Relationships between trig functions (Math AA HL)

Three new names for 1/sin, 1/cos, 1/tan: You already know sin, cos and tan. The HL course adds their reciprocals — same triangle, just flipped:

sec θ = 1/cos θ (goes with cos),

csc θ = 1/sin θ (also written cosec θ — goes with sin),

cot θ = 1/tan θ = cos θ / sin θ (goes with tan).

A handy memory trick: the third letters pair up — sec ↔ cos, csc ↔ sin, cot ↔ tan via cos/sin.

Why these identities are true: Start from the one identity you already trust: sin²θ + cos²θ = 1.

Divide every term by cos²θ and you get tan²θ + 1 = sec²θ.

Divide every term by sin²θ instead and you get 1 + cot²θ = csc²θ.

So all three Pythagorean identities come from the same place — you never have to memorise them as separate facts.

The same point, read two ways: Co-function relationships come from a right-angled triangle: the two acute angles add to 90°, so one angle's sine is the other's cosine.

sin(90° − θ) = cos θ and cos(90° − θ) = sin θ (in radians, 90° = π/2).

Symmetry relationships come from the unit circle: reflecting an angle changes signs but not size, e.g. sin(−θ) = −sin θ (sine is odd) and cos(−θ) = cos θ (cosine is even).

The golden rule for simplifying: When an expression mixes sec, csc, cot, tan, sin and cos and you're stuck:

Rewrite EVERYTHING in sin and cos, put it over a common denominator, then use sin²θ + cos²θ = 1.

Nine times out of ten the mess collapses to something tiny. This is exactly the move the M25 paper rewards.