sin 2A is not 2 sin A. Doubling an angle is not doubling its sine.
Choose the expansion
sin 2A
Put B = A in sin(A + B) = sin A cos B + cos A sin B and both terms become sin A cos A, giving 2 sin A cos A. A and D are the same wrong idea: test them at A = 90°, where sin 180° = 0 but 2 sin 90° = 2. Doubling the angle is nothing like doubling the value.
sin(A + B) = sinA cosB + cosA sinB
last lesson
set B = A
both terms are the same
sin 2A = 2 sinA cosA
nothing new memorised
cos 2A = cos²A − sin²A
the same substitution
If a formula slips your mind in an exam, derive it in two lines from the sum identity. That is faster and safer than guessing at a half-remembered one. Read backwards, it collapses a product: 2 sin 75° cos 75° = sin 150° = 12.
All three are the same identity, with sin²A + cos²A = 1 used to trade one function for the other. Choose whichever form matches what the question has already given you.
sin A = 513, cos A = 1213
quadrant I, both positive
sin 2A = 2(513)(1213) = 120169
straight substitution
cos 2A = 1 − 2(25169)
sine was the given one
= 1 − 50169 = 119169
check: 120² + 119² = 169² ✓
The Pythagorean check at the end costs one line and catches almost every arithmetic slip, because sin²2A + cos²2A must come to exactly 1.
If A is 300°, in quadrant IV, then A2 is 150°, in quadrant II — a different quadrant with different signs. Halve the angle first, then decide the sign. Leaving a bare ± in a final answer is leaving the question half-answered.
With cos A given, 2cos²A − 1 needs no sine at all. The answer can come out negative even though A is acute.
If cos A = 513 and A is in quadrant I, find cos 2A.
cos 2A = 2cos²A − 1
the cosine-only face
= 2(25169) − 1 = 50169 − 169169
square first, then double
= −119169
negative
A is above 45°, so 2A is above 90°
2A is in quadrant II, where cosine is negative ✓
2 × 513 = 1013 treats cos 2A as 2 cos A. Doubling the angle does not double the value; here it even changes the sign.
The same formulas, with the signs doing more of the work.
A in quadrant II: sin A = 45. Find sin 2A.
cos A = −35
negative in quadrant II
sin 2A = 2(45)(−35) = −2425
the minus carries through
A positive sin A does not make sin 2A positive; the sign of cos A matters too.
Half-angle with mixed signs: cos A = 725, 270° < A < 360°
135° < A2 < 180°, quadrant II
halve the whole range first
sin(A2) = +√1 − 7252 = √925 = 35
sine positive in II
cos(A2) = −√1 + 7252 = −√1625 = −45
cosine negative in II
A is in quadrant IV but A2 is in quadrant II, so the two answers take opposite signs.
Tangent: tan A = 2. Find tan 2A.
tan 2A = 2 tan A1 − tan²A
tan(A + B) with B = A
= 2(2)1 − 4 = 4−3
substitute
= −43
the bottom went negative
tan A = 2 means A is above 45°, so 2A is past 90° and a negative tangent is expected.
15° = 30°2
half a special angle
sin 15 = ±√1 − cos302
cos 30 = √32
15° is in quadrant I, so positive
the ± is resolved
= √2 − √34 ≈ 0.259
small, as a sine near 0° should be
Rearranged, cos 2A = 1 − 2sin²A gives sin²A = 1 − cos 2A2. Trading a squared function for a plain one at double the angle is a standard move later on, and it is exactly the half-angle formula read from the other end.
Tests give one ratio and a quadrant and ask for sin 2A, cos 2A or a half-angle value, or ask for an exact value such as sin 15° or cos 22.5°. The fastest route: find the missing ratio with its sign, choose the cos 2A form that uses what you were given, and halve the angle before choosing a half-angle sign. A 10-second check: sin²2A + cos²2A must equal 1 for your two answers.
Find the value
If cos A = 35 and A is in quadrant I, what is cos 2A?
Cosine was given, so use cos 2A = 2cos²A − 1 = 2(925) − 1 = 1825 − 2525 = −725. A doubles the cosine, which is the same error as thinking sin 2A = 2 sin A. Negative is correct here: A is about 53°, so 2A is about 106°, in quadrant II where cosine is negative.
Hint: you were handed cos A, so reach for the form with only cosine in it. And check the sign by asking which quadrant 2A lands in.
