3.15 Composition of trigonometric function and its inverse  (Page 2/3)

$$x=\pi -\theta$$

$$\Rightarrow \theta =\pi -x$$

Hence,

$$\Rightarrow {\sin }^{-1}\sin x=\pi -x;x\in [\frac{\pi }{2},\frac{3\pi }{2}]$$

In order to find expression corresponding to negative angle interval $[-3\pi /\mathrm{2,}-\pi /2]$ , we need to construct negative value diagram. We know that equivalent negative angle is obtained by deducting “-2π” to the positive angle. Thus, corresponding to expression for positive angles in four quadrants, the expression in terms of negative angles are “-θ”,“-π+θ”,“-π-θ” and “-2π+θ” in four quadrants counted in clockwise direction in the value diagram. Now, we estimate from the sine plot that an angle, corresponding to a positive acute angle, θ, in the principal interval, lies in third negative quadrant. Therefore,

$$x=-\pi -\theta$$

$$\Rightarrow \theta =-\pi -x$$

Hence,

$$\Rightarrow {\sin }^{-1}\sin x=-\pi -x;x\in [-\frac{3\pi }{2},-\frac{\pi }{2}]$$

Combining three results,

|-π-x; x∈ [-3π/2, -π/2] sin⁻¹ sinx = | x; x∈ [-π/2, π/2]| π- x; x∈ [π/2, 3π/2]

We can similarly find expressions for more such intervals.

Graph of sin⁻¹sinx

Using three expressions obtained above, we can draw plot of the composition function. We extend the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :

$$y=x-2\pi$$

The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :

$$y=x+2\pi$$

Sine inverse of sine

We see that graph of composition is continuous. Its domain is R. Its range is $[-\pi /\mathrm{2,}\pi /2]$ . The function is periodic with period 2π.

Composition with arccosine

The composition ${\cos }^{-1}\cos x$ evaluates to angle values lying in the interval [0, π].

$${\cos }^{-1}\cos x=x;x\in \left[\mathrm{0,}\pi \right]$$

Let us consider adjacent intervals such that all cosine values are included once. Such intervals are [π, 2π], [2π, 3π]etc on the right side and [-π, 0], [-2π, -π]etc on the left side of the principal interval.

Cosine function

The new interval $[\pi ,2\pi ]$ represents third and fourth quadrants. The angle x, corresponding to positive acute angle θ, lies in fourth quadrant. Then,

Value diagrams

$$x=2\pi -\theta$$

$$\Rightarrow \theta =2\pi -x$$

Hence,

$$\Rightarrow {\cos }^{-1}\cos x=2\pi -x;x\in [\pi ,2\pi ]$$

In order to find expression corresponding to negative angle interval $[-\pi ,0]$ , we estimate from the cosine plot that an angle corresponding to a positive acute angle, θ, in the principal interval lies in first negative quadrant. Therefore,

$$x=-\theta$$

$$\Rightarrow \theta =-x$$

Hence,

$$\Rightarrow {\cos }^{-1}\cos x=-x;x\in [-\pi ,0]$$

Combining three results,

|-x; x∈ [-π, 0] cos⁻¹ cosx = | x; x∈[0, π]|2π- x; x∈ [π, 2π]

We can similarly find expressions for other intervals.

Graph of cos⁻¹cosx

Using three expressions obtained above, we can draw plot of the composition function. We have extended the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :

$$y=x-2\pi$$

The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :

$$y=x+2\pi$$

Cosine inverse of cosine

We see that graph of composition is continuous. Its domain is R. Its range is $\left[\mathrm{0,}\pi \right]$ . The function is periodic with period 2π.