Principal Values MCQ Quiz - Objective Question with Answer for Principal Values - Download Free PDF

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

sin (sin^-1 x) =x for -π/2 ≤ x ≤ π/2

We have, 

= sin^-1(sin( ))  ----- Since  ∈ [, ]

∴ sin-1(sin( )) = 

Additional InformationPrincipal Values of Inverse Trigonometric Functions:

Concept:

• Basic Trigonometric Identity: sin²x + cos²x = 1 is true for all real x.
• Comparing sin²x + cos²x = 1 + sin2x implies 1 = 1 + sin2x ⇒ sin2x = 0.
• tanθ values: tan15° = 2 − √3, tan75° = 2 + √3, tan45° = 1
• Inverse Trigonometric Identity: sin⁻¹x + cos⁻¹x = π⁄2 holds for all x ∈ [−1, 1]
• tanx = √3 ⇒ x = π⁄3 + nπ. Count within interval to find number of solutions.

Calculation:

Given, sin²x + cos²x = 1 + sin2x in [0, 2π]

⇒ sin²x + cos²x = 1

⇒ 1 = 1 + sin2x

⇒ sin2x = 0

⇒ 2x = nπ

⇒ x = nπ⁄2

⇒ x ∈ [0, 2π]

⇒ x = 0, π⁄2, π, 3π⁄2, 2π

⇒ Total 5 values

⇒ Check sin2x = 0 at these

⇒ Valid for all 5

⇒ LHS = 1, RHS = 1 + 0 = 1

⇒ Equation holds

⇒ But sin²x + cos²x = 1 always, so only valid if sin2x = 0

⇒ So number of solutions = 5

⇒ But due to contradiction from LHS = RHS, actual valid solutions are those where sin2x = 0 only

⇒ Thus, a → P = 2 solutions

Now, tan15° + tan75° − tan45°

⇒ (2 − √3) + (2 + √3) − 1

⇒ 4 − 1 = 3

⇒ b → Q

tanx = √3 in [0, 3π]

⇒ tanx = √3

⇒ x = π⁄3 + nπ

⇒ General solution: x = π⁄3, 4π⁄3, 7π⁄3

⇒ All ≤ 3π

⇒ c → R = 3 solutions

Also, cos⁻¹(x) + sin⁻¹(x) = π⁄2

⇒ Identity true for all x ∈ [−1, 1]

⇒ Infinite values

⇒ d → S

∴ Correct Matches:
(a) → P
(b) → Q
(c) → R
(d) → S
Hence, correct option is (B).

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

cot (cot^-1 x) =x for x ∈ R

We have, 

Let = θ

⇒ cosθ = 7/25

⇒ sinθ = 24/25

⇒ cotθ = 7/24

∴  

= cotθ 

= 7/24

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

sin (sin^-1 x) =x for -π/2 ≤ x ≤ π/2

We have, 

= sin^-1(sin( ))  ----- Since  ∈ [, ]

∴ sin-1(sin( )) = 

Additional InformationPrincipal Values of Inverse Trigonometric Functions:

Calculation 

16(sec^–1 x)² + (cosec^–1 x)²

= 

Sum = 22π² 

Hence option 4 is correct

Concept:

The principal solutions of a trigonometric equation are those solutions that lie between 0 and 2π.

Formula:

General solution of tan(x) = tan(α) is  given as;

x = nπ + α where α ∈ (-π/2 , π/2) and n ∈ Z.

Calculation:

Given, 

⇒ tan(x) = tan(-π/6)

∴ α = -π/6

⇒ x = nπ + (-π/6) , n ∈ Z

Putting n = 1 and 2, we get - 

x = 5π/6 and 11π/6

Concept:

sin-1 (sin x) = x

Calculation

Concept:

sin (π - θ) = sin θ 

, x ∈ []

Calculation:

                       (∵ sin (π - θ) = sin θ)

Hence, option (2) is correct.

Concept: 

sin^-1 (sin θ ) = θ , if  

Calculation: 

We have, θ = 9 radian, which does not lie between  and  . But, 3π - θ  i.e,  3π - 9 lies between  and  . 

Also , sin (3π - 9) = sin 9.

∴  sin^-1 (sin 9) = sin^-1 ( sin (3π - 9) ) = 3π - 9 

The correct option is 4.

Alternate Method

Referring to the graph of the given periodic function,

For x = 9 ϵ [2.5π  3π]

⇒ sin-1 (sin x) = - (x -3π) =  3π - x 

⇒ sin-1 (sin 9) = - (9 -3π) =   3π - 9

Concept:

Calculation: 

Using the concepts above, we can find the principal value of the given expression as follows:

.

Explanation:

Given: x∈ [-π/2, π/2] 

sin-1 (sin x) = x

Concept:

cos (-θ) = cos θ 

Calculation:

x = 

x = 

x = 

∵ Principle value can be ∈ 

∴ x = 0

Concept:

Calculation: 

As we know sin-1 (-x) = - sin-1 x

So,  

Let  = θ 

⇒ sin θ =  = sin 45° 

∴ θ = 45° 

Hence,  = -θ = -45°

Concept:

Calculation:

Let us assume sin-1 2x = y

⇒ 2x = sin y

We know that maximum and minimum values of sine function lie between -1 to 1

⇒ -1 ≤ sin y ≤ 1

⇒ -1 ≤ 2x ≤ 1

Apply sin-1 for the above inequality

⇒ sin-1 (-1) ≤ sin-1 2x ≤ sin-1 (1)

⇒ (-π /2) ≤ sin-1 2x ≤ (π /2)

Hence, the principal value of sin-1 2x lies in the interval 

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

cot (cot^-1 x) =x for x ∈ R

We have, 

Let = θ

⇒ cosθ = 7/25

⇒ sinθ = 24/25

⇒ cotθ = 7/24

∴  

= cotθ 

= 7/24