Trigonometric Ratios of Compound Angles

Trigonometric Ratios of Compound Angles

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.

Trigonometric Ratios of Compound Angle A + B (Addition Formula): 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.

Trigonometric Ratios of Compound Angle A – B (Subtraction Formula): 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): tan(A + B) = (sin(A + B))/(cos(A + B)) 	 = (sinA cosB + cosA sinB)/(cosA cosB - sinA sinB) 	 = ((sinA cosB+cosA sinB)/(cosA cosB))/((cosA cosB-sinA sinB)/(cosA cosB))  [Dividing numerator and denominator by cosA cosB] 	 = ((sinA cosB)/(cosA cosB)  + (cosA sinB)/(cosA cosB))/((cosA cosB)/(cosA cosB)  - (sinA sinB)/(cosA cosB)) 	 = (tanA + tanB)/(1 - tanA tanB) Hence, the tangent formula of compound angle (A + B) is, tan(A + B) = (tanA + tanB)/(1 - tanA tanB)

Tangent Formula of Compound Angle (A - B)

Tangent Formula of Compound Angle (A - B): tan(A - B)  = (sin(A - B))/(cos(A - B)) 	  = (sinA cosB - cosA sinB)/(cosA cosB + sinA sinB) 	  = ((sinA cosB - cosA sinB)/(cosA cosB))/((cosA cosB + sinA sinB)/(cosA cosB))  [Dividing numerator and denominator by cosA cosB]                 = ((sinA cosB)/(cosA cosB)  - (cosA sinB)/(cosA cosB))/((cosA cosB)/(cosA cosB)  + (sinA sinB)/(cosA cosB))                 = (tanA - tanB)/(1 + tanA tanB) Hence, the tangent formula of compound angle (A - B) is, tan(A - B) = (tanA - tanB)/(1 + tanA tanB)

Cotangent Formula of Compound Angle (A + B)

Cotangent Formula of Compound Angle (A + B): cot(A + B) = (cos(A + B))/(sin(A + B)) 	 = (cosA cosB - sinA sinB)/(sinA cosB + cosA sinB) 	 = ((cosA cosB - sinA sinB)/(sinA sinB))/((sinA cosB + cosA sinB)/(sinA sinB))  [Dividing numerator and denominator by sinA sinB] 	 = ((cosA cosB)/(sinA sinB)  - (sinA sinB)/(sinA sinB))/((sinA cosB)/(sinA sinB)  + (cosA sinB)/(sinA sinB))	 	 = (cotA tanB - 1)/(tanB + cotA) Hence, the cotangent formula of compound angle (A + B) is, cot(A + B) = (cotA tanB - 1)/(tanB + cotA)

Cotangent Formula of Compound Angle (A - B)

Cotangent Formula of Compound Angle (A - B): cot(A - B)  = (cos(A - B))/(sin(A - B)) 	= (cosA cosB + sinA sinB)/(sinA cosB - cosA sinB) 	= ((cosA cosB + sinA sinB)/(sinA sinB))/((sinA cosB - cosA sinB)/(sinA sinB))  [Dividing numerator and denominator by sinA sinB] 	= ((cosA cosB)/(sinA sinB)  + (sinA sinB)/(sinA sinB))/((sinA cosB)/(sinA sinB)  - (cosA sinB)/(sinA sinB)) 	= (cotA tanB + 1)/(tanB - cotA) Hence, the cotangent formula of compound angle (A - B) is, cot(A - B) = (cotA tanB + 1)/(tanB - cotA)

List of Formulae for Trigonometric Ratios of Compound Angles:-


Worked Out Examples

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

Solution:  (a) sin75° = sin(45° + 30°) = sin45° cos30° + cos45° sin30°  = 1/√2  .  √3/2   +  1/√2  .  1/2  = √3/(2√2)  + 1/(2√2)  = (√3  + 1)/(2√2)       (b) tan15° = tan(45° - 30°)  = (tan45° - tan30°)/(1 + tan45° tan30°)  = (1 - 1/√3)/(1 + 1 .1/√3  )  = (√3  - 1)/(√3  + 1)  = (√3  - 1)/(√3  + 1) × (√3  - 1)/(√3  - 1)  = (√3  - 1)^2/((√3)^2- 1^2 )  =  ((√3)^2  – 2.√3.1 + 1^2  )/(3 - 1)  = (4 – 2√3  )/2  = 2 – √3

 

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 4: Prove that: (cos14° - sin14°)/(cos14° + sin14°) = cot59°  Solution:  RHS = cot59°  = cot(45° + 14°)   = (cot45° cot14°- 1)/(cot45° + cot14°)   = (cot14°- 1)/(1 + cot14°)   = ((cos14°)/(sin14°)  - 1)/(1 + (cos14°)/(sin14°))   = ((cos14° - sin14°)/(sin14°))/((cos14° + sin14°)/(sin14°))   = (cos14° - sin14°)/(cos14° + sin14°)  = LHS. 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.

  

You can comment your questions or problems regarding the trigonometric ratios of compound angles here.

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