Trigonometric Ratios of Multiple Angles

Trigonometric Ratios of Multiple Angles

Trigonometric Ratios of Multiple Angles

If A is an angle, then 2A, 3A, 4A, 5A, etc are called multiple angles of A. In this section we will learn about the trigonometric ratios of multiple angles 2A and 3A in terms of A.

 

Trigonometric Ratios of 2A in terms of A


Trigonometric ratios of sin2A in terms of A

Trigonometric ratios of cos2A in terms of A

Trigonometric ratios of tan2A in terms of A

Trigonometric ratios of cot2A in terms of A

 

Some Useful Results

Some useful results on sin2A and cos2A


Trigonometric Ratios of 3A in terms of A

Trigonometric ratios of sin3A, cos3A and tan3A in terms of A


Here is the list of formulae for trigonometric ratios of multiple angles 2A and 3A in terms of A:

List of formulae for trigonometric ratios of multiple angles 2A and 3A


Geometrical Proof of 2A Formulae:

Let O be the centre of the circle ABC and AB be a diameter. Let CAB = A. Then COB = 2A. [Because, central angle is double than inscribed angle on same arc.]

O is the centre of the circle ABC, AB is a diameter and ∠CAB = A

Here, ACB = 90°. Let CM is perpendicular to AB. Then ACM = 90° - A and hence, BCM = A.

Now, sin2A = CM/OC  = 2CM/2OC  = 2CM/AB

          = 2 . CM/AC . AC/AB

          = 2sinA cosA

          cos2A = OM/OC  = 2OM/2OC  = 2OM/AB

                       = [(AO + OM) - (AO - OM)]/AB  = [(AO + OM) - (BO - OM)]/AB

                       = (AM - BM)/AB  = AM/AB - BM/AB

                       = AM/AC . AC/AB - BM/BC . BC/AB

                       = cosA cosB – sinA sinB

                       = cos2A – sin2A

          tan2A = CM/OM  = 2CM/2OM  = 2CM/[(AO + OM) – (AO – OM)]

                       = 2CM/[(AO + OM) – (BO – OM)]  = 2CM/(AM – BM)

                       = [2CM/AM]/[(AM – BM)/AM]  = [2CM/AM]/[AM/AM – BM/AM]

                       = [2CM/AM]/[1 – BM/CM . CM/AM]

                       = 2tanA/(1 – tanA tanA)

                       = 2tanA/(1 – tan2A)

 

Worked Out Examples

Example 1: If sinA = 5/13, find the value of sin2A, cos2A and tan2A.

Solution: Here,

          sinA = 5/13

        cosA = √(1 – sin2A) = √(1 – 25/169) = √(144/169) = 12/13

Now, sin2A = 2sinA cosA

        = 2 × 5/13 × 12/13

        = 120/169

          cos2A = cos2A – sin2A

         = 144/169 – 25/169

         = 119/169

          tan2A = sin2A/cos2A

                     = [120/169]/[119/169]

         = 120/119

 

Example 2: If tanA = 3/4, find the value of sin2A, cos2A and tan2A. Solution:  Here, 	tanA = 3/4 Now,	sin2A = 2tanA/(1 + 〖tan〗^2 A) = (2 × 3/4)/(1 + 9/16)  = (3/2)/(25/16) = 3/2 × 16/25 = 24/25 cos2A = (1 - 〖tan〗^2 A)/(1 + 〖tan〗^2 A) = (1 - 9/16)/(1 + 9/16)  = (7/16)/(25/16) = 7/16 × 16/25 = 7/25 tan2A = 2tanA/(1 - 〖tan〗^2 A) = (2 × 3/4)/(1 - 9/16) = (3/2)/(7/16) = 3/2 × 16/7 = 24/7

 
 

Example 3: Express cos5A in terms of cosA.

Solution:

 cos5A = cos(2A + 3A)

             = cos2A cos3A – sin2A sin3A

 = (2cos2A – 1)(4cos3A – 3cosA) – 2sinA cosA(3sinA – 4sin3A)

 = 8cos5A – 6cos3A – 4cos3A + 3cosA – 2cosA sin2A(3 – 4sin2A)

 = 8cos5A – 6cos3A – 4cos3A + 3cosA – 2cosA(1 – cos2A)(3 – 4 + 4cos2A)

 = 8cos5A – 6cos3A – 4cos3A + 3cosA – (2cosA – 2cos3A)(4cos2A - 1)

 = 8cos5A – 6cos3A – 4cos3A + 3cosA – 8cos3A + 2cosA + 8cos5A – 2cos3A

 = 16cos5A – 20cos3A + 5cosA

 

Example 4: Find the value of sin3A, cos3A and tan3A when sinA = √3/2 Solution:  Here, sinA = √3/2 ∴  cosA = √(1- 〖sin〗^2 A) = √(1- 3/4) = √(1/4) = 1/2 ∴  tanA = sinA/cosA = (√3/2)/(1/2) = √3 Now, 	     	           sin3A = 3sinA – 4sin3A = 3 × √3/2 - 4 × (3√3)/8 = (3√3)/2 - (3√3)/2 = 0 	cos3A = 4cos3A – 3cosA = 4 × 1/8 - 3 × 1/2 = 1/2 - 3/2 = -1 	tan3A = (3tanA - 〖tan〗^3 A)/(1 - 3〖tan〗^2 A) = (3√3  - 3√3)/(1 - 3×3) = 0/(1 - 9) = 0


Example 5: Prove that: (sinθ + sin2θ)/(1 + cosθ + cos2θ) = tanθ Solution: Here, 	LHS = (sinθ + sin2θ)/(1 + cosθ + cos2θ) 	        = (sinθ + 2sinθ cosθ)/(cosθ + 1 + cos2θ) 	        = (sinθ(1 + 2cosθ))/(cosθ + 2〖cos〗^2 θ) 	        = (sinθ(1 + 2cosθ))/(cosθ(1 + 2cosθ)) 	        = sinθ/cosθ 	        = tanθ = RHS. Proved.


Example 6: Prove that: (1- cos2θ + sin2θ)/(1 + cos2θ + sin2θ) = tanθ Solution: Here, 	LHS = (1- cos2θ + sin2θ)/(1 + cos2θ + sin2θ) 	        = (2〖sin〗^2 θ + 2sinθ cosθ)/(2〖cos〗^2 θ + 2sinθ cosθ) 	        = (2sinθ(sinθ + cosθ))/(2cosθ(cosθ + sinθ)) 	        = sinθ/cosθ 	        = tanθ = RHS. Proved.

 

Example 7: Prove that: tan⁡(45°+ θ)= cos2θ/(1-sin2θ) Solution: Here, 	RHS = cos2θ/(1-sin2θ) 	        = (〖cos〗^2 θ - 〖sin〗^2 θ)/(cosθ - sinθ)^2  	        = (cosθ + sinθ)(cosθ - sinθ)/(cosθ - sinθ)^2  	        = (cosθ + sinθ)/(cosθ - sinθ) 	        = (cosθ/cosθ  + sinθ/cosθ)/(cosθ/cosθ  - sinθ/cosθ) 	        = (1 + tanθ)/(1 - tanθ)  	        = (tan45° + tanθ)/(tan45° - tanθ) 	        = tan⁡(45°+ θ) = LHS. Proved.

 

Example 8: Show that: cos6θ – sin6θ = cos2θ(1 - 1/4 sin22θ) Solution: Here, 	LHS = cos6θ – sin6θ 	        = (cos2θ)3 – (sin2θ)3 	        = (cos2θ – sin2θ)(cos4θ + cos2θ sin2θ + sin4θ) 	        = cos2θ(cos4θ + 2cos2θ sin2θ + sin4θ – cos2θ sin2θ) 	        = cos2θ[(cos2θ + sin2θ)2 – cos2θ sin2θ] 	        = cos2θ[12 - 1/4  .4 cos2θ sin2θ] 	        = cos2θ[1 - 1/4 (2sinθ cosθ)2] 	        = cos2θ[1 - 1/4 (sin2θ)2] 	        = cos2θ(1 - 1/4 sin22θ) 	        = RHS. Proved.


Example 9: Prove that: √3 cosec20° - sec20° = 4 Solution: Here, 	LHS = √3 cosec20° - sec20° 	        = √3/(sin20°) - 1/(cos20°) 	        = (√3  cos20° - sin20°)/(sin20° cos20°) 	        = (2(√3  cos20° - sin20°))/(2sin20° cos20°) 	        = (4(√3/2  cos20° - 1/2  sin20°))/(sin40°) 	        = (4(sin60° cos20° - cos60°sin20°))/(sin40°) 	        = (4sin(60°-20°))/(sin40°) 	        = (4sin40°)/(sin40°) 	        = 4 = RHS. Proved.


Example 10: Prove that: cot(45° + A) + tan(45° - A) = 2cos2A/(1 + sin2A) Solution: Here, 	LHS = cot(45° + A) + tan(45° - A) 	        = (cot45° cotA - 1)/(cot45° + cotA) + (tan45°- tanA)/(tan45° tanA + 1) 	        = (cotA - 1)/(1 + cotA) + (1 - tanA)/(tanA + 1) 	        = (1/tanA  - 1)/(1 + 1/tanA)  + (1 - tanA)/(tanA + 1) 	        = (1 - tanA)/(tanA + 1) + (1 - tanA)/(tanA + 1) 	        = (2(1 – tanA))/(tanA + 1) 	        = (2(1 – sinA/cosA))/(sinA/cosA  + 1) 	        = (2(cosA – sinA))/(cosA + sinA) 	        = (2(cosA – sinA))/(cosA + sinA) × (cosA + sinA)/(cosA + sinA) 	        = (2(〖cos〗^2 A - 〖sin〗^2 A))/(cosA + sinA)^2  	        = 2cos2A/(1 + sin2A) = RHS. Proved.

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