Trigonometric Ratios of Multiple Angles

September 25, 2020 Trigonometry

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 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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