## Trigonometric Ratios of Compound Angles

The sum or difference of two or more angles is said to be a compound angle. If A and B are two angles then A + B or A - B are known as the compound angles. We express the trigonometric ratios of compound angle A + B and A - B in terms of trigonometric ratios of the angles A and B. The trigonometric ratios of the angles A + B and A - B are known as the addition and subtraction formula respectively.

### Trigonometric Ratios of A + B (Addition Formula)

Let a revolving line start from OX and trace out an angle XOY = A and revolve further through an angle YOZ = B. Therefore, XOZ = A + B.

Let, P be any point on OZ. Draw PM perpendicular to OX and PN perpendicular to OY. From N draw NQ perpendicular to OX and NR perpendicular to PM.

Here, RPN = 90° - PNR     [  NR ⊥ MP]

= RNO     [  PN ⊥ OY]

= NOQ     [  RN OX]

= A

Again, RMQN is a rectangle. So, MR = QN and RN = MQ.

Now, sin(A + B) = MP/OP

= (MR + RP)/OP

= (QN + RP)/OP

= QN/OP + RP/OP

= QN/ON . ON/OP + RP/NP . NP/OP

= sinA cosB + cosA sinB

And, cos(A + B) = OM/OP

= (OQ - MQ)/OP

= (OQ - RN)/OP

= OQ/OP - RN/OP

= OQ/ON . ON/OP - RN/NP . NP/OP

= cosA cosB - sinA sinB

Hence, the sine formula and cos formula of compound angle (A + B) are:

sin(A + B) = sinA cosB + cosA sinB

cos(A + B) = cosA cosB - sinA sinB

### Trigonometric Ratios of A – B (Subtraction Formula)

Let a revolving line start from OX and trace out an angle XOY = A and then revolve back through an angle YOZ = B. Therefore, XOZ = A - B.

Let, P be any point on the line OZ. Draw PM perpendicular to OX and PN perpendicular to OY. From N draw NQ perpendicular to OX and NR perpendicular to MP produced.

Here, RPN = 90° - PNR      [  PR ⊥ NR]

= RNY      [  PN ⊥ OY]

= XOY      [  OX NR]

= A

Again, QMRN is a rectangle. So, QN = MR and QM = NR.

Now, sin(A - B) = MP/OP

= (MR - PR)/OP

= (QN - RP)/OP

= QN/OP - PR/OP

= QN/ON . ON/OP - PR/NP . NP/OP

= sinA cosB - cosA sinB

And, cos(A - B) = OM/OP

= (OQ + QM)/OP

= (OQ + NR)/OP

= OQ/OP + NR/OP

= OQ/ON . ON/OP + NR/NP . NP/OP

= cosA cosB + sinA sinB

Hence, the sine formula and cos formula of compound angle (A - B) are:

sin(A - B) = sinA cosB - cosA sinB

cos(A - B) = cosA cosB + sinA sinB

### Tangent Formula of Compound Angle (A + B)

Tangent Formula of Compound Angle (A - B)

Cotangent Formula of Compound Angle (A + B)

Cotangent Formula of Compound Angle (A - B)

### Worked Out Examples

Example 1: Find the value of: a. sin75°  b. tan15°

Example 2: If sinA = 3/5 and cosB = 5/13, find the value of sin(A + B)

Solution:

Here,    sinA = 3/5

cosA = √(1 – sin2A) = √(1 – 9/25) = √(16/25) = 4/5

cosB = 5/13

sinB = √(1 – cos2B) = √(1 – 25/169) = √(144/169) = 12/13

Now,  sin(A + B) = sinA cosB + cosA sinB

= 3/5 . 5/13 + 4/5 . 12/13

= 15/65 + 48/65

= 63/65

Example 3: If sinA = 1/√10 and sinB = 1/√5, show that A + B = π/4

Solution:

Here,    sinA = 1/√10

cosA = √(1 – sin2A) = √(1 – 1/10) = √(9/10) = 3/√10

sinB = 1/√5

cosB = √(1 – sin2B) = √(1 – 1/5) = √(4/5) = 2/√5

Now,  sin(A + B) = sinA cosB + cosA sinB

= 1/√10 . 2/√5 + 3/√10 . 1/√5

= 2/√50 + 3/√50

= 5/√50

= 1/√2

= sin45° = sin(π/4)

A + B = π/4.  Proved.

Example 5: Prove that: sin35° + cos35° = √2 cos10°

Solution: Here,

RHS = √2 cos10°

= √2 cos(45° - 35°)

= √2 (cos45° cos35° + sin45° sin35°)

= √2 (1/√2 . cos35° + 1/√2 . sin35°)

= cos35° + sin35°

= LHS. Proved.

Example 6: Prove that: tan10° + tan35° + tan10° tan35° = 1

Solution:

We know that, 10° + 35° = 45°

So,        tan(10° + 35°) = tan45°

or,         (tan10° + tan35°)/(1 – tan10° tan35°) = 1

or,         tan10° + tan35° = 1 – tan10° tan35°

or,         tan10° + tan35° + tan10° tan35° = 1

Hence proved.

Example 7: If A, B and C are the angles of a triangle then prove that: tanA + tanB + tanC = tanA tanB tanC.

Solution: Here, A, B and C are angles of a triangle,

So,        A + B + C = 180°

or,         A + B = 180° - C

or,         tan(A + B) = tan(180° - C)  [Taking tan to both sides]

or,         (tanA + tanB)/(1 – tanA tanB) = - tanC

or,         tanA + tanB = - tanC + tanA tanB tanC

or,         tanA + tanB + tanC = tanA tanB tanC

Hence proved.

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