Addition and Subtraction Formulas in Trigonometry:
Let us consider a unit circle with its centre at O, the origin of the coordinate system. Then, the coordinates of P, the point of intersection of the circle with the positive X-axis are (1, 0). If ∠POQ = A, ∠QOR = B and ∠POS = -B, then the coordinates of Q, R and S are (cos A, sin A), [cos (A + B), sin (A + B)] and (cos B, -sin B) respectively. Since ∠POR = A + B and ∠QOS = A + B, we have PR = QS.
| ⇒ PR2 = QS2 ⇒ {cos (A + B) – 1}2 + {sin (A + B) – 0}2 = (cos A – cos B)2 + (sin A + sin B)2 ⇒ cos2 (A + B) + 1 – 2 cos (A + B) + sin2 (A + B) = cos2 A + cos2 B – 2cos A cos B + sin2 A + sin2 B + 2 sin A sin B ⇒ 1 + 1 – 2 cos (A + B) = 2 – 2 cos A cos B + 2 sin A sin B ⇒ -2 cos (A + B) = -2 (cos A cos B – sin A sin B) ⇒ cos (A + B) = cos A cos B – sin A sin B …………(i) Replacing B by -B, we get- cos (A – B) = cos A cos (-B) – sin A sin (-B) = cos A cos B + sin A sin B [∵ cos (-B) = cos B and sin (-B) = -sin B] ……………(ii) Now, replacing A by (π/2 + A) in (i), we get- cos (π/2 + A + B) = cos (π/2 + A) cos B – sin (π/2 + A) sin B ⇒ -sin (A + B) = -sin A cos B – cos A sin B ⇒ sin (A + B) = sin A cos B + cos A sin B …………(iii) Replacing B by -B in (iii), we get- sin (A – B ) = sin A cos (-B) + cos A sin (-B) = sin A cos B – cos A sin B ………(iv) |
Corollaries:
| (1) For any values of A and B, (i) sin (A + B) . sin (A – B) = sin2 A – sin2 B (ii) cos (A + B) . cos (A – B) = cos2 A – sin2 B Proof: (i) sin (A + B) . sin (A – B) = (sin A cos B + cos A sin B) (sin A cos B – cos A sin B) = sin2 A cos2 B – cos2 A sin2 B = sin2 A (1- sin2 B) – (1 – sin2 A) sin2 B = sin A – sin2 A sin2 B – sin2 B + sin2 A sin2 B = sin2 A – sin2 B (ii) cos (A + B) . cos (A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin A sin B) = cos2 A cos2 B – sin2 A sin2 B = cos2 A (1 – sin2 B) – (1 – cos2 A) sin2 B = cos2 A – cos2 A sin2 B – sin2 B + cos2 A sin2 B = cos2 A – sin2 B |
| (2) For any values of A, B and C, (i) sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C – tan A tan B tan C) (ii) cos (A + B + C) = cos A cos B cos C (1- tan A tan B – tan B tan C – tan C tan A) Proof: (i) sin (A + B + C) = sin {A + (B + C)} = sin A cos (B + C) + cos A sin (B + C) = sin A (cos B cos C – sin B sin C) + cos A (sin B cos C + cos B sin C) = cos A cos B cos C [(sin A cos B cos C – sin A sin B sin C + cos A sin B cos C + cos A cos B sin C)/cosA cosB cos C] = cos A cos B cos C [tan A – tan A tan B tan C + tan B + tan C] = cos A cos B cos C [tan A + tan B + tan C – tan A tan B tan C] (ii) cos (A + B + C) = cos {A + (B + C)} = cos A cos (B + C) – sin A sin (B + C) = cos A (cos B cos C – sin B sin C) – sin A (sin B cos C + cos B sin C) = cos A cos B cos C – cos A sin B sin C – sin A sin B cos C – sin A cos B sin C = cos A cos B cos C [1 – (cos A sin B sin C + sin A sin B sin C + sin A cos B sin C)/cos A cos B cos C] = cos A cos B cos C [1 – tan B tan C – tan A tan B – tan C tan A] = cos A cos B cos C [1- tan A tan B – tan B tan C – tan C tan A] |
(3) For any values of A, B and C, tan (A + B + C) = (tan A + tan B + tan C – tan A tan B tan C)/(1- tan A tan B – tan B tan C – tan C tan A)
Proof:
| Important Formulas: (1) sin (A + B) = sin A cos B + cos A sin B (2) sin (A – B) = sin A cos B – cos A sin B (3) cos (A + B) = cos A cos B – sin A sin B (4) cos (A – B) = cos A cos B + sin A sin B (5) tan (A + B) = (tan A + tan B)/(1 – tan A tan B), tan A tan B ≠ 1 (6) tan (A – B) = (tan A – tan B)/(1 + tan A tan B) (7) cot (A + B) = (cot A cot B – 1)/(cot B + cot A) (8) cot (A – B) = (cot A cot B + 1)/(cot B – cot A) (9) sin (A + B). sin (A- B) = sin2 A – sin2 B (10) cos (A + B). cos (A- B) = cos2 A – sin2 B = cos2 B – sin2 A (11) sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C – tan A tan B tan C) (12) cos (A + B + C) = cos A cos B cos C (1- tan A tan B – tan B tan C – tan C tan A) (13) tan (A + B + C) = (tan A + tan B + tan C – tan A tan B tan C)/(1- tan A tan B – tan B tan C – tan C tan A) |
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