Trigonometric Functions

Lesson

This chapter continues on from a previous chapter on solving trigonometric equations.

In this chapter, we look at the periodic nature of trigonometric functions and the problem of finding *all *the solutions of an equation that fall within a given interval. We also explore the use of trigonometric identities in the process of finding solutions to equations.

The periodic character of the sine, cosine and tangent functions can be seen in their graphs.

We divide the interval $[0,2\pi]$[0,2π] into four 'quadrants': $\left[0,\frac{\pi}{2}\right)$[0,π2), $\left[\frac{\pi}{2},\pi\right)$[π2,π), $\left[\pi,\frac{3\pi}{2}\right)$[π,3π2) and $\left[\frac{3\pi}{2},2\pi\right]$[3π2,2π]. These quadrants also coincide with the unit circle definition of the trig functions.

From the graph above, we see that the sine function is positive in the first and second quadrants, cosine is positive in the first and fourth quadrants and tangent is positive in the first and third quadrants. This pattern repeats at intervals of $2\pi$2π.

Find all the solutions of the equation $\sin t=2\sin t-\frac{1}{2}$`s``i``n``t`=2`s``i``n``t`−12 in the interval $[0,2\pi]$[0,2π].

After collecting the like terms, we have $\frac{1}{2}=\sin t$12=`s``i``n``t`. Since $\frac{1}{2}$12 is positive, there should be a value of $t$`t` in both the first and second quadrants that satisfies this equation. In the first quadrant, this is $t=\frac{\pi}{6}$`t`=π6. To find the second quadrant solution, we subtract the first quadrant solution from $\pi$π. Thus, the second solution is $\frac{5\pi}{6}$5π6.

The solutions can be visualised by means of the following graphical representation.

When there are several trigonometric terms involving different multiples of the variable and different trigonometric functions, it is usually necessary to use identities to make simplifications. For example, if there is a term in $2\theta$2`θ` as well as a term in $\theta$`θ`, we need to express both as functions of $\theta$`θ`. If there is a sine term and a cosine term, we would look for a way to express both with a single trigonometric function.

Solve $\cos2x=2\sin^2x$`c``o``s`2`x`=2`s``i``n`2`x` over the interval $[0,2\pi]$[0,2π].

We first look for a way to express $\cos2x$`c``o``s`2`x` as a function of $x$`x`. This is best done through the identity $\cos2A=1-2\sin^2A$`c``o``s`2`A`=1−2`s``i``n`2`A` since this will make a good match with the right-hand term. Thus, we have

$1-2\sin^2x=2\sin^2x$1−2`s``i``n`2`x`=2`s``i``n`2`x`

Therefore,

$4\sin^2x$4sin2x |
$=$= | $1$1 |

$\sin^2x$sin2x |
$=$= | $\frac{1}{4}$14 |

$\sin x$sinx |
$=$= | $\pm\frac{1}{2}$±12 |

$x$x |
$=$= | $\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$π6,5π6,7π6,11π6 |

The four solutions can be seen in the following diagram.

Identities that may be useful include the following:

$\sin^2A+\cos^2A\equiv1$`s``i``n`2`A`+`c``o``s`2`A`≡1 (The Pythagorean identity)

$\sin2A\equiv2\sin A\cos A$`s``i``n`2`A`≡2`s``i``n``A``c``o``s``A`

$\cos2A\equiv\cos^2A-\sin^2A$`c``o``s`2`A`≡`c``o``s`2`A`−`s``i``n`2`A` (and its variants derived from the Pythagorean identity)

$\tan2A\equiv\frac{2\tan A}{1-\tan^2A}$`t``a``n`2`A`≡2`t``a``n``A`1−`t``a``n`2`A`

$\tan A\equiv\frac{\sin A}{\cos A}$`t``a``n``A`≡`s``i``n``A``c``o``s``A`

$\sec^2A\equiv1+\tan^2A$`s``e``c`2`A`≡1+`t``a``n`2`A`

Solve $\sin^2\left(x\right)-6\cos^2\left(x\right)=1$`s``i``n`2(`x`)−6`c``o``s`2(`x`)=1 over the interval $[$[$0$0, $2\pi$2π$)$).

Solve $\sec^2\left(x\right)=2\tan x$`s``e``c`2(`x`)=2`t``a``n``x` over the interval $[$[$0$0, $2\pi$2π$)$).

Solve for $x$`x` over the interval $[$[$0$0, $2\pi$2π$)$).

$\sin2x=2\cos x$`s``i``n`2`x`=2`c``o``s``x`.

Write all solutions on the same line, separated by commas.

Form and use trigonometric, polynomial, and other non-linear equations

Apply trigonometric methods in solving problems