Chapter 3: Trigonometric Functions

According to the question,

Let, y = 3[sin⁴ (3π/2 – α) + sin ⁴ (3π + α)] – 2[sin⁶ (π/2 + α) + sin⁶ (5π – α)]

We know that,

sin(3π/2 – α) = -cos α

sin(3π + α) = -sin α

sin(π/2 + α) = cos α

sin(5π – α) = sin α

Therefore,

y = 3[(– cos α)⁴ + (– sin α)⁴] – 2[cos⁶ α + sin⁶ α]

⇒ y = 3 [cos⁴α + sin⁴α] – 2[sin⁶α + cos⁶α]

⇒ y = 3[(sin²α + cos²α)² – 2sin²α cos²α] – 2[(sin²α)³ + (cos²α)³]

Since, we know that,

sin²α + cos²α = 1

Also, we know that,

a³+b³ = (a+b)(a² – ab + b²)

⇒ y = 3[1 – 2sin²α cos²α] – 2[(sin²α + cos²α)( cos⁴α + sin⁴α- sin²α cos²α)]

⇒ y = 3[1 – 2sin²α cos²α] – 2[cos⁴α + sin⁴α- sin²α cos²α]

⇒ y = 3[1 – 2sin²α cos²α] – 2[(sin²α + cos²α)² – 2sin²α cos²α – sin²α cos²α]

⇒ y = 3[1 – 2sin²α cos²α] – 2[1 – 3sin²α cos²α]

⇒ y = 3 – 6sin²α cos²α – 2 + 6 sin²α cos²α

⇒ y = 1

According to the question,

a cos2θ + b sin 2θ = c

α and β are the roots of the equation.

Using the formula of multiple angles,

We know that,

⇒ a(1 – tan²θ) + 2b tan θ – c(1 + tan²θ) = 0

⇒ (-c – a)tan²θ + 2b tan θ – c + a = 0 …(i)

We know that,

The sum of roots of a quadratic equation, ax² + bx + c = 0 is given by (-b/a)

Therefore,

tan α + tan β = –2b/–(c + a) = 2b/(c + a)

Hence, tan α + tan β = 2b/(c + a)

According to the question,

x = sec ϕ – tan ϕ and y = cosec ϕ + cot ϕ

Given that, LHS = xy + x – y + 1

Thus, LHS = xy + x – y + 1 = 0

According to the question,

cos θ = 8/17

sin θ = ±√(1 – cos²θ)

Since, θ lies in first quadrant, only positive sign can be considered.

⇒ sin θ = √(1 – 64/289) = 15/17

Let, y = cos(30° + θ) + cos (45° – θ) + cos (120° – θ)

We know that,

cos(x + y) = cos x cos y – sin x sin y

Therefore,

y = cos30° cos θ – sin30° sin θ + cos45° cos θ + sin45°sin θ +cos120° cos θ + sin120° sin θ

Substituting values of cos30°, sin30°, cos 120°, sin120° and cos 45°

According to the question,

Let y = cos⁴(π/8) + cos⁴(3π/8) + cos⁴(5π/8) + cos⁴(7π/8).

⇒ y = cos⁴(π/8) + cos⁴(3π/8) + cos⁴(π – 3π/8) + cos⁴(π – π/8).

Since we know that, cos (π – x) = – cos x, we get,

= 2 – (1/√2)²

= 2 – ½

= 3/2

According to the question,

5cos²θ + 7sin²θ – 6 = 0

We know that,

sin²θ = 1 – cos²θ

Therefore, 5cos²θ + 7(1 – cos²θ) – 6 = 0

⇒ 5cos²θ + 7 – 7cos²θ – 6 = 0

⇒ -2cos²θ + 1 = 0

⇒ cos²θ = ½

Therefore, cos θ = ±1/√2

Therefore, cos θ = cos π/4 or cos θ = cos 3π/4

Since, solution of cos x = cos α is given by

x = 2mπ ± α ∀ m ∈ Z

θ = nπ ± π/4, n ∈ Z

According to the question,

sin x – 3sin2x + sin3x = cos x – 3cos2x + cos3x

Grouping sin x and sin 3x in LHS and, cos x and cos 3x in RHS,

We get,

sin x + sin3x – 3sin2x = cos x + cos3x – 3cos2x

Applying transformation formula,

cos A + cos B = 2cos ((A + B)/2) cos((A – B)/2)

sin A + sin B = 2sin ((A + B)/2) cos((A – B)/2)

⇒

⇒ 2sin 2x cos x – 3sin 2x = 2cos 2x cos x – 3cos 2x

⇒ 2sin 2x cos x – 3sin 2x – 2cos 2x cos x + 3cos 2x = 0

⇒ 2cos x (sin 2x – cos 2x) – 3(sin 2x – cos 2x) = 0

⇒ (sin 2x – cos 2x)(2cos x – 3) = 0

⇒ cos x = 3/2 or sin 2x = cos 2x

As cos x ∈ [-1,1]

Hence, no value of x exists for which cos x = 3/2

Therefore, sin 2x = cos 2x

⇒ tan 2x = 1 = tan π/4

We know solution of tan x = tan α is given by,

x= nπ + α , n ∈ Z

Therefore, 2x = nπ + (π/4)

⇒ x = nπ/2 + (π/8), n ∈ Z

Let, r sinα = √3 – 1 and r cosα = √3 + 1

Therefore, r = √{(√3 – 1)² + (√3 + 1)²} = √8 = 2√2

And, tan α = (√3 – 1) / (√3 + 1)

Therefore, r(sinα cos θ + cosα sin θ) = 2

⇒ r sin (θ+α) = 2

⇒ sin (θ+α) = 1/√2

⇒ sin (θ+α) = sin (π/4)

⇒ θ+α = nπ + (– 1)ⁿ (π/4), n ∈ Z

⇒ θ = nπ + (– 1)ⁿ (π/4) – (π/12), n ∈ Z

Objective Type Questions

C. 2

Explanation:

According to the question,

sin θ + cosec θ = 2

Squaring LHS and RHS,

We get,

⇒ (sin θ + cosec θ)² = 2²

⇒ sin²θ + cosec²θ + 2 sin θ cosec θ = 4

⇒ sin²θ + cosec²θ + 2 sin θ (1/sin)θ = 4

⇒ sin²θ + cosec²θ + 2 = 4

⇒ sin²θ + cosec²θ = 2

Thus, option (C) 2 is the correct answer.