Inverse Trigonometric Functions: questions with answers

A is false: the two terms are angles and it is the angles that add, so gathering both arguments under one inverse -- which is where the stated value comes from -- is not a legal move. R is a true statement about a branch. So A falls and R stands.

Step 1 is a harmless observation. Step 2 is where the reading goes wrong: the bracket is the ARGUMENT of the inverse, so it has to be evaluated first and its value $-\sqrt{3}$ handed outward. Step 3 only carries step 2's misreading forward. Read correctly, the term is $\frac{5\pi}{6}$.

$A$ is an angle whose sine is $\frac{7}{25}$, so in a right triangle the opposite side is $7$ and the hypotenuse is $25$; Pythagoras supplies the adjacent side, and the tangent is opposite over adjacent, giving $\frac{7}{24}$. The number $\frac{7}{25}$ is the SINE of $A$, not its tangent, and $\frac{24}{7}$ inverts the two sides the tangent compares.

Put $\theta = \tan^{-1} \frac{8}{15}$, so $\tan\theta = \frac{8}{15}$. Since $\tan$ compares the opposite and adjacent sides, those are the two sides the given number describes; the hypotenuse then follows from Pythagoras. Reading $\sin\theta$ off the same triangle gives $\frac{8}{17}$. The answer is not $\frac{8}{15}$ -- $\sin$ and $\tan^{-1}$ are different functions and do not cancel -- and it is not $\frac{15}{17}$, which is what comes of taking the ratio over the wrong pair of sides.

The defining property of an inverse is that it undoes $f$: if $f$ sends $3$ to $2$, then $f^{-1}$ sends $2$ back to $3$, so the point $(2, 3)$ lies on the inverse graph. The two coordinates are interchanged, which on the picture is reflection in the dashed line $y = x$. Reflecting in the $x$-axis would give $(3, -2)$ and reflecting in the $y$-axis $(-3, 2)$; neither interchanges the coordinates, and both leave the point on the wrong side of the wrong line.

The two arms stop at the marked ticks $-1$ and $1$, and there is nothing drawn between them, so the missing inputs are exactly those with $-1 < x < 1$. The endpoints themselves are marked with solid dots and do belong to the graph. The empty stretch is not an asymptote -- the arms end at points they reach -- and it does not stop the drawing being a graph of a function, since a vertical line there meets the curve no times at all.

It can, but not because the two operations always undo each other -- they do not. $\sin^{-1}$ returns a value in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and nowhere else, so cancelling is legal exactly when the angle already lies there. Here $\frac{\pi}{3}$ does, so the answer is $\frac{\pi}{3}$. Replacing it by $\frac{2\pi}{3}$ would leave a perfectly good angle for one that is outside the branch.

The two folds obey different rules because the two branches are different. $\cos^{-1}$ lands in $[0, \pi]$, and the cosine repeats only after a full turn, so its answer is $\frac{3\pi}{4}$. $\tan^{-1}$ lands in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and the tangent repeats every $\pi$, so subtracting $\pi$ gives $-\frac{\pi}{4}$. Using the supplement on the tangent would give $\frac{\pi}{4}$, and reducing the cosine by $\pi$ would give $-\frac{\pi}{4}$; both leave their own branch.

$\sec$ is the reciprocal of $\cos$, so $\sec^{-1}$ inherits $\cos^{-1}$'s interval and then deletes the one angle at which $\sec$ is undefined. That gives $[0, \pi] - \left\{\frac{\pi}{2}\right\}$. The interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$ belongs to $\operatorname{cosec}^{-1}$, the other reciprocal inverse, and $[0, \pi]$ is the right interval with the deletion left out.

$\sin^{-1}$ has branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so its value here is the negative angle $-\frac{\pi}{3}$. $\cos^{-1}$ has branch $[0, \pi]$, which contains no negative angle, so its value is the obtuse angle $\frac{5\pi}{6}$. They are not related by a change of sign, and they are not both negative: the camp a function belongs to, not the sign of the argument, decides the sign of the answer.

A function and its inverse must have their two sets exchanged, and the function itself must be one-one and onto between them. Here the exchange is written correctly, so the only question left is whether $\tan$ really is one-one from $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ onto $\mathbf{R}$. on that interval it is one-one, so the pairing is sound.

The two inverses accept complementary sets: $\sin^{-1}$ takes $[-1, 1]$ and $\sec^{-1}$ takes everything else, $\mathbf{R} - (-1, 1)$. Imposed on the same inner expression they therefore give complementary answers: $\left[-\frac{1}{4}, \frac{1}{4}\right]$ for the first and $\left(-\infty, -\frac{1}{4}\right] \cup \left[\frac{1}{4}, \infty\right)$ for the second. They overlap only at the two points where $4x$ is exactly $\pm 1$.

Multiplying the branch by $2$ and then adding $-\pi$ to every part gives the endpoints of $\left[-\pi, \pi\right]$. The BRACKETS are inherited: $\cos^{-1}$ attains both ends of its branch, and neither scaling nor shifting can stop an attained value being attained, so the range stays closed. Doing the shift before the scaling would put the interval at $\left[-2\pi, 0\right]$, in the wrong place entirely.

It is, and the stated interval is why. As $x$ runs over $0 < x < \pi$, $\frac{x}{2}$ runs over $\left(0, \frac{\pi}{2}\right)$, which sits inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ -- the one interval on which $\tan^{-1}$ can return its own argument. The multiplier has to be applied to the endpoints in the right direction: taken the other way round it would suggest $(0, 2\pi)$, a quite different interval.

The branch of $\cos^{-1}$ is $[0, \pi]$, so those and only those are its values. The comparison is therefore between $3$ and the ends of that interval: $3 < \pi$. That settles the question outright, because an equation asking a function for a value it never takes has no solution, and no amount of algebra afterwards can produce one. The domain of $\cos^{-1}$ is a separate matter and says nothing about which values come out.