The identity that links sin and cos: For every angle θ, sin²θ + cos²θ = 1. (It's Pythagoras on the unit-circle point (cos θ, sin θ).) Note sin²θ means (sin θ)².
It works for any angle: True for acute, obtuse, negative — any θ at all. That's why it's an identity, not an equation to solve.
Rearrange, root, then fix the sign: Given one ratio, get the other: sin θ = ±√(1 − cos²θ). Choose the sign from the quadrant (or 'acute' usually means positive).
IB-style question — sin from cos
Given cos θ = 2/3 with θ acute, find the exact value of sin θ.
Step by step
- Rearrange.
- Root; acute ⇒ positive.
Final answer
sin θ = √5 / 3 (the audited exam value).
Don't forget the sign: √ gives a magnitude — the quadrant decides whether sin (or cos) is + or −.
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Three handy versions: From the identity: sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ. Spotting these lets you replace one squared ratio with the other.
IB-style question — substitute
Simplify 1 − sin²θ.
Step by step
- Rearrange the identity.
Final answer
cos²θ.
Swap squares freely: Whenever you see 1 − sin²θ or 1 − cos²θ, replace it with the other square — it usually unlocks the simplification.
Replace and cancel: Use the identity to replace a 1 − sin²θ (or 1 − cos²θ), or to introduce a 1, so terms cancel. This is the engine of many 'show that' trig identities.
IB-style question — a short proof
Show that (1 − cos²θ)/sin θ = sin θ (for sin θ ≠ 0).
Step by step
- Replace 1 − cos²θ with sin²θ.
- Cancel one sin θ.
Final answer
So (1 − cos²θ)/sin θ ≡ sin θ.
Look for a hidden square: If you see 1 − cos²θ or 1 − sin²θ anywhere, swap it immediately — it's almost always the key step.