Inverse Trigonometric Functions

INVERSE TRIGONOMETRIC (CIRCULAR) FUNCTIONS:

If sinθ = x , then θ may be any angle whose sine is x, and we write θ = sin-1x . It means that θ is an angle which can be determined from its sine.

Thus , tan-1 (1/√3) is an angle whose tangent is 1/√3 ,  i.e.  θ = tan-1 (1/√3) = nπ + π/6  , where π/6 is the least positive value of  θ .

The functions sin-1x , cos-1x , tan-1x , cot-1x , cosec-1x and sec-1x are called inverse circular or inverse trigonometric functions.

Each of the inverse circular function is multivalued (infact they are relations). To make each inverse circular function single valued we define principal value as follows.

If x is positive, the principal values of all the inverse circular functions lie between 0 and π/2 .

If x is negative, the principal values of sin-1x , cosec-1x and tan-1x lie between -π/2 and 0 . Those of cos-1x, sec-1x and cot-1x lie between π/2 and π . From now onwards we take only principal values.

sinθ = x ⇒ θ = sin-1x

Where θ ∈ [-π/2 , π/2] and x ∈ [-1, 1].

cosecθ = x ⇒ θ = cosec-1x

Where θ ∈ [ -π/2 , 0 ) ∪ (0 , π/2  ] and x ∈ (-∞ , -1] ∪ [1 , ∞)

tanθ = x ⇒ θ = tan-1x

where  θ ∈ (-π/2 , π/2 )and x ∈ (-∞, ∞)

cos θ = x ⇒ θ = cos-1x

where θ ∈ [0, π] and x ∈ [-1, 1]

secθ = x ⇒ θ = sec-1x

where θ ∈ [0 , π/2) ∪(π/2 , π] and x ∈ (-∞, -1] ∪ [1 , ∞)

cotθ = x ⇒ θ = cot-1x

where θ ∈ (0, π) and x ∈ (-∞, ∞)

Remarks : 

(i) sin-1(sinθ) = θ if and only if -π/2 ≤ θ ≤ π/2

and ,  sin(sin-1x) = x , where -1 ≤ x ≤ 1

$ \large sin^{-1}(sinx) = \left\{\begin{array}{lll} x \; , 0 \leq x \le \pi/2 \\ \pi – x \; , \pi/2 < x \leq 3\pi/2 , \\ x-2\pi \; , 3\pi/2 < x \le 2\pi \end{array} \right. $

⇒  f(x)= sin-1(sinx) is periodic with period 2π .

(ii) cosec-1(cosecθ) = θ if and only if – π/2  ≤ θ <  0 or 0 < θ ≤ π/2 and cosec(cosec-1x) = x
where -∞ < x ≤ -1 or 1 ≤ x < ∞

$ \large cosec^{-1}(cosecx) = \left\{\begin{array}{lll} x \; , 0 \leq x \le \pi/2  , x \ne  0 \\ \pi – x \; , \pi/2 < x \leq 3\pi/2 , \\ x-2\pi \; , 3\pi/2 < x \le 2\pi \end{array} \right. $

⇒  f(x)= cosec-1(cosecx) is periodic with period 2π .

(iii) tan-1(tanθ) = θ if and only if -π/2 < θ < π/2

and tan (tan-1x) = x  , where – ∞ < x < ∞

$ \large tan^{-1}(tanx) = \left\{\begin{array}{llll} x \; , 0 \leq x \le \pi/2   \\ x – \pi  \; , \pi/2 < x < \pi , \\ \pi -x \; , \pi \le  x <  3\pi /2  \\ x-2\pi \; , 3\pi/2 < x \le 2\pi \end{array} \right. $

⇒  f(x)= tan-1(tanx) is periodic with period π .

(iv) cos-1(cosθ) = θ if and only if 0 ≤ θ ≤ π

and ,  cos(cos-1x) = x , where -1 ≤ x ≤ 1

$ \large cos^{-1}(cosx) = \left\{\begin{array}{ll} x \; , 0 \leq x \le \pi \\ 2\pi – x \; , \pi < x \leq 2\pi , \end{array} \right. $

⇒  f(x)= cos-1(cosx) is periodic with period 2π .

(v) sec-1(secθ) = θ if and only if 0 ≤ θ < π/2 , or π/2 < θ ≤ π

and ,  sec(sec-1x) = x , where -∞ < x ≤ -1 or , 1 ≤ x < ∞

$ \large sec^{-1}(secx) = \left\{\begin{array}{ll} -x \; ,  -\pi \leq x \le 0  , x \ne -\pi/2  \\  x \; , 0 < x \leq \pi , x \ne \pi/2 \end{array} \right. $

⇒  f(x)= sec-1(secx) is periodic with period 2π .

(vi) cot-1(cotθ) = θ if and only if 0 < θ < π

and cot(cot-1x) = x  , where – ∞ < x < ∞

$ \large cot^{-1}(cotx) = \left\{\begin{array}{lll} x + \pi \; , -\pi < x < 0   \\ x   \; , 0 < x < \pi , \\ x -\pi \; , \pi < x < 2\pi \end{array} \right. $

⇒  f(x)= cot-1(cotx) is periodic with period π .

Illustration : Evaluate :

(i) $\large sin^{-1}(sin\frac{3\pi}{4})$

(ii) $\large cot^{-1}(cot4)$

(iii) $\large cos^{-1}(cos10)$

(iv) $\large tan^{-1}(tan5)$

(v) $\large cos(tan^{-1}2)$

(vi) $\large sin(cos^{-1}(3/5))$

Sol: (i) $\large sin^{-1}(sin\frac{3\pi}{4})$

$\large = sin^{-1}(sin(\pi – \frac{\pi}{4}))$

$\large = sin^{-1}(sin\frac{\pi}{4}) = \frac{\pi}{4}$

(ii) $\large cot^{-1}(cot4)$

$\large = cot^{-1}(cot(\pi + (4-\pi)))$

$\large = cot^{-1}(cot(4-\pi))$

= 4 – π

(iii) $\large cos^{-1}(cos10)$

$\large = cos^{-1}(cos(4\pi-10))$

= 4π – 10

(iv) $\large tan^{-1}(tan5)$

$\large tan^{-1}(tan(2\pi + (5-2\pi))$

$\large = tan^{-1}(tan(5-2\pi))$

= 5 – 2π

(v) Let θ = tan-12

tanθ = 2

$\large cos\theta = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}} $

(vi) Let θ = cos-1(3/5)

cosθ = 3/5

$\large sin\theta = \sqrt{1-cos^2 \theta} $

$\large sin\theta = \sqrt{1-(\frac{3}{5})^2} = \frac{4}{5}$

Some Important Results:

(i) $\large sin^{-1}x + cos^{-1}x = \frac{\pi}{2} \; , |x| \le 1 $

(ii) $\large tan^{-1}x + cot^{-1}x = \frac{\pi}{2} \; , |x| \in R $

(iii) $\large sec^{-1}x + cosec^{-1}x = \frac{\pi}{2} \; , |x| \ge 1 $

(iv) $\large sin^{-1}(-x) = – sin^{-1}x $

(iv) $\large cosec^{-1}(-x) = – cosec^{-1}x $

(v) $\large tan^{-1}(-x) = – tan^{-1}x $

(vi) $\large cos^{-1}(-x) = \pi – cos^{-1}x $

(vii) $\large sec^{-1}(-x) = \pi – sec^{-1}x $

(viii) $\large cot^{-1}(-x) = \pi – cot^{-1}x $

(ix) $\large sin^{-1}x = tan^{-1}\frac{x}{\sqrt{1-x^2}} = cosec^{-1}(\frac{1}{x}) \; , x > 0 $

(x) $\large sin^{-1}x = cos^{-1}\sqrt{1-x^2} = cot^{-1}\frac{\sqrt{1-x^2}}{x} = sec^{-1}\frac{1}{\sqrt{1-x^2}} \; , x > 0$

(xi) $\large cos^{-1}x = sin^{-1}\sqrt{1-x^2} = tan^{-1}\frac{\sqrt{1-x^2}}{x} = cosec^{-1}\frac{1}{\sqrt{1-x^2}} \; , x > 0$

(xii) $\large tan^{-1}x = sin^{-1}\frac{x}{\sqrt{1+x^2}} = cosec^{-1}\frac{\sqrt{1+x^2}}{x} \; , x \in R \sim \{0\}$

(xiii) $\large cot^{-1}x = cos^{-1}\frac{x}{\sqrt{1+x^2}} = sec^{-1}\frac{\sqrt{1+x^2}}{x} \; , x \in R \sim \{0\}$

(xiv) $\large cos^{-1}x + cos^{-1}y = cos^{-1}[x y – \sqrt{1-x^2} \sqrt{1-y^2}] \; , x \ge 0 , y \ge 0 $

(xv) $ \large cos^{-1}x – cos^{-1}y = \left\{\begin{array}{ll} cos^{-1}[x y + \sqrt{1-x^2} \sqrt{1-y^2}] \; , x \ge 0 , y \ge 0 , x \le y \\ – cos^{-1}[x y + \sqrt{1-x^2} \sqrt{1-y^2}] \; , x \ge 0 , y \ge 0 , x > y \end{array} \right. $

(xvi) $ \large sin^{-1}x + sin^{-1}y = \left\{\begin{array}{ll} sin^{-1}[x \sqrt{1-y^2} + y \sqrt{1-x^2} ] \; , x \ge 0 , y \ge 0 , x^2 + y^2 \le 1 \\ \pi – sin^{-1}[x \sqrt{1-y^2} + y \sqrt{1-x^2} ] \; , x \ge 0 , y \ge 0 , x^2 + y^2 > 1 \end{array} \right. $

(xvii) $\large sin^{-1}x – sin^{-1}y = sin^{-1}[x \sqrt{1-y^2} – y \sqrt{1-x^2} ] \; , x \ge 0 , y \ge 0 $

(xviii) $ \large tan^{-1}x + tan^{-1}y = \left\{\begin{array}{lll} tan^{-1} \frac{x+y}{1-x y} \; , x \ge 0 , y \ge 0 , x y < 1 \\ \pi + tan^{-1} \frac{x+y}{1-x y} \; , x > 0 , y > 0 , x y > 1 \\ \frac{\pi}{2} \; , x > 0 , y > 0 , x y = 1\end{array} \right. $

(xix) $ \large tan^{-1}x – tan^{-1}y = tan^{-1} \frac{x-y}{1 + xy} \; , x \ge 0 , y \ge 0$