Trigonometry: Trigonometric functions
Trigonometric equations 1
We have seen that the trigonometric functions #\sin(x)#, #\cos(x)# and #\tan(x)# on a limited domain have an inverse function. With this inverse function we can solve equations of the form #\sin(x)=c#, #\cos(x)=c# and #\tan(x)=c#. For limited values of #c# we can give exact solutions, for other values we do this with the aid of the calculator.
Solving trigonometric equations
Step-by-step |
Example |
|
We solve a trigonometric equation of the form #\sin(ax+b)=c#, #\cos(ax+b)=c# or #\tan(ax+b)=c# with numbers #a#, #b# and #c#. |
#\sin(2x+\pi)=\frac{1}{\sqrt{2}}# |
|
Step 1 |
Use the inverse function to determine one solution of #ax+b#. This solution is repeated every period; we indicate this by adding #k \cdot 2 \pi#, in which #k# is an integer. Note: the calculator does not give exact solutions. If an exact solution is required, we must use the table with special values. |
#2x+\pi=\frac{\pi}{4}+k \cdot 2\pi# |
Step 2 |
Use the graphs of the trigonometric functions or the symmetry of the unit circle to determine if there is a second solution within one period. This solution is also repeated every period. |
#\begin{array}{c}2x+\pi=\frac{\pi}{4}+k \cdot 2\pi \\ \lor \\ 2x+\pi=\frac{3\pi}{4}+k\cdot 2\pi\end{array}# |
Step 3 |
Reduce the left side of the equation to #x#. |
#\begin{array}{c}x=-\frac{3\pi}{8}+k \cdot \pi \\ \lor \\x=-\frac{\pi}{8}+k\cdot \pi\end{array}# |
#\begin{array}{rcl}
\cos(2\cdot x)&=&{{1}\over{\sqrt{2}}} \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}}\\
2\cdot x={{\pi}\over{4}}+2\cdot \pi\cdot k &\lor& 2\cdot x={{7\cdot \pi}\over{4}}+2\cdot \pi\cdot k \\ &&\phantom{xxx}\blue{\text{taken the inverse }\cos \text{ on both sides}}\\
x={{\pi}\over{8}}+\pi\cdot k &\lor& x={{7\cdot \pi}\over{8}}+\pi\cdot k \\ &&\phantom{xxx}\blue{\text{reduced}}\\
\end{array}#
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