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Inverse Trigonometric Functions is an integral part of the NCERT Class 11 Mathematics that is used to find the angles of a triangle.
- Inverse trigonometric functions are popular with the name anti-trigonometric functions, arcus functions, or cyclometric functions.
- The inverse of a function makes use of trigonometric ratios.
- The functions are named using the arc- prefix, which goes as arcsine and arccosine.
- Domain and range of a function can be used to calculate the principal value of inverse trigonometric functions.
- Sine, cosine, tangent, cotangent, secant, and cosecant functions are six basic functions whose inverse function is calculated.
- Range of inverse trigonometric functions is subsets of domains of a function.
- The function is used in the fields of navigation, geometry and physics.
The important inverse trigonometric functions used for calculation are as follows:
- sin-1(-x) = -sin-1x
- tan-1(-x) = -tan-1x
- cosec-1(-x) = -cosec-1x
- cos-1(-x) = π - cos-1x
- sec-1(-x) = π - sec-1x
- cot-1(-x) = π - cot-1x
The sum and difference formulas for inverse trigonometric functions are as follows:
- sin-1x + sin-1y = sin-1(x.√(1 - y2) + y√(1 - x2))
- sin-1x - sin-1y = sin-1(x.√(1 - y2) - y√(1 - x2))
- cos-1x + cos-1y = cos-1(xy - √(1 - x2).√(1 - y2))
- cos-1x - cos-1y = cos-1(xy + √(1 - x2).√(1 - y2))
- tan-1x + tan-1y = tan-1(x + y)/(1 - xy), if xy < 1
- tan-1x + tan-1y = tan-1(x - y)/(1 + xy), if xy > - 1
Inverse Trigonometric Functions MCQs
Ques. What is the range of arctan?
- (-π/2,π/2)
- [-π/2,π/2]
- (-π/2,π)
- (-π/2, 0)
Click here for the answer
Ans. (a) (-π/2,π/2)
Explanation: The domain of the arctan is in the range of (-π/2,π/2).
Ques. Find the value of x for sin(x) = 10?
- 1
- No value
- 2
- Multiple values
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Ans. (b) No value
Explanation: Given, sin (x) = 10
⇒ x = sin-1(10), which is not possible.
⇒ There is no required value of x for which sin x = 10.
∴ So the domain of sin-1x is -1 to 1 for the required values of x.
Ques. Find the value of sin-1(sin π/2)?
- π
- π/2
- π/3
- π/6
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Ans. (b) π/2
Explanation: From identity sin-1(sin x) = x, we have
⇒ sin-1(sin π/2) = π/2
Ques. Calculate the value of sin(tan-1 x), where |x| < 1?
- x/√(1 – x2)
- 1/√(1 – x2)
- x/√(1 + x2)
- 1
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Ans. (c) x/√(1 + x2)
Explanation: Consider tan-1x = θ.
⇒ tan θ = x = x/1
⇒ Since we can write the sin θ and cos θ values as:
⇒ sin θ = x/√(1 + x2)
⇒ cos θ = 1/√(1 + x2)
⇒ Now, sin(tan-1 x) = sin θ = x/√(1 + x2).
Ques. The value of sin (2 tan–1 (0.416)) is equal to
- 0.71
- 1.5
- 0.96
- sin 1.5
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Ans. (a) 0.71
Explanation: To calculate sin (2tan–1 (0.416)), assume tan–1 (0.416) = θ
⇒ tan θ = 0.416
⇒ tan θ = 5/12
Thus by Pythagoras theorem, we get;
⇒ sin θ = 5/13 and cos θ = 12/13.
⇒ Now, sin (2tan–1 (0.416)) = 2 × (5/13) × (12/13)
⇒ Now, sin (2tan–1 (0.416)) = 0.71
Ques. What is the range of arccot?
- (-π/2,π/2)
- [-π/2,π/2]
- (0,π)
- (-π/2, 0)
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Ans. (c) (0,π)
Explanation: The domain of the arctan is in the range of (0,π).
Ques. Determine the value of sin (cos-1 /5)?
- 12/17
- 13/17
- 15/17
- 1/17
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Ans. (c)15/17
Explanation: Let cos-1 8/17 = x
⇒ cos x = 8/17
⇒ It is given that sin x = √(1 – cos2 x)
⇒ sin x = √(1 – 8/17)
⇒ sin x = sin (cos-1 15/17)
∴ sin x = 15/17
Ques. What is the domain of arccsc?
- (-∞ ,-1] ∪ [1,∞ )
- [1,∞ )
- (-∞ ,-1]
- (-∞ ,-1) ∪ (1,∞ )
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Ans. (a) (-∞ ,-1] ∪ [1,∞ )
Explanation: The domain of the arccsc is in the domain of (-∞ ,-1] ∪ [1,∞ )
Ques. Determine the value of sin-1(sin 12)?
- 12
- -2π
- 0
- -2π + 12
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Ans. (d) -2π + 12
Explanation: As it is know that sin x = sin(π – 12)
⇒ sin 12 = sin(π – 12)
⇒ sin 12 = sin(2π + 12)
⇒ sin 12 = sin(-2π – 12)
∴ sin-1(sin (-2π + 12)) = -2π + 12
Ques. Calculate the required value of the function sin [cot–1 (cos (tan–1 1))] is
- 3
- 1
- √(2/3)
- √(2)
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Ans. (c) √(2/3)
Explanation: The function is given assin [cot–1 (cos (tan–1 1))]
⇒sin[cot-1 {cos (tan-1 (tan π/4))}]
⇒sin[cot-1 (cos π/4)]
⇒ sin[cot-1(1/√2)]
So, sin [sin-1(√(⅔))] = √(⅔)
Ques. Determine the domain of y = cos–1 (x2 – 6) is
- [3, 5]
- [0, π]
- [-√7, -√5] ∪ [√5, √7]
- [-√7, -√5] ∪ (√5, √7)
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Ans. (c) [-√7, -√5] ∪ [√5, √7]
Explanation: It is given that: y = cos–1 (x2 – 6 )
⇒ cos y = x2 – 6
⇒ As we know, –1 ≤ cos y ≤ 1
= So, – 1 ≤ x2 – 6 ≤ 1
⇒ Adding 6 on both sides, we get; 5 ≤ x2 ≤ 7
⇒ Taking square root on both sides, we get; √5 ≤ x ≤ √7
∴ x∈ [-√7, -√5] ∪ [√5, √7]
Ques. Determine the domain of sin–1(3x) is
- [0, 1]
- [– 1, 1]
- [-⅓ , ⅓ ].
- [–3, 3]
Click here for the answer
Ans. (a) [-⅓ , ⅓ ].
Explanation: Assume sin–1(3x) = θ.
⇒ It will make 3x = sin θ.
⇒ Since, – 1 ≤ sin θ ≤ 1
⇒ We can write the function as – 1 ≤ 3x ≤ 1
⇒ -1/3 ≤ x ≤ 1/3.
∴ The domain of sin-1(3x) is [-⅓ , ⅓ ].
Ques. It is given that sin–1x + sin–1y = π/2, then determine the value of cos–1x + cos–1y is
- π/3
- Π
- 0
- π/2
Click here for the answer
Ans. (d) π/2
Explanation: Since it is given that: sin–1 x + sin–1 y = π/2
⇒ [(π/2) – cos-1x] + [(π/2) – cos-1y] = π/2
⇒ (π/2) + (π/2) – (π/2) = cos-1x + cos-1y
∴ cos–1x + cos–1y = π/2
Ques. What is the range of arccos?
- (-1,1)
- [-1,1]
- (1,0)
- (0, 0)
Click here for the answer
Ans. (b) [-1,1]
Explanation: The domain of the arccos is in the range of [-1,1]
Ques. Find the value of x for sin(x) = 120?
- 1
- 5
- 2
- No values
Click here for the answer
Ans. (d) No value
Explanation: Given, sin (x) = 120
⇒ x =sin-1(120), which is not possible.
⇒ There is no required value of x for which sin x = 120.
So the domain of sin-1x is -1 to 1 for the required values of x.
Ques. Find the value of sin-1(sin π)?
- Π
- π/2
- π/3
- π/6
Click here for the answer
Ans. (a) π
Explanation: From identity sin-1(sin x) = x, we have
∴ sin-1(sin π) = π
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