Trigonometric Functions

Maths · Class 11

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?

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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.

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