Principal value branch

The principal value of \(\sin^{-1}x\) lies in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Remember the range in exams.

The principal value of \(\cos^{-1}x\) is in \(\left[0,\pi\right]\). This is a common exam fact.

The principal value of \(\tan^{-1}x\) lies in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). The endpoints are not included.

The principal value of \(\cot^{-1}x\) is taken in \(\left(0,\pi\right)\). Remember that both endpoints are excluded.

Since \(\sin 0=0\) and \(0\) lies in the principal range of \(\sin^{-1}x\), which is \([-\frac{\pi}{2},\frac{\pi}{2}]\), we have \(\sin^{-1}0=0\). Note that although \(\sin\pi=0\), \(\pi\) is not in the principal range so arcsin does not return \(\pi\). Also \(\sin(\frac{\pi}{2})=1\) and \(\sin(-\frac{\pi}{2})=-1\), so those choices are incorrect. Exam tip: always recall the principal value range for \(\sin^{-1}x\) when evaluating inverse trig values.

We have \(\sin\left(-\frac{\pi}{2}\right)=-1\). Hence \(\sin^{-1}\left(-1\right)=-\frac{\pi}{2}\).

The principal value of arccos, \(\cos^{-1}x\), lies in \([0,\pi]\). Since \(\cos0=1\) and 0 is within this principal range, \(\cos^{-1}1=0\). Closest distractors fail: \(\frac{\pi}{2}\) gives \(\cos=0\), \(\pi\) gives \(\cos=-1\), and \(-\frac{\pi}{2}\) is outside the principal range, so none are correct. Exam tip: memorize that the principal value range of arccos is \([0,\pi]\).

By definition, \(\cos^{-1}(1)=\theta\) where \(\cos\theta=1\) and \(\theta\) lies in the principal range \([0,\pi]\). The only angle in \([0,\pi]\) with \(\cos\theta=1\) is \(\theta=0\). So \(\cos^{-1}(1)=0\).

Since \(\cos\pi=-1\) and the principal value (range) of \(\cos^{-1}x\) is \([0,\pi]\), we have \(\cos^{-1}(-1)=\pi\). The option \(-\pi\) is misleading because although \(\cos(-\pi)=-1\), arccos returns the principal value in \([0,\pi]\), not a negative angle. Options 0 and \(\frac{\pi}{2}\) are incorrect because \(\cos0=1\) and \(\cos(\frac{\pi}{2})=0\). Exam tip: always remember the range of arccos is \([0,\pi]\).

Since \(\tan 0=0\), the angle whose tangent is 0 is 0, so \(\tan^{-1}0=0\). Also remember the principal value range of \(\tan^{-1}x\) is \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\), so values like \(\pi\) or \(\tfrac{\pi}{2}\) are not valid principal outputs. The closest distractor \(\tfrac{\pi}{4}\) is wrong because \(\tan(\tfrac{\pi}{4})=1\), not 0. Exam tip: always use the definition \(\tan(\theta)\) and the principal range of \(\tan^{-1}\) to find inverse values quickly.

Reason: \(\tan\frac{\pi}{4}=1\), and the principal value of \(\tan^{-1}x\) lies in \((-\frac{\pi}{2},\frac{\pi}{2})\), so \(\tan^{-1}1=\frac{\pi}{4}\). Why other choices are wrong: \(0\) is wrong because \(\tan0=0\); \(\frac{\pi}{6}\) is wrong because \(\tan\frac{\pi}{6}=1/\sqrt{3}\), not 1; \(\frac{\pi}{2}\) is not valid as \(\tan\) is undefined there. Exam tip: Memorise standard tan values and remember arctan's principal range to pick the correct branch quickly.

Since \(\tan\left(-\frac{\pi}{4}\right)=-1\) and \(-\frac{\pi}{4}\) lies in the principal range, the answer is \(-\frac{\pi}{4}\).

Since \(\cot(\pi/4)=1\) and the principal value of \(\cot^{-1}x\) is taken in the interval \(0<\cot^{-1}x<\pi\), we get \(\cot^{-1}1=\frac{\pi}{4}\). The closest distractor \(\frac{\pi}{2}\) is incorrect because \(\cot(\pi/2)=0\). Exam tip: remember the principal-value range of arccot is \((0,\pi)\).

Since \(\sec\theta=\dfrac{1}{\cos\theta}\) and \(\sec 0=1\), the principal value of \(\sec^{-1}1\) is \(0\). The common principal range for \(\sec^{-1}x\) is \([0,\pi]\) excluding \(\pi/2\), so \(\sec^{-1}1=0\). Option \(\pi\) is incorrect because \(\sec\pi=-1\). Options \(\tfrac{\pi}{2}\) and \(-\tfrac{\pi}{2}\) are invalid since \(\sec\) is undefined where \(\cos\theta=0\). Exam tip: always check the principal range and where the original trig function is defined before selecting the inverse value.

If \(\csc y=1\) then \(\sin y=1\). The general solution of \(\sin y=1\) is \(y=\frac{\pi}{2}+2k\pi\) (\(k\in\mathbb{Z}\)). The principal value for \(cosec^{-1}x\) is usually taken in \([ -\frac{\pi}{2},\frac{\pi}{2}]\setminus\{0\}\), and the only solution in this interval is \(\frac{\pi}{2}\). Therefore \(cosec^{-1}1=\frac{\pi}{2}\). About distractors: \(0\) is not allowed (\(\csc\) undefined), \(-\frac{\pi}{2}\) gives \(\csc=-1\), and at \(\pi\) \(\sin\pi=0\) so \(\csc\) is undefined. Exam tip: convert \(\csc y=\) value to \(\sin y=\) value and then pick the principal value from the defined range.

Let \(\theta=\csc^{-1}(-1)\). Then \(\csc\theta=-1\), i.e., \(\frac{1}{\sin\theta}=-1\). So \(\sin\theta=-1\). The principal range for \(\csc^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). In this range, \(\sin\theta=-1\) occurs at \(\theta=-\frac{\pi}{2}\). Therefore, \(\csc^{-1}(-1)=-\frac{\pi}{2}\).

The angle \(\frac{\pi}{6}\) lies in the principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Hence the value is \(\frac{\pi}{6}\).

The angle \(\frac{2\pi}{3}\) lies in the principal range \(\left[0,\pi\right]\) of \(\cos^{-1}x\). Hence the value is \(\frac{2\pi}{3}\).

The principal-value rule for inverse tangent is essential: tan⁻¹(tan θ) equals θ only when θ lies in the principal range (−π/2, π/2). The angle π/6 lies inside this interval. Since tan(π/6) = 1/√3, applying the principal inverse tangent gives tan⁻¹(1/√3) = π/6. Therefore option A is correct. Although 5π/6 has the same tangent as π/6 because tangent is periodic, it is outside the principal range and cannot be returned by tan⁻¹. Option C has the opposite sign, while tan(π/3) = √3, not 1/√3, so option D is also incorrect. The inverse function is not unrestricted cancellation; its principal range must always be checked.

Direct answer: option A, \(\frac{\pi}{6}\). The inverse sine asks for the principal angle whose sine is the given number. Since \(\sin\frac{\pi}{6}=\frac12\), the principal value is \(\sin^{-1}(\frac12)=\frac{\pi}{6}\). Option A is correct. B, \(\frac{\pi}{4}\), has sine \(\frac{\sqrt2}{2}\), not \(\frac12\). C, \(\frac{\pi}{3}\), has sine \(\frac{\sqrt3}{2}\). D, \(\frac{\pi}{2}\), has sine 1. Inverse trigonometric notation here means inverse function, not reciprocal. The principal range of inverse sine is \([-\frac\pi2,\frac\pi2]\), and \(\frac\pi6\) lies in it. Memory cue: the 30-degree angle has sine one-half.

Let \(\theta=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\). Then \(\sin\theta=\frac{\sqrt{3}}{2}\) with \(\theta\) in the principal range \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). We know \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), and \(\frac{\pi}{3}\) lies in the principal range. So \(\theta=\frac{\pi}{3}\). Therefore, \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\).

Direct answer: option A, \(\frac{\pi}{3}\). We need the principal angle whose cosine is \(\frac12\). The standard value is \(\cos\frac{\pi}{3}=\frac12\), so \(\cos^{-1}(\frac12)=\frac\pi3\). A is correct. B, \(\frac\pi6\), has cosine \(\frac{\sqrt3}{2}\), not one-half. C, \(\frac{2\pi}{3}\), has cosine \(-\frac12\), so its sign is wrong. D, \(\frac\pi2\), has cosine 0. The principal range of inverse cosine is \([0,\pi]\), and \(\frac\pi3\) lies in that range. Do not confuse sine and cosine standard values. Memory cue: cosine one-half corresponds to 60 degrees.

The principal value of arccos is taken in the interval \([0,\pi]\). Since \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\), we have \(\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}\). The other choices are incorrect: \(\cos\frac{\pi}{3}=\tfrac{1}{2},\;\cos\frac{\pi}{4}=\tfrac{\sqrt{2}}{2}\), and \(\cos\frac{5\pi}{6}=-\tfrac{\sqrt{3}}{2}\). Exam tip: memorize standard trig values and remember arccos range \([0,\pi]\).

Reason: \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\), and the principal value of arctan lies in \((-\frac{\pi}{2},\frac{\pi}{2})\). Hence \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\). Checking close distractors: \(\tan\frac{\pi}{4}=1\), so \(\frac{\pi}{4}\) is incorrect; \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\tan\frac{\pi}{2}\) is undefined. Exam tip: Memorise standard angle values and remember the principal range for inverse trig functions when answering quickly.

Since \(\cos\frac{2\pi}{3}=-\frac{1}{2}\) and \(\frac{2\pi}{3}\) is in the principal range, the answer is \(\frac{2\pi}{3}\).