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$$\text{Domain of sine}=[-\frac{\pi}{2},\frac{\pi}{2}]$$
$$\text{Range of sine}=[-\mathrm{1,}1]$$
This redefinition renders sine function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :
$$\text{Domain of arcsine}=[-\mathrm{1,}1]$$
$$\text{Range of arcsine}=[-\frac{\pi}{\mathrm{2,}}\frac{\pi}{2}]$$
Therefore, we define arcsine function as :
$$f:[-\mathrm{1,1}]\to [-\frac{\pi}{2},\frac{\pi}{2}]\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arcsin(x)}$$
The arcsin(x) .vs. x graph is shown here.
The arccosine function is inverse function of trigonometric cosine function. From the plot of cosine function, it is clear that an interval between 0 and $\pi $ includes all possible values of cosine function only once. Note that end points are included. The redefinition of domain of trigonometric function, however, does not change the range.
$$\text{Domain of cosine}=\left[\mathrm{0,}\pi \right]$$
$$\text{Range of cosine}=[-\mathrm{1,}1]$$
This redefinition renders cosine function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :
$$\text{Domain of arccosine}=[-\mathrm{1,1}]$$
$$\text{Range of arccosine}=\left[\mathrm{0,}\pi \right]$$
Therefore, we define arccosine function as :
$$f:[-\mathrm{1,1}]\to \left[\mathrm{0,}\pi \right]\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arccos(x)}$$
The arccos (x) .vs. x graph is shown here.
The arctangent function is inverse function of trigonometric tangent function. From the plot of tangent function, it is clear that an interval between $-\pi /2$ and $\pi /2$ includes all possible values of tangent function only once. Note that end points are excluded. The redefinition of domain of trigonometric function, however, does not change the range.
$$\text{Domain of tangent}=\left(-\frac{\pi}{\mathrm{2,}}\frac{\pi}{2}\right)$$
$$\text{Range of tangent}=R$$
This redefinition renders tangent function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :
$$\text{Domain of arctangent}=R$$
$$\text{Range of arctangent}=\left(-\pi /\mathrm{2,}\pi /2\right)$$
Therefore, we define arctangent function as :
$$f:R\to \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arctan (x)}$$
The arctan(x) .vs. x graph is shown here.
The arccosecant function is inverse function of trigonometric cosecant function. From the plot of cosecant function, it is clear that union of two disjointed intervals between “ $-\pi /2$ and 0” and “0 and $\pi /2$ ” includes all possible values of cosecant function only once. Note that zero is excluded, but “ $-\pi /2$ “ and “ $\pi /2$ ” are included . The redefinition of domain of trigonometric function, however, does not change the range.
$$\text{Domain of cosecant}=[-\pi /\mathrm{2,}\pi /2]-\left\{0\right\}$$
$$\text{Range of cosecant}=\left(-\infty ,-1]\cup [\mathrm{1,}\infty \right)=R-\left(-\mathrm{1,}1\right)$$
This redefinition renders cosecant function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :
$$\text{Domain of arccosecant}=R-\left(-\mathrm{1,}1\right)$$
$$\text{Range of arccosecant}=[-\pi /\mathrm{2,}\pi /2]-\left\{0\right\}$$
Therefore, we define arccosecant function as :
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