Trigonometry is not just useful for solving triangles, we can also solve equations and draw graphs of trigonometric functions, just as we can for any other functions and these have a huge number of practical applications. Trigonometric functions are sometimes called circular functions as they can be derived from triangles inside a unit circle.

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Below is a quiz on Trigonometric functions. If you can answer all these questions, then you can skip this chapter as you already have the requisite knowledge.

Quiz: Trigonometric functions

1. Convert to radians:

(a)[latex]180°[/latex]        (b)  [latex]90°[/latex]       (c) [latex]60°[/latex]      (d) [latex]315°[/latex]

 

2.  Convert to degrees:

(a) [latex]\dfrac{π}{6}[/latex]        (b)  [latex]\dfrac{2π}{3}[/latex]         (c) [latex]\dfrac{7π}{12}[/latex]       (d) [latex]1.2[/latex]

 

3. Sketch the graph of [latex]y=3\sin{2x}[/latex] over the domain [latex]0≤x≤2π[/latex]

 

4. Solve the equation [latex]2\cos{2x}=1[/latex] for [latex]0≤x≤π[/latex]

 

5. Solve the equation [latex]\cos{2x}+\cos{x}+1=0[/latex] for [latex]0≤x≤2π[/latex]

 

Answers

1.    (a) [latex]\pi[/latex]      (b) [latex]\dfrac{\pi}{2}[/latex]      (c) [latex]\dfrac{\pi}{3}[/latex]      (d) [latex]\dfrac{7\pi}{4}[/latex]

2.  (a) [latex]30°[/latex]   (b) [latex]120°[/latex]   (c) [latex]105°[/latex]   (d) [latex]68.75°[/latex]

3. 

4.   [latex]\dfrac{π}{6}\,\ ,\dfrac{5π}{6}[/latex]        5.   [latex]\dfrac{π}{2}\,\ ,\dfrac{2π}{3}\,\ , \dfrac{3π}{2}\,\ , \dfrac{4π}{3}[/latex] 

If you need to review this topic continue reading.

Angular measurement: Definition of a radian

Though angles have commonly been measured in degrees they may also be measured in units known as radians.

One radian is the angle created by bending the radius length around the arc of a circle.

Converting between radians and degrees

Because the circumference of a circle is given by the formula [latex]C = 2πr[/latex], we know [latex]2π[/latex] radians [latex]\left(2π^c \right)[/latex]  is a complete rotation and the same as [latex]360[/latex] degrees. Similarly half a rotation or [latex]180[/latex] degrees [latex]= \pi[/latex] radians ([latex]180° = π^c[/latex] ).

Angles that represent fractional parts of a circle can be expressed in terms of [latex]π[/latex].

For other angles rearranging [latex]π^c = 180°[/latex] gives: [latex]1^{c} =\dfrac{180°}{\pi}[/latex] and [latex]1°=\dfrac{\pi^{c}}{180}.[/latex]

Note that it is customary to leave out the radian symbol. [latex]θ=1^c[/latex] is usually just written as [latex]θ=1[/latex]. If there is no degree symbol on a given angle. then we always assume that the angle is measured in radians.

Note also that angles measured in radians are not always written in terms of [latex]π[/latex]. For example, an angle of [latex]60°[/latex] could be written in radians as [latex]\dfrac{π}{3}[/latex] or as [latex]\dfrac{3.14159...}{3}=1.0472[/latex] radians.

Now try the following exercises.

Circular functions

The trigonometric ratios that have been defined in right-angled triangles can be extended to angles greater than 90° by considering angles as rotations within a unit circle. The centre of the unit circle is at point (0,0) and it has a radius of one unit.

Angles are considered as rotations from the positive x-axis.

An angle of [latex]135°[/latex] or [latex]\dfrac{3\pi}{4}[/latex] is shown in Fig. 11.3.

Angles greater than [latex]180°[/latex] and negative angles can also be defined in terms of the unit circle. Anticlockwise rotations are considered positive and clockwise rotations are negative, for example, the angle below could be defined as an angle of [latex]210°[/latex] or as an angle of [latex]-150°[/latex].

If [latex]P(x,y)[/latex] is any point on the unit circle, and [latex]\theta[/latex] is the angle [latex]POQ[/latex] in the triangle [latex]POQ[/latex] as shown:

Then, from the trigonometric properties of a right triangle:

[latex]\cos(\theta)=\dfrac{Adjacent}{Hypotenuse}=\dfrac{OQ}{OP}=\dfrac{x}{1}=x[/latex]

[latex]\sin(\theta)=\dfrac{Opposite}{Hypotenuse}=\dfrac{PQ}{OP}=\dfrac{y}{1}=y[/latex]

[latex]\tan(\theta)=\dfrac{Opposite}{Adjacent}=\dfrac{PQ}{OQ}=\dfrac{y}{x.}[/latex]

In summary, in the unit circle:

[latex]\cos(\theta) =x\qquad\sin(\theta)=y\qquad\tan(\theta)=\dfrac{y}{x}.[/latex]

Examples: Circular functions

From the above diagram and the rules:   [latex]\cos(\theta) =x\qquad\sin(\theta)=y\qquad\tan(\theta)=\dfrac{y}{x}[/latex], we can deduce the following:

[latex]\sin{0}° =0[/latex]

[latex]\cos180° =-1[/latex]

[latex]\sin{(-90°)} =-1[/latex]

[latex]\sin{63°} =0.89[/latex]

[latex]\tan{(-180°)} =0[/latex]

[latex]\tan{63°} =\dfrac{0.89}{0.45}=1.98[/latex]

[latex]\cos{(-297°)} =\cos{63°}=0.45[/latex]

Cosine, sine and tangent values can also be calculated using a scientific calculator. (Hint: make sure your calculator is in degrees or radian mode accordingly).

Exact values of circular functions

Using the table of exact values and the symmetry of the unit circle it is also possible to find the exact values for multiples of  [latex]30°\ ,\ 45°\ ,\ 60°[/latex].

Examples: Exact values

1.  Evaluate [latex]\sin330°[/latex].

 

In Fig. 11.7, plotting [latex]330°[/latex] on a unit circle shows that [latex]\sin330°[/latex] is closely related to [latex]\sin30°.[/latex] The [latex]y[/latex]-coordinates differ only by sign because the distances from the [latex]x[/latex]-axis are the same.Therefore,  [latex]\sin30°=\dfrac{1}{2}[/latex] (from the table above) and [latex]\sin330°=-\dfrac{1}{2}[/latex] .2. Evaluate [latex]\cos150°[/latex]. 

Plotting [latex]150°[/latex] on a unit circle as in Fig. 11.8 shows that [latex]\cos150°[/latex] is closely related to [latex]\cos30°[/latex]. The [latex]x[/latex]-coordinates differ only by sign because the distances from the [latex]x[/latex]-axis are the same.Therefore, [latex]\cos30°=\dfrac{\sqrt{3}}{2}[/latex] (from the table above) and [latex]\cos150°=-\dfrac{\sqrt{3}}{2}[/latex] .3. Evaluate [latex]\tan \dfrac{4\pi}{3}[/latex]. 

Plotting [latex]\dfrac{4\pi}{3}[/latex] on a unit circle as in Fig. 11.9 shows that [latex]\tan\dfrac{4\pi}{3}[/latex] is closely related to [latex]\tan\dfrac{\pi}{3}.[/latex] Tan values of diametrically opposite angles are the same.Therefore, [latex]\tan\dfrac{\pi}{3}=\sqrt{3}[/latex] (from the table) and [latex]\tan\dfrac{4\pi}{3}=\sqrt{3}[/latex] .

Note: By drawing a sketch diagram, it is always possible to use the symmetry of the unit circle and the values in the table to find the exact value of any multiple of [latex]30°, 45°, 60°[/latex]. The only difference will be a change of sign. Figure 11.10 may assist in determining whether a negative sign will be required.

Rotating anticlockwise from the first quadrant: All trigonometric functions are +ve in the first quadrant Sine is +ve for angles in the second quadrant Tangent is +ve for angles in the third quadrant Cosine is +ve for angles in the fourth quadrant.

Graphs of sine and cosine functions

The functions [latex]y=\sin x[/latex] and [latex]y=\cos x[/latex] have a domain of [latex]\mathbb{R}[/latex] and a range of [latex][-1,1][/latex].

The graphs of these functions are periodic graphs, that is, the shape of the graph repeats every set period.

The graphs of both functions have an amplitude of [latex]1[/latex] and a period of [latex]2\pi[/latex] radians (meaning it repeats every [latex]2\pi[/latex] units).

[Remember[latex]\pi\approx3.142[/latex], so [latex]2\pi\approx6.284[/latex]]

Translation and change of amplitude and period

The graphs of both [latex]y=a\sin nx[/latex] and [latex]y=a\cos nx[/latex] have an amplitude[latex]\left|a\right|[/latex] and a period of [latex]\dfrac{2\pi}{n}[/latex].

Examples - Graphs of sine and cosine functions

1.  Dilation

(a)  [latex]y=3\sin x[/latex]

In this case, [latex]a=3[/latex] and [latex]n=1[/latex], therefore the graph has an amplitude of [latex]3[/latex] and period of [latex]2\pi[/latex] .

(b)  [latex]y=3\cos2x[/latex] 

In this case, [latex]a=3[/latex] and [latex]n=2[/latex], therefore the graph has an amplitude of [latex]3[/latex] and period of [latex]\dfrac{2\pi}{2}=\pi[/latex] .2. Vertical translationThe graph of [latex]y=a\sin nx+k[/latex] is the graph of [latex]y=a\sin nx[/latex] translated up [latex]k[/latex] units (or down [latex]k[/latex] units if [latex]k[/latex] is negative).The graphs of [latex]y=\sin x+2[/latex] (red) and [latex]y=\sin x[/latex] (blue) as shown in Fig. 11.13. 

Similarly, the graph of [latex]y=a\cos nx+k[/latex] is the graph of [latex]y=a\cos nx[/latex] translated up [latex]k[/latex] units (or down [latex]k[/latex] units if [latex]k[/latex] is negative).3.  Horizontal translationReplacing the [latex]x[/latex] with [latex](x-\phi)[/latex] shifts the graphs of [latex]y=\sin x[/latex] and [latex]y=\cos x[/latex] horizontally [latex]\phi[/latex] units to the right.Similarly, replacing the [latex]x[/latex] with [latex](x+\phi)[/latex] would shift the graphs of [latex]y=\sin x[/latex] and [latex]y=\cos x[/latex] horizontally [latex]\phi[/latex] units to the left.[latex]y=\sin(x-\frac{\pi}{4})[/latex]The graph of [latex]y=\sin(x-\frac{\pi}{4})[/latex] (red) superimposed on the graph of [latex]y=\sin x[/latex] (blue) is shown in Fig. 11.14. 

4. ReflectionChanging the sign of a in the equations [latex]y=a\sin nx[/latex] and [latex]y=a\cos nx[/latex] results in reflection about the x-axis.Consider [latex]y=-2\cos x[/latex]. The graph of [latex]y=-2\cos x[/latex] (red) superimposed on the graph of [latex]y=3\cos x[/latex] (blue) is shown in Fig. 11.15. 

Now try the following exercise.

Exercises: Graphs of sine and cosine functions

1. Sketch the graphs of the following functions for one complete cycle, stating the amplitude and the period.

(a) [latex]y=2\cos x[/latex]

(b) [latex]y=2\sin3x[/latex]

(c) [latex]y=\frac{1}{2}\sin2x[/latex]

(d) [latex]y=3\cos\frac{x}{2}[/latex]

(e) [latex]y=-\sin4x[/latex]

 

2. Sketch the graphs of the following functions for [latex]0≤x≤2π[/latex].

(a) [latex]y=2\sin(x-\pi)[/latex]

(b) [latex]y=3\cos(x+\frac{\pi}{2})[/latex]

 

3. Sketch the graphs of the following functions for [latex]0≤x≤2π[/latex].

(a) [latex]y=2\sin3(x-\frac{\pi}{4})[/latex]

(b) [latex]y=3\cos(4x-2\pi)[/latex]

(c)  [latex]y=2\sin(2x+\frac{\pi}{3})[/latex]

 

Answers

1.  a)

Amplitude [latex]= 2[/latex] , Period [latex]= 2\pi[/latex]

b)

Amplitude [latex]= 2[/latex], Period [latex]= \dfrac{2\pi}{3}[/latex]

 

c)

Amplitude [latex]= \dfrac{1}{2}[/latex] , Period [latex]= \pi[/latex]

 

d)               Amplitude [latex]= 3[/latex] , Period [latex]= 4\pi[/latex]

e)

Amplitude [latex]= 1[/latex], Period [latex]= \dfrac{\pi}{2}[/latex]

2. a)

b)

3.  a)

b)

c)

 

Trigonometric equations

Often, the value of the trigonometric function is given and the corresponding angle(s), within a given domain, are required.

Examples: Trigonometric equations

1. Given [latex]\sinθ = 0.3[/latex] find all values of [latex]θ[/latex] in the domain [latex]0° ≤ θ≤ 360°[/latex].

Solution:

Figure 11.16 shows that there are two angles in the given domain (one complete positive rotation) that will have a sine value of [latex]0.3[/latex].

Using the calculator to find the first value:

\begin{align*}
\sin \theta &= 0.3 \\
\implies \theta &= \sin ^{-1}\left(0.3 \right) \\
&=17.46 ^ \circ \, .
\end{align*}

Using symmetry, the angle in the 2nd. quadrant is [latex]180 – 17.46 = 162.54°[/latex]

Therefore, the solutions are [latex]\theta= 17.46°, 162.54°[/latex]

2. Solve  [latex]2\cosα+1 = 0[/latex] over the domain [latex]0  ≤  α ≤  2π[/latex]

Solution:

\begin{align*}
2\cos \alpha+1 &=0 \\
\implies 2\cos \alpha &=-1 \\
\implies \cos \alpha &=-\frac{1}{2}\,.
\end{align*}

Figure 11.17 shows that there will be two angles in the given domain (one complete positive rotation) that will have a cosine value of [latex]-0.5[/latex].

Using the table of exact values to find the first value:

      [latex]\cosα = -0.5[/latex]

[latex]⇒      α= \cos^{-1} (-0.5)[/latex]

[latex]⇒      α = \dfrac{2π}{3}[/latex] or [latex]\dfrac{4π}{3}.[/latex]

Try solving the equations in the following exercise.

Exercises: Trigonometric equations

1.  a)   If [latex]\sin\phi = 0.25[/latex] find [latex]\phi[/latex] for   [latex]0°≤\phi≤180°[/latex]

b)  If [latex]\tan⁡\phi = 0.8[/latex]  find  [latex]\phi[/latex]  for   [latex]0° ≤ \phi  ≤ 360°[/latex]

c)  If [latex]\cos\phi = 0.4[/latex]  find  [latex]\phi[/latex]  for   [latex]-180° ≤ ∅  ≤ 360°[/latex]

d)  If [latex]\cos\phi = -0.4[/latex]  find  [latex]\phi[/latex]  for   [latex]-180° ≤ \phi  ≤ 360°[/latex]

e)  If [latex]\tan\phi = -1.5[/latex]   find  [latex]\phi[/latex]  for   [latex]0° ≤ ∅  ≤ 360°[/latex]

f)  If [latex]\cos\phi = -0.3[/latex]  find  [latex]\phi[/latex]  for   [latex]0° ≤ ∅  ≤ 360°[/latex]

Answers

1. a)  [latex]14.5°\ ,\ 165.5°[/latex]                b)  [latex]38.7°\ ,\ 218.7°[/latex]             c)  [latex]66.4°\ ,\ -66.4°[/latex]

d)  [latex]113.6° \ ,\ - 113.6°[/latex]       e)  [latex]123.7°\ ,\ 303.7° [/latex]          f)  [latex]107.5°\ ,\ 252.5°[/latex]

 

 

 

Trigonometric identities

\begin{align*}
\sin\theta&=\frac{Opp}{Hyp}, \quad    \cos\theta=\frac{Adj}{Hyp}, \quad \tan\theta=\frac{Opp}{Adj}.
\end{align*}

Therefore

\begin{align*}
\frac{\sin\theta}{\cos\theta}&=\frac{\frac{Opp}{Hyp}}{\frac{Adj}{Hyp}}\\
&=\frac{Opp}{Hyp}\times \frac{Hyp}{Adj}\\
&=\frac{Opp}{Adj}\\
&=\tan\theta.
\end{align*}

Another identity, sometimes referred to as the fundamental trigonometric identity, can be derived from the unit circle.

Fundamental Identities

[latex]\tan\theta =\dfrac{\sin\theta}{\cos\theta}[/latex].

[latex]\sin^{2}\theta+\cos^{2}\theta =1[/latex].

Double Angle Formulas

[latex]\sin2\theta =2\sin\theta\cos\theta[/latex].

\begin{align*}
\cos2\theta &=\cos ^{2}\theta-\sin ^{2}\theta \hspace{14cm} \\
&=1-2\sin ^{2}\theta \\
&=2\cos^ {2}\theta-1.
\end{align*}

[latex]\tan2\theta =\dfrac{2\tan\theta}{1-\tan^{2}\theta}[/latex].

Sums and Differences

[latex]\sin(x+y) =\sin x\cos y+\cos x\sin y[/latex].

[latex]\sin(x-y) =\sin x\cos y-\cos x\sin y[/latex].

[latex]\cos(x+y) =\cos x\cos y-\sin x\sin y[/latex].

[latex]\cos(x-y) =\cos x\cos y+\sin x\sin y[/latex].

[latex]\tan(x+y) =\dfrac{\tan x+\tan y}{1-\tan x\tan y}[/latex].

[latex]\tan(x-y) =\dfrac{\tan x-\tan y}{1+\tan x\tan y}[/latex].

Definitions of Other Trigonometric Functions

[latex]\csc x =\dfrac{1}{\sin x}[/latex].

[latex]\sec x =\dfrac{1}{\cos x}[/latex].

[latex]\cot x =\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x}[/latex].