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Inverse Trigonometry Class notes

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The document consists of inverse trigonometry functions, their definition and the concept to solve its problem. It contains various properties of inverse trigonometry functions.

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Subject:- MATHEMATICS
LEVEL:-10+2 and all Entrances

Topic:- Inverse Trigonometric Functions


Inverse Function:
If 𝑓: 𝑋 → 𝑌 is a function which is both one-one and onto, then its inverse function 𝑓 −1 : 𝑌 →
𝑋 is defined as 𝑦 = 𝑓(𝑥) ⟺ 𝑓 −1 (𝑦) = 𝑥, ∀𝑥 ∈ 𝑋, ∀𝑦 ∈ 𝑌



Inverse Trigonometric Functions:
Trigonometric functions are not one-to-one and onto on their natural domains and ranges,
so in order to make them one-one and onto, we restrict their domain and codomain. And
thus we obtain inverses of trigonometric functions. Inverse of 𝑓 is denoted by 𝑓 −1 .

Let 𝑦 = 𝑓(𝑥) = sin 𝑥 be a function. Then, its inverse is 𝑥 = 𝑠𝑖𝑛−1 𝑦.



Domain and Range of Inverse Trigonometric Functions:
𝒚 Domain 𝒙 Range 𝒚 (principal
value
i) 𝑠𝑖𝑛−1 𝑥 −1 ≤ 𝑥 ≤ 1 −𝜋/2 ≤ 𝑦 ≤ 𝜋/2

ii) 𝑐𝑜𝑠𝑒𝑐 −1 𝑥 𝑥 ≤ −1 𝑜𝑟 𝑥 ≥ 1 −𝜋/2 ≤ 𝑦 ≤ 𝜋/2, 𝑦
≠0
iii) 𝑐𝑜𝑠 −1 𝑥 −1 ≤ 𝑥 < 1 0≤𝑦≤𝜋

iv) 𝑠𝑒𝑐 −1 𝑥 𝑥 ≤ −1 𝑜𝑟 𝑥 ≥ 1 0 ≤ 𝑦 ≤ 𝜋, 𝑦 ≠ 𝜋/2

v) 𝑡𝑎𝑛−1 𝑥 −∞ < 𝑥 < ∞ −𝜋/2 < 𝑦 < 𝜋/2

vi) 𝑐𝑜𝑡 −1 𝑥 −∞ < 𝑥 < ∞ 0<𝑦<𝜋

, Properties of Inverse Trigonometric Functions :
𝜋 𝜋
(i) 𝑠𝑖𝑛−1 (𝑠𝑖𝑛𝜃) = 𝜃 and sin(𝑠𝑖𝑛−1 𝑥) = 𝑥, − ≤ 𝜃 ≤ , −1 ≤ 𝑥 ≤ 1
2 2

𝑐𝑜𝑠 −1 (𝑐𝑜𝑠𝜃) = 𝜃 and cos(𝑐𝑜𝑠 −1 𝑥) = 𝑥, 0 ≤ 𝜃 ≤ 𝜋, −1 ≤ 𝑥 ≤ 1
𝜋 𝜋
𝑡𝑎𝑛−1 (𝑡𝑎𝑛𝜃) = 𝜃 and tan(𝑡𝑎𝑛−1 𝑥) = 𝑥, − ≤ 𝜃 ≤ , 𝑥 ∈ ℝ
2 2

𝑐𝑜𝑡 −1 (𝑐𝑜𝑡𝜃) = 𝜃 and cot(𝑐𝑜𝑡 −1 𝑥) = 𝑥, 0 < 𝜃 < 𝜋, 𝑥 ∈ ℝ

𝑠𝑒𝑐 −1 (𝑠𝑒𝑐𝜃) = 𝜃 and sec(𝑠𝑒𝑐 −1 𝑥) = 𝑥

𝑐𝑜𝑠𝑒𝑐 −1 (𝑐𝑜𝑠𝑒𝑐𝜃) = 𝜃 and 𝑐𝑜𝑠𝑒𝑐(𝑐𝑜𝑠𝑒𝑐 −1 𝑥) = 𝑥


1 1
(ii) 𝑠𝑖𝑛−1 𝑥 = 𝑐𝑜𝑠𝑒𝑐 −1 ( ) or 𝑐𝑜𝑠𝑒𝑐 −1 𝑥 = 𝑠𝑖𝑛−1 ( )
𝑥 𝑥
1 1
𝑐𝑜𝑠 −1 𝑥 = 𝑠𝑒𝑐 −1 ( ) or 𝑠𝑒𝑐 −1 𝑥 = 𝑐𝑜𝑠 −1 ( )
𝑥 𝑥
1 1
𝑡𝑎𝑛−1 𝑥 = 𝑐𝑜𝑡 −1 ( ) or 𝑐𝑜𝑡 −1 𝑥 = 𝑡𝑎𝑛−1 ( ), if 𝑥 > 0
𝑥 𝑥
1 1
𝑡𝑎𝑛−1 𝑥 = 𝑐𝑜𝑡 −1 ( ) − 𝜋 or 𝑐𝑜𝑡 −1 𝑥 = 𝑡𝑎𝑛−1 ( ) − 𝜋, if 𝑥 < 0
𝑥 𝑥


(iii) 𝑠𝑖𝑛−1 (−𝑥) = −𝑠𝑖𝑛−1 (𝑥)
𝑐𝑜𝑠 −1 (−𝑥) = 𝜋 − 𝑐𝑜𝑠 −1 (𝑥)
𝑡𝑎𝑛−1 (−𝑥) = −𝑡𝑎𝑛−1 (𝑥)
𝑐𝑜𝑡 −1 (−𝑥) = 𝜋 − 𝑐𝑜𝑡 −1 (𝑥)
𝑠𝑒𝑐 −1 (−𝑥) = 𝜋 − 𝑠𝑒𝑐 −1 (𝑥)
𝑐𝑜𝑠𝑒𝑐 −1 (−𝑥) = 𝜋 − 𝑐𝑜𝑠𝑒𝑐 −1 (𝑥)

(iv) 2𝑠𝑖𝑛−1 𝑥 = 𝑠𝑖𝑛−1 2𝑥√(1 − 𝑥 2 ) = 𝑐𝑜𝑠 −1 (1 − 2𝑥 2 )
2𝑐𝑜𝑠 −1 𝑥 = 𝑐𝑜𝑠 −1 (2𝑥 2 − 1) = 𝑠𝑖𝑛−1 (2𝑥√1 − 𝑥 2 )
2𝑥 2𝑥 1−𝑥 2
2𝑡𝑎𝑛−1 𝑥 = 𝑡𝑎𝑛−1 = 𝑠𝑖𝑛−1 = 𝑐𝑜𝑠 −1
1−𝑥 2 1+𝑥 2 1+𝑥 2
−1 −1 3
3𝑠𝑖𝑛 𝑥 = 𝑠𝑖𝑛 (3𝑥 − 4𝑥 )
3𝑐𝑜𝑠 −1 𝑥 = 𝑐𝑜𝑠 −1 (4𝑥 3 − 3𝑥
3𝑥−𝑥 3
3𝑡𝑎𝑛−1 𝑥 = 𝑡𝑎𝑛−1
1−3𝑥 2

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