Lesson 11 of 13 · 8 min
Double and triple angle formulas
NCERT §3.4
Suppose a slower ride leaves Meera's spoke at an angle A with sin A = 4/5 after one stage. If the next stage turns it through the same angle again, can we find where she is after 2A, or 3A, without measuring?
The lesson in notes
In short
Putting y = x in the sum formulas gives the double angle results: sin 2x = 2 sin x cos x and cos 2x = cos² x − sin² x.
Using sin² x + cos² x = 1, cos 2x also equals 2 cos² x − 1 and 1 − 2 sin² x. Rearranged, cos² x = (1 + cos 2x)/2 and sin² x = (1 − cos 2x)/2.
In terms of tan x: cos 2x = (1 − tan² x)/(1 + tan² x) and sin 2x = 2 tan x/(1 + tan² x), both for x ≠ nπ + π/2; tan 2x = 2 tan x/(1 − tan² x) for 2x ≠ nπ + π/2, n ∈ Z.
Triple angles: sin 3x = 3 sin x − 4 sin³ x and cos 3x = 4 cos³ x − 3 cos x, each proved by writing 3x = 2x + x.
tan 3x = (3 tan x − tan³ x)/(1 − 3 tan² x), for 3x ≠ nπ + π/2, n ∈ Z.
Worked example: if sin x = 4/5 with x acute, cos x = 3/5, so sin 2x = 2 × 4/5 × 3/5 = 24/25, cos 2x = 9/25 − 16/25 = −7/25, tan 2x = −24/7, and sin 3x = 12/5 − 4 × 64/125 = 44/125. The negative cos 2x shows that 2x is obtuse.
Half angles: writing x/2 for x gives cos x = 2 cos² (x/2) − 1 = 1 − 2 sin² (x/2), so sin² (x/2) = (1 − cos x)/2 and cos² (x/2) = (1 + cos x)/2, with the sign fixed by the quadrant of x/2. For example sin² 22.5° = (1 − 1/√2)/2 = (2 − √2)/4, so sin 22.5° = √(2 − √2)/2.
Check at x = π/6: sin 3x = sin π/2 = 1, and 3 × 1/2 − 4 × 1/8 = 3/2 − 1/2 = 1.