Principal value branch

Here \(\cos\theta=\frac{12}{13}\) and \(\theta\in\left[0,\pi\right]\). In this case, \(\sin\theta=\frac{5}{13}\) is positive.

The angle \(-\frac{\pi}{4}\) lies in the principal range of \(\sin^{-1}x\). Hence the value is \(-\frac{\pi}{4}\).

Since \(\cos0=1\) and \(\cos^{-1}1\) is the angle in the principal range \([0,\pi]\) whose cosine equals 1, we get \(\cos^{-1}1=0\). The other choices are incorrect: \(\cos(\pi/2)=0\), \(\cos\pi=-1\), and \(-\pi\) is outside the principal range of arccos. Exam tip: When evaluating \(\cos^{-1}(\cos\theta)\), reduce or map \(\theta\) into the principal range \([0,\pi]\) and use that equivalent angle.

Let θ = cos⁻¹x. By definition, θ lies in the principal range [0, π] and cos θ = x. Using the identity cos(π − θ) = −cos θ, we obtain cos(π − θ) = −x. Since π − θ also lies in [0, π], it is the principal angle whose cosine is −x. Therefore cos⁻¹(−x) = π − θ = π − cos⁻¹x, so option A is correct. The expression −cos⁻¹x is generally outside the principal range and incorrectly treats arccos as an odd function. Option C does not follow from the cosine subtraction identity, and option D would incorrectly claim that arccos has the same value for x and −x. The domain condition ensures both inverse values are real.

Since \(\tan\frac{3\pi}{4}=-1), (\tan^{-1}(-1)=-\frac{\pi}{4}\). The answer must lie in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

Since \(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) and it lies in the principal range. Therefore \(-\frac{\pi}{3}\) is correct.

Since \(\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\) and \(\frac{\pi}{4}\in\left[0,\pi\right]\). Remember the table of special values.

Since \(\sin\left(-\frac{5\pi}{6}\right)=-\frac{1}{2}\), \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\). The answer must lie in the principal range.

The principal value of \(\tan^{-1}x\) is always chosen from \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). In this interval, \(\tan\theta\) is one-to-one, so every real \(x\) corresponds to exactly one angle. Option C is incorrect because \(\tan\theta\) is undefined at \(\theta=\pm\frac{\pi}{2}\). Exam tip: the endpoint angles are never included in the principal-value range of \(\tan^{-1}x\).

Since \(\sin\frac{7\pi}{6}=-\frac{1}{2}\), \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\). Keep the answer in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).

Since \(\cos\left(-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\), \(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}). The answer of (\cos^{-1}\) is not negative.

From \(\csc y=\frac{2}{\sqrt{3}}\), \(\sin y=\frac{\sqrt{3}}{2}\), so \(y=\frac{\pi}{3}\). Solve \(\csc^{-1}\) by relating it to \(\sin^{-1}\).

\(\sin\frac{\pi}{6}=\frac{1}{2}\) and \(\frac{\pi}{6}\) lies in the principal range. Always choose the value inside the principal range.

The output of \(\sin^{-1}x\) lies in \([-\frac{\pi}{2},\frac{\pi}{2}]\). Therefore \(\theta\) must lie in this range.

The principal output of (\cos^{-1}x) lies in ([0,\pi]). For equality, take (\theta) in that interval.

The principal range of (\tan^{-1}x) is ((-\frac{\pi}{2},\frac{\pi}{2})). Since (\tan) is periodic, the principal interval is necessary.

\(cosec\frac{\pi}{6}=2\), so the principal value is \(\frac{\pi}{6}\). Think of \(cosec^{-1}x\) through \(\sin^{-1}\frac{1}{x}\).

\(\cos\frac{4\pi}{3}=-\frac{1}{2}\), so \(\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}\). Give the answer of \(\cos^{-1}x) in ([0,\pi]\).

\(\tan^{-1}x=-\frac{\pi}{3}\) means \(x=\tan\left(-\frac{\pi}{3}\right)\). Hence \(x=-\sqrt{3}\).

\(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) and it lies in the principal range. Keep the answer of \(\sin^{-1}x) in ([-\frac{\pi}{2},\frac{\pi}{2}]\).