Introduction to inverse trigonometric functions

The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) and \(\sin\frac{\pi}{6}=\frac{1}{2}\). Always remember the principal range in exams.

The range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\) and in this range \(\cos\frac{2\pi}{3}=-\frac{1}{2}\). Do not take a negative angle as principal value.

The range of \(\tan^{-1}x\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) and \(\tan\left(-\frac{\pi}{4}\right)=-1\). Keep the quadrant in mind.

For every real \(x\), \(\tan^{-1}x\) lies strictly between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). Hence, its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). As \(x\) becomes very large positive or negative, the value approaches \(\frac{\pi}{2}\) or \(-\frac{\pi}{2}\), respectively, but never attains them; therefore, option B is incorrect. Exam tip: the endpoints of the range of \(\tan^{-1}x\) are always excluded.

\(y=\sin^{-1}x\) means that \(y\) is the principal-value angle whose sine is \(x\); hence \(x=\sin y\). The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so option A is correct. Option D reverses the relation between \(\sin\) and \(\sin^{-1}\). Exam tip: read \(\sin^{-1}x\) as the inverse sine function, not as \(1/\sin x\).

For \(y=\cos^{-1}x\), the domain is \(x\in[-1,1]\), and its principal-value range is \(\left[0,\pi\right]\). At \(x=1\), \(y=0\), while at \(x=-1\), \(y=\pi\); hence both endpoints are included. The interval \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) is associated with the principal range of \(\sin^{-1}x\). Exam tip: remember that \(\cos^{-1}x\) takes values from \(0\) to \(\pi\).

The principal value range of \(\cos^{-1}x\) is \([0,\pi]\), for \(-1\le x\le1\). In contrast, \(\sin^{-1}x\) has range \([-\pi/2,\pi/2]\). Exam tip: learn each inverse function’s domain together with its principal range.

If \(y=\sin^{-1}x\), then \(\sin y=x\) and \(y\in[-\pi/2,\pi/2]\). For example, \(\sin^{-1}(1/2)=\pi/6\), not 2. In exams, do not confuse an inverse function with a reciprocal.

\(\tan(3\pi/4)=-1\). Since the principal-value range of \(\tan^{-1}\) is \((-\pi/2,\pi/2)\), \(\tan^{-1}(-1)=-\pi/4\), not \(3\pi/4\). Exam tip: always check the principal-value range before cancelling inverse functions.

From (\csc y=-2), (\sin y=-\frac{1}{2}), so the principal value is (y=-\frac{\pi}{6}). Remember the range of (\csc^{-1}).

Since (\tan\frac{\pi}{3}=\sqrt{3}) and (\frac{\pi}{3}) lies in the principal range. Remember special angles well.

To be defined, \(\sec^{-1}x\) must have \(\sec\theta=x\) for some angle \(\theta\). Since \(\sec\theta=\frac{1}{\cos\theta}\) and \(\cos\theta\neq 0\), we need \(\cos\theta\) to be defined and nonzero. Also for real values, \(\sec\theta\) can never have magnitude less than 1, so \(|x|\ge 1\). Therefore the domain is \(x\in(-\infty,-1]\cup[1,\infty)\).

The direct answer is option C: \(\tan^{-1}5\) is a real number. In inverse trigonometric functions, the input must belong to the function's domain. Since tangent can take every real value, \(\tan^{-1}x\) is defined for every real x; therefore 5 is allowed. Option A, \(\sin^{-1}2\), is not real because sine values lie only from -1 to 1, and 2 is outside that interval. Option B, \(\cos^{-1}(-3/2)\), is not real because cosine also has range [-1,1], while \(-3/2\) is less than -1. Option D, \(\sec^{-1}(1/2)\), is not real under the usual real inverse-secant definition because secant values have absolute value at least 1; \(|1/2|<1\). Thus C alone is defined in the reals. Memory cue: check the input range before using an inverse trigonometric function.

The domain of \(\tan^{-1}x\) is \(\mathbb{R}\) and its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). The endpoints of the open interval are not included.

By making (\cos x) one-one on \(\left[0,\pi\right]\), \(\cos^{-1}x\) becomes decreasing on \(\left[-1,1\right]\). The graph direction also confirms this.

For \(y=\sin^{-1}x\), the principal range is chosen so that the inverse sine function is one-to-one. On its domain \(x\in\left[-1,1\right]\), \(\sin^{-1}x\) is strictly increasing, so it increases throughout \(\left[-1,1\right]\).

Restricting (\sin x) to (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]) makes it one-one and gives values (\left[-1,1\right]). This defines (\sin^{-1}x).

The function (\cos x) is made one-one on (\left[0,\pi\right]). Therefore the range of (\cos^{-1}x) is (\left[0,\pi\right]).

Here (\sin^{-1}1=\frac{\pi}{2}) and (\cos^{-1}1=0), so the sum is (\frac{\pi}{2}). Remember endpoint values.

Since (\cos^{-1}0=\frac{\pi}{2}) and (\sin^{-1}0=0), the difference is (\frac{\pi}{2}). Inverse trigonometric values at zero are very useful.

Here (\tan^{-1}0=0) and (\cot^{-1}0=\frac{\pi}{2}), so the sum is (\frac{\pi}{2}). Do not write (\cot^{-1}0) as (0).

From (\sin^{-1}x=\frac{\pi}{4}), (x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}). Convert the inverse form into the normal trigonometric form.

From (\cos^{-1}x=\frac{2\pi}{3}), (x=\cos\frac{2\pi}{3}=-\frac{1}{2}). Decide the sign from the quadrant of the angle.

From \(\tan^{-1}x=-\frac{\pi}{6}\), \(x=\tan\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}\). The range of \(\tan^{-1}\) also includes negative angles.

From (\sec y=-\sqrt{2}), (\cos y=-\frac{1}{\sqrt{2}}), so the principal value is (y=\frac{3\pi}{4}). For (\sec^{-1}), decide the quadrant from the sign.