2eMoreExamples

More Examples using Inverse Trig Functions

Notice that when we consider the function math y=\text {Sin } \big( x \big) math

We are feeding values of x into the function and what comes out of the function is going to y math x \; \Longrightarrow \; \textbf{sin } \; \Longrightarrow \; y math

Using the implied domain for Sin, we can show the domain and range as a chain: math \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] \; \Longrightarrow \; \textbf{sin } \; \Longrightarrow \; [-1, 1] math

More complicated functions can be treated in the same way to investigate the domain and range


 * Example 1 **

Find the implied domain and range of the following function: math y=3\text{Cos}^{-1} \left(2x \right) math


 * Solution: **

This becomes math x \; \Longrightarrow \; \times 2 \; \Longrightarrow \; \textbf{Cos}^{-1} \; \Longrightarrow \; \times 3 \; \Longrightarrow \; y math

We start by filling in each side of the Arcos term with the **Principal Domain** and **range** math x \; \Longrightarrow \; \times 2 \; \Longrightarrow \; [-1,\; 1] \; \Longrightarrow \; \textbf{Cos}^{-1} \; \Longrightarrow \; [0,\; \pi] \; \Longrightarrow \; \times 3 \; \Longrightarrow \; y math

Now we work backwards and forwards from there math \left[ -\frac{1}{2},\, \frac{1}{2} \right] \, \Longrightarrow \, \times 2 \, \Longrightarrow \, [-1,\, 1] \, \Longrightarrow \, \textbf{Cos}^{-1} \, \Longrightarrow \, [0,\, \pi] \, \Longrightarrow \times 3 \, \Longrightarrow \; [0,\, 3\pi] math

Thus we can see that for this function:

math \text {Domain: } \; x \in \left[ -\frac{1}{2},\; \frac{1}{2} \right] math

math \text {Range: } \; y \in [0,\; 3\pi] math


 * Example 2 **

Find the implied domain and range of the following function: math y = \text { Sin} \left( 2 \text {Cos}^{-1} (x) \right) math


 * Solution: **

This becomes: math x \; \Longrightarrow \; \textbf{Cos}^{-1} \; \Longrightarrow \; \times 2 \; \Longrightarrow \; \textbf{Sin} \; \Longrightarrow \; y math

Start by filling in each side of the Arcos term with the **Principal Domain** and **range** math [-1, 1] \; \Longrightarrow \; \textbf{Cos}^{-1} \; \Longrightarrow \; [0, \pi] \; \Longrightarrow \; \times 2 \; \Longrightarrow \; \textbf{Sin } \; \Longrightarrow \; y math

Fill in the next step: math [-1, 1] \; \Longrightarrow \; \textbf{Cos}^{-1} \; \Longrightarrow \; [0, \pi] \; \Longrightarrow \; \times 2 \; \Longrightarrow \; [0, 2\pi] \; \Longrightarrow \; \textbf{Sin} \; \Longrightarrow \; y math

BUT [0,2 p ] cant go into Sin(x). ... {2 p is outside of the **Principal Domain** for **Sin(x)**} The largest value that Sin(x) can accept is p /2, so adjust the maximum input for Sin(x) in our chain math [?,\;?] \; \Longrightarrow \; \textbf{Cos}^{-1} \; \Longrightarrow \; [?, \;?] \; \Longrightarrow \; \times 2 \; \Longrightarrow \; \left[ 0, \dfrac{\pi}{2} \right] \; \Longrightarrow \; \textbf{Sin} \; \Longrightarrow \; y math

Now work forwards and backwards from that point: math [?,\;?] \Longrightarrow \, \textbf{Cos}^{-1} \Longrightarrow \left[ 0, \dfrac{\pi}{4} \right] \Longrightarrow \times 2 \Longrightarrow \left[ 0, \dfrac{\pi}{2} \right] \Longrightarrow \textbf{Sin} \Longrightarrow [0, 1] math

Keep working backwards, decide what input is needed for **Arcos**, to get those values as output: math \left[ \dfrac{1}{\sqrt{2}},\;1 \right] \Longrightarrow \textbf{Cos}^{-1} \Longrightarrow \left[ 0, \dfrac{\pi}{4} \right] \Longrightarrow \times 2 \Longrightarrow \left[ 0, \dfrac{\pi}{2} \right] \Longrightarrow \textbf{Sin} \Longrightarrow [0,\;1] math

We can now write down the domain and range for this function

math \text {Domain: } x \in \left[ \dfrac{1}{\sqrt{2}},\;1 \right] math

math \text {Range: } y \in [0,\;1] math

Return To Inverse Trig Functions

Return to Summary of Trigonometry

.