E4 - Reciprocal and inverse trigonometric functions

Reciprocal ratios and their domains

Why does $\sin^{-1}(1/2)$ give an angle, while $(\sin(\pi/6))^{-1}$ gives $2$? The small raised $-1$ has two different jobs in these notations. First, look at the reciprocal job: $\cos(2\pi/3)=-1/2$, so its reciprocal is $-2$.

The reciprocal trigonometric functions keep the angle as input and reverse the ratio's numerator and denominator.

These definitions are not supplied as formulae in the booklet: learn them. With radian inputs and $k\in\mathbb Z$, secant excludes $x=\pi/2+k\pi$; cosecant and cotangent exclude $x=k\pi$.

Cotangent has a subtle boundary: at $x=\pi/2$, it is $\cos x/\sin x=0$, even though $1/\tan x$ cannot be evaluated there because tangent itself is undefined. The cosine-over-sine definition gives cotangent's complete domain. Simplifying an expression must not silently discard these zeros.

For exact values, $\operatorname{cosec}(5\pi/6)=1/(1/2)=2$, and $\cot(3\pi/4)=(-\sqrt2/2)/(\sqrt2/2)=-1$. Reciprocation preserves the sign of a nonzero ratio. Secant and cosecant cannot have values strictly between $-1$ and $1$, because their denominators have magnitude at most one.

Change only the angle to its reflection in the horizontal axis. Cosine, and therefore secant, stays unchanged.

At $3\pi/2$, cosine is zero, so secant is undefined; infinity is not its value. Cosecant is similarly undefined at every sine zero.

Change only the angle in the secant calculation.

Reciprocal graphs and transformations

A cosine value near zero produces a secant value with large magnitude. A cosine value of $1$ or $-1$ is unchanged by reciprocation. Those two facts determine the shape of the graph.

For secant, zeros of cosine at $\pi/2+k\pi$ become vertical asymptotes. At $x=k\pi$, secant is $(-1)^k$. Positive branches have minimum $1$; negative branches have maximum $-1$. Secant is even and has period $2\pi$.

For cosecant, zeros of sine at $k\pi$ become asymptotes. At $x=\pi/2+k\pi$, cosecant is $(-1)^k$. Cosecant is odd and has period $2\pi$.

Cotangent has asymptotes $x=k\pi$, zeros $x=\pi/2+k\pi$ and range $\mathbb R$. Each branch decreases from arbitrarily large positive values to arbitrarily large negative values. It is odd with period $\pi$.

These features follow from the definitions; they are not printed in the formula booklet. Keep the attained range endpoints separate from the excluded asymptote positions.

For $y=d+a\sec(bx+c)$ or cosecant, with $a,b\ne0$, the period is $2\pi/|b|$ and the range is $y\le d-|a|$ or $y\ge d+|a|$. Solve the whole argument's denominator-zero condition for the domain. A negative $a$ reflects branches vertically. For transformed cotangent, the period is $\pi/|b|$, the range is $\mathbb R$, and the sign of $ab$ determines whether branches decrease or increase. These transformation results are not in the formula booklet: learn how to derive them.

Change only the argument's constant by subtracting $\pi/2$. The vertical range and period stay fixed, but exclusions move.

For comparison, $y=\cot2x$ has period $\pi/2$, exclusions $x=k\pi/2$, zeros $x=\pi/4+k\pi/2$, and decreasing branches. Sketches should show these separate branches and labelled asymptotes; do not join across an excluded input.

Change only the coefficient of $x$ in the last secant transformation.