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, \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\)

= sin^-1(sin( \(\frac{-\pi}{3}\)))  ----- Since \(\frac{-\pi}{3}\) ∈ [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)]

∴ sin-1(sin( \(\frac{-\sqrt{3}}{2}\))) = \(\frac{-\pi}{3}\)

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, \(\cot \left[\cos^{−1}\left(\frac{7}{25}\right)\right]\)

Let \(\cos^{−1}\left(\frac{7}{25}\right)\)= θ

⇒ cosθ = 7/25

⇒ sinθ = 24/25

⇒ cotθ = 7/24

∴  \(\cot \left[\cos^{−1}\left(\frac{7}{25}\right)\right]\)

= 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, \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\)

= sin^-1(sin( \(\frac{-\pi}{3}\)))  ----- Since \(\frac{-\pi}{3}\) ∈ [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)]

∴ sin-1(sin( \(\frac{-\sqrt{3}}{2}\))) = \(\frac{-\pi}{3}\)

Additional InformationPrincipal Values of Inverse Trigonometric Functions:

Calculation 

16(sec^–1 x)² + (cosec^–1 x)²

\(\operatorname{Sec}^{-1} \mathrm{x}=\mathrm{a} \in[0, π]-\left\{\frac{π}{2}\right\}\)

\(\operatorname{cosec}^{-1} \mathrm{x}=\frac{π}{2}-\mathrm{a}\)

= \(16\left[\mathrm{a}^{2}+\left(\frac{π}{2}-\mathrm{a}\right)^{2}\right]=16\left[2 \mathrm{a}^{2}-π \mathrm{a}+\frac{π^{2}}{4}\right]\)

\(\max ]_{a=π}=16\left[2 π^{2}-π^{2}+π \frac{2}{4}\right]=20 π^{2}\)

\(\min ]_{a=\frac{π}{4}}=16\left[\frac{2 \times π^{2}}{16}-\frac{π^{2}}{4}+\frac{π^{2}}{4}\right]=2 π^{2}\)

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, \(\tan x=-\frac{1}{\sqrt{3}}\)

⇒ 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 

\({\sin ^{ - 1}}(sinx)= x\), x ∈ [\(\rm \frac {-\pi}{2},\rm \frac {\pi}{2}\)]

Calculation:

\(\rm {\sin ^{ - 1}}(sin \frac{{3\pi }}{5})\)

\(\rm = {\sin ^{ - 1}}[sin \rm (\pi - \frac{{2\pi }}{5})]\)

\(= {\sin ^{ - 1}}(sin \frac{{2\pi }}{5})\)                       (∵ sin (π - θ) = sin θ)

\(= \frac{{2\pi }}{5}\)

Hence, option (2) is correct.

Concept: 

sin^-1 (sin θ ) = θ , if \(\rm -\frac{π}{2}\leq θ\leq \frac{π}{2}\) 

Calculation: 

We have, θ = 9 radian, which does not lie between \(\rm -\frac{π}{2}\) and \(\rm \frac{π}{2}\) . But, 3π - θ  i.e,  3π - 9 lies between \(\rm -\frac{π}{2}\) and \(\rm \frac{π}{2}\) . 

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:

\(\rm \sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)+\cos^{-1}\left(\cos \left(\dfrac{7\pi}{6}\right)\right)\)

\(\rm =-\sin^{-1}\dfrac{\sqrt{3}}{2}+\cos^{-1}\left(\cos \left(\pi+\dfrac{\pi}{6}\right)\right)\)

\(\rm =-\sin^{-1}\dfrac{\sqrt{3}}{2}+\cos^{-1}\left(-\cos \dfrac{\pi}{6}\right)\)

\(\rm =-\dfrac{\pi}{3}+\left(\pi-\dfrac{\pi}{6}\right)\)

\(\rm =\dfrac{\pi}{2}\).

Explanation:

Given: x∈ [-π/2, π/2] 

sin-1 (sin x) = x

Concept:

cos (-θ) = cos θ 

\(\cos\left({\pi\over2}\right) = 0\)

Calculation:

x = \(\rm \tan^{-1}\left[\cos\left(-{\pi\over2}\right)\right]\)

x = \(\rm \tan^{-1}\left[\cos\left({\pi\over2}\right)\right]\)

x = \(\rm \tan^{-1}\left[0\right]\)

∵ Principle value can be ∈ \(\left[-{\pi\over2},{\pi\over2}\right]\)

∴ x = 0

Concept:

Calculation: 

As we know sin-1 (-x) = - sin-1 x

So,  \(\rm \sin^{-1} \left ( \frac {-1}{\sqrt 2}\right) = - \sin^{-1} \left ( \frac {1}{\sqrt 2}\right)\)

Let \( \sin^{-1} \left ( \frac {1}{\sqrt 2}\right) \) = θ 

⇒ sin θ = \(\frac{1}{\sqrt 2}\) = sin 45° 

∴ θ = 45° 

Hence, \(\rm \sin^{-1} \left ( \frac {-1}{\sqrt 2}\right)\) = -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 \(\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\)

Explanation:

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

cot (cot^-1 x) =x for x ∈ R

We have, \(\cot \left[\cos^{−1}\left(\frac{7}{25}\right)\right]\)

Let \(\cos^{−1}\left(\frac{7}{25}\right)\)= θ

⇒ cosθ = 7/25

⇒ sinθ = 24/25

⇒ cotθ = 7/24

∴  \(\cot \left[\cos^{−1}\left(\frac{7}{25}\right)\right]\)

= cotθ 

= 7/24