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Mathematics · Ch 4 — Inverse Trigonometric Functions

Graph of the Inverse Cosecant Function

4.6.3

Graph of the Inverse Cosecant Function

y=cosec−1xy=\text{cosec}^{-1}x has domain R∖(−1,1)\mathbb{R}\setminus(-1,1) and range [−π2,π2]∖{0}\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\} — that is, cosec−1:R∖(−1,1)→[−π2,π2]∖{0}\text{cosec}^{-1}:\mathbb{R}\setminus(-1,1)\to\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\}.

Fig. 4.21 shows the restricted cosecant curve on its principal domain, and Fig. 4.22 its reflection in y=xy=x: two flattening branches, one for x≥1x\ge1 approaching the horizontal asymptote y=0+y=0^+ near x=π2x=\tfrac{\pi}2's side and one for x≤−1x\le-1 symmetric below, with a genuine gap in the graph (matching the excluded y=0y=0 in the range) directly abo …

Figure 4.21Graph of y = cosec x on the restricted domain [-pi/2, pi/2] excluding 0: a branch over (0, pi/2] falling from infinity to 1, and a branch over [-pi/2, 0) falling from -1 to minus infinity; vertical asymptote at x = 0.
Fig. 4.21 — Graph of y = cosec x on the restricted domain [-pi/2, pi/2] excluding 0: a branch over (0, pi/2] falling from infinity to 1, and a branch over [-pi/2, 0) falling from -1 to minus infinity; vertical asymptote at x = 0.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Two branches meeting the horizontal lines y=1y=1 and y=−1y=-1 at x=π/2x=\pi/2 and x=−π/2x=-\pi/2 respectively, each shooting to ±∞\pm\infty as x→0x\to0. …

Figure 4.22Graph of y = cosec^{-1} x on domain R minus (-1,1): right branch from (1, pi/2) decreasing toward 0, left branch from (-1, -pi/2) increasing toward 0; horizontal asymptote y = 0, range [-pi/2, pi/2] minus {0}.
Fig. 4.22 — Graph of y = cosec^{-1} x on domain R minus (-1,1): right branch from (1, pi/2) decreasing toward 0, left branch from (-1, -pi/2) increasing toward 0; horizontal asymptote y = 0, range [-pi/2, pi/2] minus {0}.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Two flattening curve branches for x≥1x\ge1 and x≤−1x\le-1 approaching the horizontal asymptotes y=π/2y=\pi/2 and y=−π/2y=-\pi/2, with a gap in the graph (and in the range, at y=0y=0) directly above the interval (−1,1)(-1,1). …