Chapter 3: Trigonometric Functions

We know that,

tan 2A = tan (A + B + A – B)

And also,

According to the question,

cosα + cosβ = 0 = sinα + sinβ …(i)

Since, LHS = cos 2α + cos 2β

We know that,

cos 2x = cos²x – sin²x

Therefore,

LHS = cos²α – sin²α + (cos²β – sin²β)

⇒ LHS = cos²α + cos²β – (sin²α + sin²β)

Also, since,

a² + b² = (a+b)² – 2ab

⇒ LHS = (cosα + cosβ)² – 2cosα cosβ –(sinα + sinβ)² +2sinα sinβ

From equation (i),

⇒ LHS = 0 – 2cosα cosβ -0 + 2sinα sinβ

⇒ LHS = -2(cosα cosβ – sinα sinβ)

∵ cos (α + β) = cosα cosβ – sinα sinβ

Therefore, LHS = -2 cos (α + β) = RHS

Hence, cos 2α + cos 2β = –2cos (α + β)

According to the question,

We know that,

We know that,

tan(x-y) = (tan x – tan y) / (1 + tan x . tan y)

Therefore, tan θ = tan ( α – π/4)

⇒ θ = α – π/4

⇒ α = θ + π/4 …(i)

To prove,

sinα + cosα = √2 cos θ

∵ LHS = sinα + cosα

From equation (i)

⇒ LHS = sin(θ + π/4) + cos(θ + π/4)

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

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

Therefore, LHS = sin θ cos(π/4) + sin(π/4)cos θ + cos θ cos(π/4) – sin(π/4)sin θ

∵ sin(π/4)=cos(π/4) = 1/√2

⇒ LHS = sin θ (1/√2) + (1/√2) cos θ + cos θ (1/√2) – sin θ (1/√2)

⇒ LHS = 2 cos θ (1/√2)

⇒ LHS = √2 cos θ = RHS

Therefore, sinα + cosα = √2 cos θ

According to the question,

sin θ + cos θ = 1

As, sin θ + cos θ = 1

Since we know,

If sin θ = sinα ⇒ θ = nπ + (-1)ⁿα

We get,

θ + π/4 = nπ + (-1)ⁿ(π/4)

⇒ θ = nπ + (π/4)((-1)ⁿ – 1)

According to the question,

We have,

tan θ = -1

And cos θ =1/√2 .

⇒ θ = – π/4

So, we know that,

θ lies in IV quadrant.

θ = 2π – π/4 = 7π/4

So, general solution is θ = 7π/4 + 2 n π, n∈ Z

According to the question,

⇒ 1 = 2 cosec θ sin θ cos θ

We know that,

sin θ cosec θ = 1

⇒ 1 = 2 cos θ

⇒ cos θ = 1/2 = cos(π/3)

Hence,

The solution of cos x = cos α can be given by,

x = 2mπ ± α ∀ m ∈ Z

⇒ θ = 2nπ ± π/3, n ∈ Z

According to the question,

2sin²θ = 3cos θ

We know that,

sin²θ = 1 – cos²θ

Given that,

2 sin² θ = 3 cos θ

2 – 2 cos² θ = 3 cos θ

2 cos² θ + 3 cos θ – 2= 0

(cos θ + 2)(2 cos θ – 1) = 0

Therefore,

cos θ = ½ = cos π/3

θ = π/3 or 2π – π/3

θ = π/3, 5π/3

Therefore, 2(1 – cos²θ) = 3cos θ

⇒ 2 – 2cos²θ = 3cos θ

⇒ 2cos²θ + 3cos θ – 2 = 0

⇒ 2cos²θ + 4cos θ – cos θ – 2 = 0

⇒ 2cos θ (cos θ+ 2) +1 (cos θ + 2) = 0

⇒ (2cos θ + 1)(cos θ + 2) = 0

Since, cos θ ∈ [-1,1] , for any value θ.

So, cos θ ≠ – 2

Therefore,

2 cos θ – 1 = 0

⇒ cos θ = ½

= π/3 or 2π – π/3

θ = π/3, 5π/3

According to the question,

sec x cos 5x = -1

⇒ cos 5x = -1/sec x

We know that,

sec x = 1/cos x

⇒ cos 5x + cos x = 0

By transformation formula of T-ratios,

We know that,

⇒ 2 cos 3x cos 2x = 0

⇒ cos 3x = 0 or cos 2x = 0

∵ 0 < x ≤ π/2

Therefore, 0< 2x ≤ π or 0< 3x ≤ 3π/2

Therefore, 2x = π/2

⇒ x = π/4

3x = π/2

⇒ x = π/6

Or 3x = 3π/2

⇒ x = π/2

Hence, x = π/6, π/4, π/2.

According to the question,

sin (θ + α) = a and sin(θ + β) = b

LHS = cos 2(α – β) – 4ab cos (α – β)

Using cos 2x = 2cos²x – 1,

Let us solve,

⇒ LHS = 2cos²(α – β) – 1 – 4ab cos(α – β)

⇒ LHS = 2cos (α – β) {cos (α – β) – 2ab} – 1

Since,

cos (α – β) = cos {(θ + α) – (θ + β)}

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

⇒ cos (α – β) = cos(θ + α)cos(θ + β) + sin(θ + α)sin(θ + β)

Since,

sin(θ + α) = a

⇒ cos(θ + α) = √(1 – sin²(θ + α) = √(1 – a²)

Similarly,

cos(θ + β) = √(1 – b²)

Therefore,

cos(α – β) = √(1-a²)√(1-b²) + ab

Therefore,

LHS = 2{ab + √(1 – a²)(1 – b²)}{ab + √(1 – a²)(1 – b²) -2ab} – 1

⇒ LHS = 2{√(1 – a²)(1 – b²) + ab}{√(1 – a²)(1 – b²) – ab}-1

Using (x + y)(x – y) = x² – y²

⇒ LHS = 2{(1-a²)(1-b²) – a²b²} – 1

⇒ LHS = 2{1 – a² – b² + a²b²} – 1

⇒ LHS = 2 – 2a² – 2b² – 1

⇒ LHS = 1 – 2a² – 2b² = RHS

Therefore,

We get,

cos 2(α – β) – 4ab cos (α – β) = 1 – 2a² – 2b²