ArcSine & ArcCosine

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restricting the domain

To complete the picture on our family of Trigonometric functions, we need some inverse functions. Since none of our Trigonometric functions are one-to-one functions, we will restrict their domains for the purposes of obtaining inverse functions.

Arcsine

The sine function is not one-to-one.

This function doesn’t have an inverse.

We can’t even restrict our domain to one period, because each hill and valley in the graph would still show us that a single funciton value comes from multiple domain numbers.

Therefore, our plan is to restrict the domain to half of one period of sine. We need a strategic interval of length $\pi $. The usual choice is $\left [-\frac {\pi }{2},\frac {\pi }{2}\right ]$.

It is only half a period, but it does cover the whole range of sine. And, more importantly, it covers every funciton value exactly once.

We can now reverse all of the pairs and obtain the inverse function known as arcsine.

$Arcsine$ only returns angles between $-\frac {\pi }{2}$ and $\frac {\pi }{2}$. These are in the fourth and first quadrants. Arcsine is an inverse function for sine for these angles.

$\sin \left (\frac {\pi }{6}\right ) = \frac {1}{2}$

$\arcsin \left (\frac {1}{2}\right ) = \frac {\pi }{6}$

$\arcsin \left (\sin \left (\frac {\pi }{6}\right )\right ) = \frac {\pi }{6}$

When angles are outside $\left [-\frac {\pi }{2},\frac {\pi }{2}\right ]$, then the Arcsine brings them back into this interval.

$\sin \left (\frac {5\pi }{6}\right ) = \frac {1}{2}$

$\arcsin \left (\frac {1}{2}\right ) = \frac {\pi }{6}$

$\arcsin \left (\sin \left (\frac {5\pi }{6}\right )\right ) = \frac {\pi }{6}$

We begin with the angle $\frac {5\pi }{6}$ whose sine is $\frac {1}{2}$. Then Arcsine looks for an angle whose sine is $\frac {1}{2}$. But Arcsine only looks in quadrants I and IV. So, it finds $\frac {\pi }{6}$.

Characteristics

We can deduce many characteristics about arcsine from sine.

$\blacktriangleright $ The domain is $[-1, 1]$.

$\blacktriangleright $ The range is $\left [ -\frac {\pi }{2}, \frac {\pi }{2} \right ]$

$\blacktriangleright $ Arcsine has one zero and that is $0$.

$\blacktriangleright $ Arcsine is an increasing function.

$\blacktriangleright $ Arcsine is a continuous function.

$\blacktriangleright $ Arcsine has a maximum of $\frac {\pi }{2}$, which occurs at $1$.

$\blacktriangleright $ Arcsine has a minimum of $-\frac {\pi }{2}$, which occurs at $-1$.

If $0 < \theta < \frac {\pi }{2}$, then $\arcsin (\sin (\theta ))$ is in quadrant

I II III IV

If $\frac {\pi }{2} < \theta < \pi $, then $\arcsin (\sin (\theta ))$ is in quadrant

I II III IV

If $\pi < \theta < \frac {3\pi }{2}$, then $\arcsin (\sin (\theta ))$ is in quadrant

I II III IV

If $\frac {3\pi }{2} < \theta < 2\pi $, then $\arcsin (\sin (\theta ))$ is in quadrant

I II III IV

If $0 < \theta < \frac {\pi }{2}$, then $\arcsin (\sin (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $\theta - 2\pi $

If $\frac {\pi }{2} < \theta < \pi $, then $\arcsin (\sin (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $\theta - 2\pi $

If $\pi < \theta < \frac {3\pi }{2}$, then $\arcsin (\sin (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $\theta - 2\pi $

If $\frac {3\pi }{2} < \theta < 2\pi $, then $\arcsin (\sin (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $\theta - 2\pi $

Arccosine

The cosine function is not one-to-one.

This function doesn’t have an inverse.

We can’t even restrict our domain to one period, because of the hills and valleys in the graph.

Therefore, our plan is to restrict the domain to half of one period of cosine. We need a strategic interval of length $\pi $. The usual choice is $[0,\pi ]$.

It is only half a period, but it does cover the whole range of cosine.

We can now reverse all of the pairs and obtain the inverse function known as arccosine.

$Arccosine$ only returns angles between $0$ and $\pi $. It is an inverse function for these angles.

$\cos \left (\frac {2\pi }{3}\right ) = -\frac {1}{2}$

$\arccos \left (-\frac {1}{2}\right ) = \frac {2\pi }{3}$

$\arccos \left ( \cos \left ( \frac {2\pi }{3} \right ) \right ) = \frac {2\pi }{3}$

When angles are outside $\left [ 0,\pi \right ]$, then the Arcsine brings them back into this interval.

$\cos \left (\frac {7\pi }{6}\right ) = -\frac {\sqrt {3}}{2}$

$\arccos \left (\frac {-\sqrt {3}}{2}\right ) = \frac {5\pi }{6}$

$\arccos \left (\cos \left (\frac {7\pi }{6}\right )\right ) = \frac {5\pi }{6}$

We begin with the angle $\frac {5\pi }{6}$ whose cosine is $-\frac {\sqrt {3}}{2}$. Then Arccosine looks for an angle whose cosine is $-\frac {\sqrt {3}}{2}$. But Arccosine only looks in quadrants I and II. So, it finds $\frac {5\pi }{6}$.

Characteristics

We can deduce many characteristics about arccosine from cosine.

$\blacktriangleright $ The domain is $[-1, 1]$.

$\blacktriangleright $ The range is $[0, \pi ]$

$\blacktriangleright $ Arccosine has one zero and that is $1$.

$\blacktriangleright $ Arccosine is an decreasing function.

$\blacktriangleright $ Arccosine is a continuous function.

$\blacktriangleright $ Arccosine has a maximum of $\pi $, which occurs at $-1$.

$\blacktriangleright $ Arccosine has a minimum of $0$, which occurs at $1$.

If $0 < \theta < \frac {\pi }{2}$, then $\arccos (\cos (\theta ))$ is in quadrant

I II III IV

If $\frac {\pi }{2} < \theta < \pi $, then $\arccos (\cos (\theta ))$ is in quadrant

I II III IV

If $\pi < \theta < \frac {3\pi }{2}$, then $\arccos (\cos (\theta ))$ is in quadrant

I II III IV

If $\frac {3\pi }{2} < \theta < 2\pi $, then $\arccos (\cos (\theta ))$ is in quadrant

I II III IV

If $0 < \theta < \frac {\pi }{2}$, then $\arccos (\cos (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $2\pi - \theta $

If $\frac {\pi }{2} < \theta < \pi $, then $\arccos (\cos (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $2\pi - \theta $

If $\pi < \theta < \frac {3\pi }{2}$, then $\arccos (\cos (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $2\pi - \theta $

If $\frac {3\pi }{2} < \theta < 2\pi $, then $\arccos (\cos (\theta ))$ equals

$\theta $ $\pi - \theta $ $\pi + \theta $ $2\pi - \theta $

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More Examples of Inverse Trig Functions

2025-05-17 23:33:43