ONE MORE STEP TOWARD BETTER TOMORROW : TRIGONOMETRIC EQUATIONS

Monday, 25 March 2013

TRIGONOMETRIC EQUATIONS

Some equations which involve trigonometric functions of the unknown may be readily solved by using simple algebraic ideas (as Equation 1 below), while others may be impossible to solve exactly, only approximately (e.g., Equation 2 below):

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EXAMPLE 1: Find all solutions of the equation

.
Solution: We can graphically visualize all the angles u which satisfy the equation by noticing that

is the y-coordinate of the point where the terminal side of the angle u (in standard position) intersects the unit circle (see Figure 1):
We can see that there are two angles in

that satisfy the equation:

and

. Since the period of the sine function is

, it follows that all solutions of the original equation are:

Solve for x in the following equation.

Example 1:
$2sin\left( x\right) -1=0$
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.

$\begin{eqnarray*}&& \\ 2\sin \left( x\right) -1 &=&0 \\ && \\ \sin \left( x\right) &=&\displaystyle \frac{1}{2} \\ && \\ && \end{eqnarray*}$

We know that the $\sin \left( \displaystyle \frac{\pi }{6}\right) =\displaystyle \frac{1}{2},$ therefore $x=\displaystyle \frac{\pi }{6}.$ The sine function is positive in quadrants I and II. The $\sin \left( \pi -\displaystyle \frac{\pi }{6}\right) =\sin \displaystyle \frac{5\pi }{6}$ is also equal to $\displaystyle \frac{1}{2}.$ Therefore, two of the solutions to the problem are $x=\displaystyle \frac{\pi }{6}$ and $x=\displaystyle \frac{5\pi }{6}.$

The period of the sin $\left( x\right)$ function is $2\pi .$ This means that the values will repeat every $2\pi$ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{\pi }{6}\pm n\left( 2\pi \right)$ and $x= \displaystyle \frac{5\pi }{6}\pm n\left( 2\pi \right)$ where n is an integer. The approximate solutions are $x=0.523598775598\pm n\left( 2\pi \right)$ and $x=2.61799387799\pm n\left( 2\pi \right)$ where n is an integer.

These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.

Numerical Check:

Check answer . $x=\displaystyle \frac{\pi }{6}$

Left Side: $\qquad 2\sin \left( x\right) -1=2\sin \left( \displaystyle \frac{\pi }{6} \right) -1=2\left( \displaystyle \frac{1}{2}\right) -1=0$
Right Side: 0

Since the left side equals the right side when you substitute $\displaystyle \frac{\pi }{6 }$ for x, then $\displaystyle \frac{\pi }{6 }$ is a solution.

Check answer . $x=\displaystyle \frac{5\pi }{6}$

Left Side: $\qquad 2\sin \left( x\right) -1=2\sin \left( \displaystyle \frac{5\pi }{6} \right) -1=2\left( \displaystyle \frac{1}{2}\right) -1=0$
Right Side: 0

Since the left side equals the right side when you substitute $\displaystyle \frac{5\pi }{ 6}$ for x, then $\displaystyle \frac{5\pi }{ 6}$ is a solution.

Graphical Check:
Graph the equation

f (x) = 2 sin(x) - 1

Note that the graph crosses the x-axis many times indicating many solutions. Note that it crosses at $\displaystyle \frac{ \pi }{6}\approx 0.5236$ . Since the period is $2\pi \approx 6.2831$ , it crosses again at 0.5236+6.283=6.81 and at0.5236+2(6.283)=13.09, etc. The graph crosses at $\displaystyle \frac{5\pi }{6}\approx 2.618$ . Since the period is $2\pi \approx 6.283$ , it will cross again at 2.618+6.283=8.9011 and at 2.618+2(6.283)=15.18, etc.

Monday, 25 March 2013

TRIGONOMETRIC EQUATIONS

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