Inverse Trigonometric FunctionsClass 12 Mathematics Important Points

Trigonometric functions are not one-to-one and onto over their natural domains, so their inverses do not exist. To define their inverses, we restrict their domains to an interval where they are one-to-one and onto.

For each inverse trigonometric function, the range corresponding to the restricted domain is called the Principal Value Branch. The value of the function within this range is called the principal value.

The function $y = \sin^{-1} x$ has a domain of $[-1, 1]$ and its principal value range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The function $y = \cos^{-1} x$ has a domain of $[-1, 1]$ and its principal value range is $[0, \pi]$.

The function $y = \tan^{-1} x$ has a domain of all real numbers $\mathbf{R}$ and its principal value range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

For $y = \cot^{-1} x$, Domain is $\mathbf{R}$, Range is $(0, \pi)$. For $y = \sec^{-1} x$, Domain is $\mathbf{R} - (-1, 1)$, Range is $[0, \pi] - \{\frac{\pi}{2}\}$. For $y = \operatorname{cosec}^{-1} x$, Domain is $\mathbf{R} - (-1, 1)$, Range is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

The notation $\sin^{-1} x$ should not be confused with $(\sin x)^{-1}$. In fact, $(\sin x)^{-1} = \frac{1}{\sin x} = \operatorname{cosec} x$. The same applies to all other trigonometric functions.

For a function and its inverse, $\sin(\sin^{-1} x) = x$ is valid for all $x$ in the domain of $\sin^{-1} x$, which is $x \in [-1, 1]$. Similar results hold for other functions within their respective domains.

The property $\sin^{-1}(\sin x) = x$ is only valid when $x$ lies within the principal value range of $\sin^{-1}$, i.e., $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. If $x$ is outside this range, you must first convert $\sin x$ to an equivalent value where the angle is within the principal range.

To find $\cos^{-1}(\cos \frac{13\pi}{6})$, note that $\frac{13\pi}{6}$ is not in $[0, \pi]$. We write $\cos(\frac{13\pi}{6}) = \cos(2\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6})$. Since $\frac{\pi}{6} \in [0, \pi]$, the value is $\frac{\pi}{6}$.

For sine, tangent, and cosecant, the negative sign comes out directly: $\sin^{-1}(-x) = -\sin^{-1}x$ for $x \in [-1, 1]$, $\tan^{-1}(-x) = -\tan^{-1}x$ for $x \in \mathbf{R}$, and $\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}x$ for $|x| \geq 1$.

For cosine, secant, and cotangent, the property is: $\cos^{-1}(-x) = \pi - \cos^{-1}x$ for $x \in [-1, 1]$, $\sec^{-1}(-x) = \pi - \sec^{-1}x$ for $|x| \geq 1$, and $\cot^{-1}(-x) = \pi - \cot^{-1}x$ for $x \in \mathbf{R}$.

The identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ holds for $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$. This is derived by substituting $x = \sin\theta$.

The identity $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$ is valid for $x \in [-\frac{1}{2}, \frac{1}{2}]$. This is proved by substituting $x = \sin\theta$ and using the formula for $\sin(3\theta)$.

The identity $3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)$ is valid for $x \in [\frac{1}{2}, 1]$. This is proved by substituting $x = \cos\theta$ and using the formula for $\cos(3\theta)$.

To simplify expressions involving $\sqrt{a^2 - x^2}$, substitute $x = a\sin\theta$ or $x = a\cos\theta$. For example, to simplify $\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}$, let $x = a\sin\theta$.

To simplify expressions containing $\sqrt{a^2 + x^2}$ or $a^2+x^2$, substitute $x = a\tan\theta$ or $x = a\cot\theta$. For example, to simplify $\tan^{-1}\frac{\sqrt{1+x^2}-1}{x}$, let $x = \tan\theta$.

To simplify expressions with $\sqrt{x^2 - a^2}$, substitute $x = a\sec\theta$ or $x = a\operatorname{cosec}\theta$. For example, to simplify $\cot^{-1}(\frac{1}{\sqrt{x^2-1}})$, let $x = \sec\theta$.