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

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$x=\pi -\theta$

$⇒\theta =\pi -x$

Hence,

$⇒{\mathrm{sin}}^{-1}\mathrm{sin}x=\pi -x;\phantom{\rule{1em}{0ex}}x\in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]$

In order to find expression corresponding to negative angle interval $\left[-3\pi /2,-\pi /2\right]$ , 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,

$x=-\pi -\theta$

$⇒\theta =-\pi -x$

Hence,

$⇒{\mathrm{sin}}^{-1}\mathrm{sin}x=-\pi -x;\phantom{\rule{1em}{0ex}}x\in \left[-\frac{3\pi }{2},-\frac{\pi }{2}\right]$

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 :

$y=x+2\pi$

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

## Composition with arccosine

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

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

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.

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

$x=2\pi -\theta$

$⇒\theta =2\pi -x$

Hence,

$⇒{\mathrm{cos}}^{-1}\mathrm{cos}x=2\pi -x;\phantom{\rule{1em}{0ex}}x\in \left[\pi ,2\pi \right]$

In order to find expression corresponding to negative angle interval $\left[-\pi ,0\right]$ , 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,

$x=-\theta$

$⇒\theta =-x$

Hence,

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

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 :

$y=x+2\pi$

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

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