Inverse Trigonometric Functions and Trigonometric Equations

The following type of question arises frequently when working with the trigonometric functions: for which angles x is sin x = y? There are many correct answers to these questions, since the trigonometric functions are periodic. The inverse trigonometric functions provide answers to such questions that lie within a restricted domain. The inverse sine function, for example, produces the angle x between - ${\frac{\pi}{2}}$ and ${\frac{\pi}{2}}$ that satisfies sin x = y. This solution is denoted by arcsin x or sin-1x.

The inverse trigonometric functions and a number of other functions are available in the dialog box that comes up when you click the Math Name button on the Math toolbar. They can also be entered from the keyboard in mathematics mode.


\begin{example}
To find
\index{Inverse@Inverse!trigonometric functions@trigon...
...\end{itemize}\par
This gives\textsf{\ }$\arctan 100=\,1.56079666$
\end{example}

You can also find an angle satisfying tan x = 100 by applying Solve + Numeric to the equation. This technique does not necessarily find the solution between - ${\frac{{\pi }}{{2}}}$ and ${\frac{{\pi }}{{2}}}$. In this case, in fact, it gives the solution x = 102.09176, which is 1.56079666 + 32π. You can specify the interval for the solution, as follows.

$\blacktriangleright$ Solve + Numeric

tan x = 100
x$\left(\vphantom{ -\frac{\pi }{2},\frac{\pi }{2}}\right.$ - ${\frac{{\pi }}{{2}}}$,${\frac{{\pi }}{{2}}}$$\left.\vphantom{ -\frac{\pi }{2},\frac{\pi }{2}}\right)$
, Solution is : $\left\{\vphantom{ x=1.56079666}\right.$x = 1.56079666$\left.\vphantom{ x=1.56079666}\right\}$

Using this technique, you can find solutions to a variety of trigonometric equations in specified intervals. Following are some examples of equations that you can solve with Solve + Exact and Solve + Numeric.

Equation Solve + Exact Solve + Numeric
sin t = sin 2t $\left\{\vphantom{ t=0}\right.$t = 0$\left.\vphantom{ t=0}\right\}$$\left\{\vphantom{ t=\frac{1}{3}\pi }\right.$t = ${\frac{{1}}{{3}}}$π$\left.\vphantom{ t=\frac{1}{3}\pi }\right\}$ t = 5.236
8 tan x - 13 + 5 tan2x = 3 x = arctan$\left(\vphantom{ -\frac{4}{5}\pm \frac{4}{5}%
\sqrt{6}}\right.$ - ${\frac{{4}}{{5}}}$±${\frac{{4}}{{5}%
}}$$\sqrt{{6}}$$\left.\vphantom{ -\frac{4}{5}\pm \frac{4}{5}%
\sqrt{6}}\right)$ x = - 2.2824
tan2x - cot2x = 1 x = ±arcsin$\left(\vphantom{ \frac{1}{2}\sqrt{-2\pm 2\sqrt{5}}}\right.$${\frac{{1}}{{2}}}$$\sqrt{{-2\pm 2\sqrt{5}}}$$\left.\vphantom{ \frac{1}{2}\sqrt{-2\pm 2\sqrt{5}}}\right)$ x = 2.237

Note that Solve + Exact gives two or four solutions in these examples, whereas Solve + Numeric returns only one solution. In general, applying Evaluate Numerically to the exact solutions gives numerical solutions different from those produced by Solve + Numeric. This is a good place to experiment with a plot to visualize the complete solution.dtbpFU3.0441in2.0384in0pt y = tan2x - cot2x = 1, y = 1solvtrig.wmf

Solutions can be found in a specified range with Solve + Numeric, as demonstrated in the following example. Enter the equation and the range in different rows of a one-column matrix.

$\blacktriangleright$ Solve + Numeric

8 tan x - 13 + 5 tan2x = 3
x$\left(\vphantom{ -\frac{\pi }{2},\frac{\pi }{2}}\right.$ - ${\frac{{\pi }}{{2}}}$,${\frac{{\pi }}{{2}}}$$\left.\vphantom{ -\frac{\pi }{2},\frac{\pi }{2}}\right)$
, Solution is : $\left\{\vphantom{ x=.85916}\right.$x = .85916$\left.\vphantom{ x=.85916}\right\}$