Introduction to inverse trigonometric functions

Because \(\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\). Standard angle values are very useful in exams.

Because \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) and \(\frac{3\pi}{4}\in[0,\pi]\). For \(\cos^{-1}\), do not take a negative principal value.

Let \(\theta=\sin^{-1}\frac{5}{13}\), then \(\sin\theta=\frac{5}{13}\) and \(\theta\) is in the principal range. Therefore \(\cos\theta=\frac{12}{13}\).

Let \(\theta=\cos^{-1}\frac{4}{5}\), then \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\). Hence \(\tan\theta=\frac{3}{4}\).

For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).

For \(\cos^{-1}\left(\frac{x}{3}\right)\), we need \(-1\le\frac{x}{3}\le1\). Therefore \(-3\le x\le3\).

When (x>0), (\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}). Keep the principal range of (\cot^{-1}x) in mind.

Because \(\operatorname{cosec}\left(-\frac{\pi}{6}\right)=-2\) and \(-\frac{\pi}{6}\) lies in the principal range. For negative values choose the correct principal angle.

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.

Values of (\sec y) do not lie in \(\left(-1,1\right)\), so the domain of \(\sec^{-1}x\) is \(\mathbb{R}\setminus\left(-1,1\right)\). Zero is also not in the domain.