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Mathematics · Ch 4 — Inverse Trigonometric Functions

Graph of the Inverse Sine Function

4.3.4

Graph of the Inverse Sine Function

sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1}:[-1,1]\to\left[-\tfrac{\pi}2,\tfrac{\pi}2\right] receives x∈[−1,1]x\in[-1,1] and returns y∈[−π2,π2]y\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]. As xx increases from −1-1 to 11, yy increases from −π2-\tfrac{\pi}2 to π2\tfrac{\pi}2; connecting the plotted points (x,y)(x,y) with a smooth curve produces Fig. 4.6.

Equivalently, Fig. 4.6 is obtained by reflecting the restricted sine curve of Fig. 4.7 (on [−π2,π2]\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]) in the line y=xy=x — interchanging its xx- and yy-axes, as shown against the reflection line in Fig. 4.8. The graph passes through the origin and is symmetric about the origin, confirming once more that y=sin⁡−1xy=\sin^{-1}x is an odd function. …

Figure 4.6Fig. 4.6 - Graph of the inverse sine function y = sin^-1 x with domain [-1, 1] and range [-pi/2, pi/2], an increasing curve through the origin from (-1, -pi/2) to (1, pi/2).
Fig. 4.6 — Fig. 4.6 - Graph of the inverse sine function y = sin^-1 x with domain [-1, 1] and range [-pi/2, pi/2], an increasing curve through the origin from (-1, -pi/2) to (1, pi/2).

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. A rising S-shaped curve through the origin, flattest at the centre and steepening near the endpoints, running from (−1,−π/2)\left(-1,-\pi/2\right) to (1,π/2)\left(1,\pi/2\right). …

Figure 4.7Fig. 4.7 - Graph of y = sin x restricted to the principal branch [-pi/2, pi/2], an increasing S-curve from (-pi/2, -1) through the origin to (pi/2, 1).
Fig. 4.7 — Fig. 4.7 - Graph of y = sin x restricted to the principal branch [-pi/2, pi/2], an increasing S-curve from (-pi/2, -1) through the origin to (pi/2, 1).

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. The single monotonically increasing arch of sine used to build the inverse, from (−π/2,−1)\left(-\pi/2,-1\right) through the origin to (π/2,1)\left(\pi/2,1\right). …

Figure 4.8Fig. 4.8 - Graph of the inverse sine function y = sin^-1 x, domain [-1, 1] and range [-pi/2, pi/2], increasing through the origin from (-1, -pi/2) to (1, pi/2).
Fig. 4.8 — Fig. 4.8 - Graph of the inverse sine function y = sin^-1 x, domain [-1, 1] and range [-pi/2, pi/2], increasing through the origin from (-1, -pi/2) to (1, pi/2).

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. The Fig. 4.7 arch shown together with the diagonal y=xy=x that the inverse is reflected across. …

Figure 4.9Fig. 4.9 - Graphs of y = sin x on [-pi/2, pi/2], y = sin^-1 x on [-1, 1], and the line y = x together, showing the sine and inverse-sine curves are mirror images in the line y = x.
Fig. 4.9 — Fig. 4.9 - Graphs of y = sin x on [-pi/2, pi/2], y = sin^-1 x on [-1, 1], and the line y = x together, showing the sine and inverse-sine curves are mirror images in the line y = x.

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. Both y=sin⁡xy=\sin x (restricted) and y=sin⁡−1xy=\sin^{-1}x drawn on one pair of axes together with the line y=xy=x, visually confirming each is the other's mirror image. …