instead of being memorized.
Why can come out negative even when and are both acute?
If the denominator goes negative, signalling that has crossed into a quadrant where tangent is negative.
Why is not a coincidence?
Because , and the cofunction identity says ; complementary angles always swap sine and cosine values.
Edge cases
What happens to the sum formulas when ?
They collapse correctly: and , confirming the identities respect the "do-nothing" rotation.
What happens to when ?
is undefined, and indeed the denominator becomes there, so the formula correctly blows up to infinity rather than giving a finite wrong answer.
What does the tangent formula do if itself is undefined (e.g. )?
The formula can't be used directly — you must fall back to , since dividing by was never valid in its derivation.
Is still valid for negative or obtuse angles?
Yes — the unit-circle/distance proof uses coordinates valid for all angles, so the identity holds in every quadrant, not just acute cases.
What is when , and does it expose the linearity trap?
, while the false rule would give ; the correct formula matches, exposing why is not additive.
What happens to in the denominator when either angle is ?
The product becomes , so — but only in this degenerate case, never in general.