Problems using Compound Angle Formulae

We will learn how to solve various types of problems using compound angle formulae. While solving the problems we need to keep in mind all the formulas of trigonometric ratios of compound angles and use the formula according to the question.

1. If ABCD is a cyclic quadrilateral, then show that cos A + cos B + cos C + cos D = 0.

Solution:

Since ABCD is a cyclic quadrilateral,

A + C = π ⇒ C = π - A

B + D = π ⇒ D = π - B

Therefore, cos A + cos B + cos C + cos D

= cos A + cos B + cos (π - A) + cos (π - B)

= cos A + cos B - cos A - cos B, [Since, cos (π - A) = - cos A and cos (π - B) = - cos B]

= 0

2. Show that, cos^2A + cos^2 (120° - A) + cos^2 (120° + A) = 3/2

Solution:

L. H. S. = cos^2 A + (cos 120° cos A + sin 120° sin A)^2 + (cos 120° cos A - sin 120° sin A)^2

= cos^2 A + 2(cos^2 120° cos^2 α + sin^2 120° sin^2 α), [Since, (a + b)^2 + (a - b)^2 = 2(a^2 + b^2)]

= cos^2 A + 2[(-1/2)^2 cos^2 A + (√3/2)^2 sin^2 A], [Since, cos 120° = cos (2 ∙ 90° - 60°) = - cos 60°= -1/2 and sin 120°

= sin (2 ∙ 90° - 60°) = sin 60° = √3/2]

= cos^2 A + 2[1/4 cos^2 A + 3/4 sin^2 A]

= 3/2(cos^2 A + sin^2 A)

= 3/2                                 Proved.

 

3. If A, B, and C are angles of a triangle, then prove that tan A/2 = cot (B + C)/2

Solution:

Since A, B, and C are angles of a triangle, A + B + C = π

⇒ B + C = π - A

⇒ (B + C)/2 = π/2 - A/2

Therefore, cot (B + C)/2 = cot (π/2 - A/2) = tan A/2                                Proved.

 

Proof the problems using compound angle formulae.

4. If tan x - tan y = m and cot y - cot x = n, prove that,
                                1/m + 1/n = cot (x - y).

Solution:

We have, m = tan x - tan y

⇒ m = sin x / cos x - sin y/cos y = (sin x cos y - cos x sin y)/cos x cos y

⇒ m = sin (x - y)/cos x cos y                

Therefore, 1/m = cos x cos y/sin (x - y)                           (1)

Again, n = cot y - cot x = cos y/sin y - cos x/sin x = (sin x cos y - cos x sin y)/sin y sin x

⇒ n = sin (x - y)/sin y sin x                   

Therefore, 1/n = sin y sin x/sin (x - y)                              (2)

Now, (1) + (2) gives,

1/m + 1/n = (cos x cos y + sin y sin x)/sin (x - y) = cos (x - y)/sin (x - y)

⇒ 1/m + 1/n = cot (x - y).                                 Proved.

 

5. If tan β = sin α cos α/(2 + cos^2 α) prove that 3 tan (α - β) = 2 tan α.

Solution:    

We have, tan (α - β) = (tan α - tan β)/1 + tan α tan β                                                                                 

⇒ tan (α - β) = [(sin α/cos α) - sin α cos α/(2 + cos^2 α)]/[1 + (sin α / cos α) ∙ sin α cos α/(2 + cos^2 α)],  [Since, tan β = sin α cos α/(2 + cos^2 α)]                                                               

= (2 sin α + sin α cos^2 α - sin αcos^2 α)/(2 cos α + cos^3 α + sin^2 α cos α)                                                   

= 2 sin α/cos α (2 + cos^2 α + sin^2 α)

= 2 sin α/3 cos α                                                                                      

⇒ 3 tan (α - β) = 2 tan α                                 Proved.

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● Compound Angle

• Proof of Compound Angle Formula sin (α + β)
• Proof of Compound Angle Formula sin (α - β)
• Proof of Compound Angle Formula cos (α + β)
• Proof of Compound Angle Formula cos (α - β)
• Proof of Compound Angle Formula sin 22 α - sin 22 β
• Proof of Compound Angle Formula cos 22 α - sin 22 β
• Proof of Tangent Formula tan (α + β)
• Proof of Tangent Formula tan (α - β)
• Proof of Cotangent Formula cot (α + β)
• Proof of Cotangent Formula cot (α - β)
• Expansion of sin (A + B + C)
• Expansion of sin (A - B + C)
• Expansion of cos (A + B + C)
• Expansion of tan (A + B + C)
• Compound Angle Formulae
• Problems using Compound Angle Formulae
• Problems on Compound Angles

11 and 12 Grade Math

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