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Inverse Trigonometric Functions question

2022 · 28 Jul · Shift 1 · Q26
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  5. /2022 · 28 Jul · Shift 1 · Q26

Inverse Trigonometric Functions question

2022 · 28 Jul · Shift 1 · Q26

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)cos−1(x)−2sin−1(x)=cos−1(2x) is equal to :
  1. A
    0
  2. B
    1
  3. C
    12\frac{1}{2}21​
  4. D
    −12-\frac{1}{2}−21​
View written solutionFree

Correct answer: A

  1. Write the given equation

We need to solve

cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos^{-1}(x)-2\sin^{-1}(x)=\cos^{-1}(2x)cos−1(x)−2sin−1(x)=cos−1(2x)

with principal values.

  1. Determine the domain

For the inverse trigonometric functions to be defined:

  • sin⁡−1(x)\sin^{-1}(x)sin−1(x) requires x∈[−1,1]x\in[-1,1]x∈[−1,1]
  • cos⁡−1(x)\cos^{-1}(x)cos−1(x) requires x∈[−1,1]x\in[-1,1]x∈[−1,1]
  • cos⁡−1(2x)\cos^{-1}(2x)cos−1(2x) requires 2x∈[−1,1]⇒x∈[−12,12]2x\in[-1,1] \Rightarrow x\in\left[-\tfrac12,\tfrac12\right]2x∈[−1,1]⇒x∈[−21​,21​]

So overall,

x∈[−12,12].x\in\left[-\frac12,\frac12\right].x∈[−21​,21​].
  1. Use the identity relating sin⁡−1x\sin^{-1}xsin−1x and cos⁡−1x\cos^{-1}xcos−1x

For principal values,

sin⁡−1(x)+cos⁡−1(x)=π2.\sin^{-1}(x)+\cos^{-1}(x)=\frac\pi2.sin−1(x)+cos−1(x)=2π​.

Hence,

cos⁡−1(x)=π2−sin⁡−1(x).\cos^{-1}(x)=\frac\pi2-\sin^{-1}(x).cos−1(x)=2π​−sin−1(x).

Substitute into the equation:

(π2−sin⁡−1(x))−2sin⁡−1(x)=cos⁡−1(2x).\left(\frac\pi2-\sin^{-1}(x)\right)-2\sin^{-1}(x)=\cos^{-1}(2x).(2π​−sin−1(x))−2sin−1(x)=cos−1(2x).

So,

π2−3sin⁡−1(x)=cos⁡−1(2x).\frac\pi2-3\sin^{-1}(x)=\cos^{-1}(2x).2π​−3sin−1(x)=cos−1(2x).

Now use

cos⁡−1(2x)=π2−sin⁡−1(2x)\cos^{-1}(2x)=\frac\pi2-\sin^{-1}(2x)cos−1(2x)=2π​−sin−1(2x)

(valid for principal values, since 2x∈[−1,1]2x\in[-1,1]2x∈[−1,1]). Thus,

π2−3sin⁡−1(x)=π2−sin⁡−1(2x).\frac\pi2-3\sin^{-1}(x)=\frac\pi2-\sin^{-1}(2x).2π​−3sin−1(x)=2π​−sin−1(2x).

Therefore,

3sin⁡−1(x)=sin⁡−1(2x).3\sin^{-1}(x)=\sin^{-1}(2x).3sin−1(x)=sin−1(2x).
  1. Let
θ=sin⁡−1(x).\theta=\sin^{-1}(x).θ=sin−1(x).

Then

x=sin⁡θ,x=\sin\theta,x=sinθ,

where

θ∈[−π6,π6]\theta\in\left[-\frac\pi6,\frac\pi6\right]θ∈[−6π​,6π​]

because x∈[−12,12]x\in\left[-\tfrac12,\tfrac12\right]x∈[−21​,21​].

The equation becomes

sin⁡−1(2x)=3θ.\sin^{-1}(2x)=3\theta.sin−1(2x)=3θ.

Taking sine on both sides is valid because both sides lie in principal range:

  • sin⁡−1(2x)∈[−π2,π2]\sin^{-1}(2x)\in\left[-\frac\pi2,\frac\pi2\right]sin−1(2x)∈[−2π​,2π​]
  • 3θ∈[−π2,π2]3\theta\in\left[-\frac\pi2,\frac\pi2\right]3θ∈[−2π​,2π​]

So,

2x=sin⁡(3θ).2x=\sin(3\theta).2x=sin(3θ).

But x=sin⁡θx=\sin\thetax=sinθ, hence

2sin⁡θ=sin⁡3θ.2\sin\theta=\sin 3\theta.2sinθ=sin3θ.

Using

sin⁡3θ=3sin⁡θ−4sin⁡3θ,\sin 3\theta=3\sin\theta-4\sin^3\theta,sin3θ=3sinθ−4sin3θ,

we get

2sin⁡θ=3sin⁡θ−4sin⁡3θ.2\sin\theta=3\sin\theta-4\sin^3\theta.2sinθ=3sinθ−4sin3θ.

Thus,

0=sin⁡θ−4sin⁡3θ0=\sin\theta-4\sin^3\theta0=sinθ−4sin3θ sin⁡θ(1−4sin⁡2θ)=0.\sin\theta(1-4\sin^2\theta)=0.sinθ(1−4sin2θ)=0.

Since x=sin⁡θx=\sin\thetax=sinθ,

x(1−4x2)=0.x(1-4x^2)=0.x(1−4x2)=0.

So the possible solutions are

x=0,x=12,x=−12.x=0,\quad x=\frac12,\quad x=-\frac12.x=0,x=21​,x=−21​.
  1. Verify each solution in the original equation
  • For x=0x=0x=0:
cos⁡−1(0)−2sin⁡−1(0)=π2−0=π2,\cos^{-1}(0)-2\sin^{-1}(0)=\frac\pi2-0=\frac\pi2,cos−1(0)−2sin−1(0)=2π​−0=2π​, cos⁡−1(0)=π2.\cos^{-1}(0)=\frac\pi2.cos−1(0)=2π​.

Valid.

  • For x=12x=\frac12x=21​:
cos⁡−1(12)−2sin⁡−1(12)=π3−2⋅π6=0,\cos^{-1}\left(\frac12\right)-2\sin^{-1}\left(\frac12\right)=\frac\pi3-2\cdot\frac\pi6=0,cos−1(21​)−2sin−1(21​)=3π​−2⋅6π​=0, cos⁡−1(1)=0.\cos^{-1}(1)=0.cos−1(1)=0.

Valid.

  • For x=−12x=-\frac12x=−21​:
cos⁡−1(−12)−2sin⁡−1(−12)=2π3−2(−π6)=π,\cos^{-1}\left(-\frac12\right)-2\sin^{-1}\left(-\frac12\right)=\frac{2\pi}3-2\left(-\frac\pi6\right)=\pi,cos−1(−21​)−2sin−1(−21​)=32π​−2(−6π​)=π, cos⁡−1(−1)=π.\cos^{-1}(-1)=\pi.cos−1(−1)=π.

Valid.

So all three are solutions.

  1. Find the sum of all solutions
0+12−12=0.0+\frac12-\frac12=0.0+21​−21​=0.
  1. Compare with the given options

The sum is

0.0.0.

So the correct option is A.

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