Title: 4.2 TrigonometricFunctions
Course: HELM Workbooks 2023-2024

Trigonometric functions for any size angle

The radian

First, we introduce an alternative to measuring angles in degrees. Look at the circle shown in the figure below. It has radius \(r\) and we have shown an arc \(AB\) of length \(\ell\) (measured in the same units as \(r\)). As you can see the arc subtends an angle \(\theta\) at the centre \(O\) of the circle.

[Image: Arc subtending angle theta]

The angle \(\theta\) in radians is defined as \(\displaystyle{\theta=\frac{\text{length of arc}\ AB}{\text{radius}}=\frac{\ell}{r}}\)

So, for example, if \(r=10 \text{ cm}\) , \(\ell=20 \text{ cm}\) , the angle \(\theta\) would be \(\dfrac{20}{10}=2 \text{ radians}\). The relation between the value of an angle in radians and its value in degrees is readily obtained as follows. Referring to figure (b) imagine that the arc \(AB\) extends to cover half the complete perimeter of the circle. The arc length is now \(\pi r\) (half the circumference of the circle) so the angle \(\theta\) subtended by \(AB\) is now \(\theta=\displaystyle{\frac{\pi r}{r}}=\pi \text{ radians}\)

But clearly this angle is \(180^{\circ}\). Thus \(\pi \text{ radians}\) is the same as \(180^\circ\). Note conversely that since \(\pi \text{ radians}\)\(=180^\circ\) then \(1 \text{ radian}\) \(=\dfrac{180}{\pi} \text{ degrees}\) (about \(57.3^\circ\)).

Key Point

\(\qquad 180^\circ= \pi\ \ \text{radians}\)

\(\qquad 360^\circ=2\pi\ \ \text{radians}\)   \(1 \text{ radian}\) \(=\dfrac{180}{\pi} \text{ degrees}\) (\(\approx 57.3^\circ\))

\(\qquad 1^\circ=\dfrac{\pi}{180}\ \ \text{radians}\)

\(\qquad x^\circ=\dfrac{\pi x}{180}\ \ \text{radians}\)   \(y \text{ radians}\) \(=\dfrac{180y}{\pi} \text{ degrees}\)

Exercise

Convert the following angles in radians to degrees:

1. \(\displaystyle{[value]}\) \(\text{rad} =\) \(\text{degrees}\)
2. \(\displaystyle{[value]}\) \(\text{rad} =\) \(\text{degrees}\)
3. \(\displaystyle{[value]}\) \(\text{rad} =\) \(\text{degrees}\)
4. \(\displaystyle{[value]}\) \(\text{rad} =\) \(\text{degrees}\)
Feedback: 1. \( \displaystyle{[value]} \text{ rad } = \dfrac{180}{\pi} \times \displaystyle{[value]} = [value] \text{ degrees}\) 2. \(\displaystyle{[value]} \text{ rad } = \dfrac{180}{\pi} \times \displaystyle{[value]} = [value] \text{ degrees}\) 3. \( \displaystyle{[value]} \text{ rad } = \dfrac{180}{\pi} \times \displaystyle{[value]} = [value] \text{ degrees}\) 4. \(\displaystyle{[value]} \text{ rad } = \dfrac{180}{\pi} \times \displaystyle{[value]} = [value] \text{ degrees}\)

General definitions of trigonometric functions

We now define the trigonometric functions in a more general way than in terms of ratios of sides of a right-angled triangle. To do this we consider a circle of unit radius whose centre is at the origin of a Cartesian coordinate system and an arrow (or radius vector) \(OP\) from the centre to a point \(P\) on the circumference of this circle. We are interested in the angle \(\theta\) that the arrow makes with the positive \(x\)-axis. See the figure below

[Image: Clockwise angle from origin to point P on unit circle]

Imagine that the vector \(OP\) rotates in anti-clockwise direction. With this sense of rotation the angle \(\theta\) is taken as positive whereas a clockwise rotation is taken as negative. See examples below.

[Image: Angle theta of 90 degrees, so the vector points up]

\(\theta = 90^\circ = \dfrac{\pi}{2} \text{ rad}\), a quarter rotation anticlockwise

[Image: Angle theta of 315 degrees, so the vector P points southeast]

\(\theta = 315^\circ = \dfrac{7\pi}{4}\text{ rad}\), a seven eighths rotation anticlockwise

[Image: Angle theta of -45 degrees, so the vector P points southeast]

\(\theta = -45^\circ = -\dfrac{\pi}{4} \text{ rad}\), a one eighths rotation clockwise

The sine and cosine of an angle

For \(0\le\theta\le \dfrac{\pi}{2}\) (called the first quadrant) we have the following situation with our unit radius circle. In the figure below we let \(O\) be the origin point, \(P\) a point on the circumference, and \(Q\) and \(R\) the horizontal and vertical projections onto the axes of \(P\), respectively.

[Image: Vector rotated approximately 45 degrees or pi over 4 radians]

The projection of \(OP\) along the positive \(x\)-axis is \(OQ\). But, in the right-angled triangle \(OPQ\) we have:

\(\cos\theta=\frac{OQ}{OP}\qquad \text{or} \qquad OQ=OP\cos\theta\)

and since \(OP\) has unit length \(\cos\theta =OQ \qquad (1) \)

Similarly in this right-angled triangle:

\(\sin\theta=\frac{PQ}{OP}\qquad\text{or}\qquad PQ=OP\sin\theta\)

but \(PQ=OR\) and \(OP\) has unit length, so \(\sin\theta=OR\qquad \)

Equation (1) tells us that we can interpret \(\cos\theta\) as the projection of \(OP\) along the positive \(x\)-axis and \(\sin \theta\) as the projection of \(OP\) along the positive \(y\)-axis.

We shall use these interpretations as the definitions of \(\sin\theta\) and \(\cos\theta\) for any values of \(\theta\).

Key Point

For a radius vector \(OP\) of a circle of unit radius making an angle \(\theta\) with the positive \(x\)-axis
\(\cos\theta=\) projection of \(OP\) along the positive \(x\)-axis
\(\sin\theta=\) projection of \(OP\) along the positive \(y\)-axis

Exercise

For each of the functions, identify its graph.

\(A: y = \displaystyle[value]\)

\(B: y = \displaystyle[value] \)

\(C: y = \displaystyle[value] \)

\(D: y = \displaystyle[value] \)

[value] | [value] | [value] | [value]
Feedback: The general form of the sinusoidal function is \( y = A \sin(\omega t +\alpha), \omega>0. \) We are given the functions: \(A: y = [value]\) \(B: y = [value] \) \(C: y = [value] \) \(D: y =[value] \) We start with the function \( y = [value] \) which has amplitude \( A = [value] \), \( \omega = [value]\) and period \( T = \frac{2 \pi}{\omega} = [value] = [value]. \) The graph of \( y = [value]\) is obtained by moving the graph of \( y = [value] \) horizontally to the left by [value] units. The function \( y = [value] \) has \( A = [value] \), \( \omega = [value]\) and period \( T = \frac{2 \pi}{\omega} = [value] = [value]. \) Finally, the function \( y = [value] \) has \( A = [value] \), \( \omega = [value]\) and period \( T = \frac{2 \pi}{\omega} = [value] = [value]. \)

Sine and cosine in the four quadrants

First quadrant \((0^\circ\le\theta\le 90^\circ)\)

Below are some figures highlighting the behaviour of \(\cos \theta\) and \(\sin \theta\) as \(\theta\) moves in the first quadrant.

[Image: Angle theta of 0 degrees, so P on x axis]

\(\theta = 0^\circ\). Thus \(OQ = OP =1\) and so \(\cos 0^\circ =1\). Also \(OR=0\) hence \(\sin 0^\circ = 0\).

[Image: P now makes a nonzero angle in the first quadrant, projections Q and R to the x and y axes are shown.]

\(0^\circ < \theta < 90^\circ\). We have \(\cos \theta = OQ\) so \(0<\cos\theta<1\). Similarly \(\sin \theta = OR\), so \(0<\sin \theta<1\).

[Image: P makes angle 90 degree, so points up along the y axis]

\(\theta = 90^\circ\). Now \(OQ = 0\) so \(\cos90^\circ = 0\), and \(OR = OP = 1\), so \(\sin 90^\circ = 1\)

It follows from the figures above that \(\cos\theta\) decreases from \(1\) to \(0\) as \(OP\) rotates from the horizontal position to the vertical, i.e. as \(\theta\) increases from \(0^\circ\) to \(90^\circ\). \(\sin\theta=OR\) increases from \(0\) (when \(\theta=0^\circ\)) to \(1\) (when \(\theta=90^\circ\)).

Second quadrant \((90^\circ\le \theta\le 180^\circ)\)

Referring to the figures below, remember that it is the projections along the positive \(x\) and \(y\) axes that are used to define \(\cos\theta\) and \(\sin\theta\) respectively. It follows that as \(\theta\) increases from \(90^\circ\) to \(180^\circ, \ \cos\theta\) decreases from \(0\) to \(-1\) and \(\sin\theta\) decreases from \(1\) to \(0\).

[Image: 90 degree angle, so vector points up]

\(\theta = 90^\circ\), so \(\cos \theta = 0\) and \(\sin \theta = 1\).

[Image: Angle between 90 and 180 with projections R and Q to y and x axes.]

\(90^\circ < \theta < 180^\circ\). \(\cos \theta = OQ\) (negative), and \(\sin \theta = OR\) (positive).

[Image: angle 180 degrees, so P is at (-1,0)]

\(\theta = 180^\circ\), then \(\cos \theta = OQ = OP = -1\) and \(\sin \theta = OR = 0\).

Considering for example an angle of \(135^\circ\), referring to the figure below, by symmetry we have:
\(\sin 135^\circ=OR=\sin 45^\circ=\dfrac{1}{\sqrt{2}}\qquad \)
\( \cos 135^\circ=OQ_2=-OQ_1=-\cos 45^\circ=-\dfrac{1}{\sqrt{2}}\)

[Image: Vectors and its mirror in the y-axis]

Key Point

\( \sin (180^\circ-x)\equiv\sin x \qquad \text{and}\qquad \cos (180^\circ-x)\equiv-\cos x \)

Third quadrant \( (180^\circ\le \theta\le 270^\circ) \)

The figure below shows how \(\cos \theta\) and \(\sin \theta\) change as \(\theta\) moves in the third quadrant.

[Image: Angle of 180 degrees, so P is at (-1,0)]

\(\theta = 180^\circ\) so \(\cos \theta = -1\) and \(\sin \theta = 0\).

[Image: Vector in third quadrant with projections OQ and OP to x and y axes.]

\(180^\circ <\theta < 270^\circ\), then \(\cos \theta = OQ\) (negative) and \(\sin \theta = OR\) (negative)

[Image: Vector at 270 degrees, so P at (0,-1).]

\(\theta = 270^\circ\)

We let \(Q\) and \(R\) be the horizontal and vertical projections of \(P\), respectively. \(Q'\) and \(R'\) are their reflections in \(O\). From the projection definition, using \(\theta =180^\circ +x\) (see figure below) we obtain for all \(x\) the relations: \[\begin{alignedat}{4} \sin\theta &= \sin(180^\circ+x) &= OR &= -OR' &= -\sin x \\ \cos\theta &= \cos(180^\circ+x) &= OQ &= -OQ' &= -\cos x \end{alignedat}\]

Hence,

\[\tan(180^\circ+x) = \dfrac{\sin (180^\circ+x)}{\cos (180^\circ+x)}=\dfrac{-\sin x}{-\cos x}=+\tan x\]

for all \(x\).

[Image: vector in third quadrant and its reflection in the origin.]

Key Point

\[\begin{aligned} \sin(180^\circ+x) &\equiv -\sin x \\ \cos(180^\circ+x) &\equiv -\cos x \\ \tan (180^\circ+x) &\equiv +\tan x \end{aligned}\]

Fourth quadrant \( (270^\circ\le\theta\le 360^\circ) \)

The figures below illustrate how \(\cos \theta\) and \(\sin \theta\) change for \(\theta\) in the fourth quadrant.

[Image: Theta=270, so that P points down at (0,-1)]

\(\theta = 270^\circ\), we find that \(\cos 270^\circ = 0\) and \(\sin 270^\circ = -1\).

[Image: P in fourth quadrant with projections Q and R to x and y axes.]

\(270^\circ < \theta <360^\circ\)  (or alternatively \(-90^\circ <\theta < 0\). Then \(\cos \theta = OQ > 0\) and \(\sin\theta = OR < 0\)

[Image: Angle 360 degrees, we have gone full circle, the vector now points at (1,0).]

\(\theta = 360^\circ\). This is the same as \(\theta = 0\), so \(\cos 360^\circ = 1\) and \(\sin 360^\circ = 0\).

From the figures above the results in the Key Point should be clear.

Key Point

\(\cos(-x)\equiv\cos x \qquad\quad \sin(-x)\equiv-\sin x\qquad\quad\tan (-x)\equiv-\tan x\)

Rotation beyond the fourth quadrant \((360^\circ<\theta)\)

If the vector \(OP\) continues to rotate around the circle of unit radius then in the next complete rotation \(\theta\) increases from \(360^\circ\) to \(720^\circ\). However, a \(\theta\) value of, say, \(405^\circ\) is indistinguishable from one of \(45^\circ\) (just one extra complete revolution is involved).

So \(\sin(405^\circ)=\sin(45^\circ)=\dfrac{1}{\sqrt{2}} \text{ and } \cos(405^\circ)=\cos(45^\circ)=\dfrac{1}{\sqrt{2}}\)

In general \(\sin(360^\circ+x^\circ)=\sin(x^\circ), \cos(360^\circ+x^\circ)=\cos(x^\circ)\)

Key Point

If \(n\) is any integer, then

\[\begin{aligned} \sin (x^\circ+360 n^\circ) &\equiv \sin(x^\circ) \\ \cos (x^\circ+360 n^\circ) &\equiv \cos(x^\circ) \end{aligned}\]

or, since \(360^\circ\equiv2\pi \text{ radians}\), \[\begin{aligned} \sin (x + 2\pi n) &\equiv \sin(x) \\ \cos (x + 2\pi n) &\equiv \cos(x) \end{aligned}\]

We say that the functions \(\sin(x)\) and \(\cos(x)\) are periodic with period (in radian measure) of \(2\pi\).

Graphs of sin \( \pmb \theta \) and cos \(\pmb\theta\)

Since we have defined both \(\sin\theta\) and \(\cos\theta\) in terms of the projections of the radius vector \(OP\) of a circle of unit radius it follows immediately that

\(-1\le\sin\theta\le +1\qquad\text{and}\qquad -1\le \cos\theta\le +1\) for any value of \(\theta\).

We can draw the graphs of these two trigonometric functions. See the figure below. We have labelled the horizontal axis using radians and have shown two periods in each case. Notice the graph of \(\sin(\theta)\) repeats every \(2\pi\) radians, with maximum values at \(\frac{\pi}{2}+2k\pi\) and minimum values at \(\frac{3\pi}{2}+2k\pi\) for \(k \in \mathbb{N}\). Also notice the graph of \(\cos(\theta)\) is obtained by shifting the graph of \(\sin(\theta)\) by \(\frac{\pi}{2}\) radians to the left.

[Image: Graph of sine and cosine]

We have extended the graphs to negative values of \(\theta\) using the relations \(\sin(-\theta)=-\sin\theta,\ \ \cos(-\theta)=\cos\theta.\) Both graphs could be extended indefinitely to the left (\(\theta\to -\infty\)) and right (\(\theta\to +\infty)\).

The graph of \(\tan \theta\)

The graph of \(\tan\theta\) against \(\theta\), for \(-2\pi\le\theta\le 2\pi\) is then as in the figure below. Note that whereas \(\sin\theta\) and \(\cos\theta\) have period \(2\pi\), \(\tan\theta\) has period \(\pi\).

Notice the graph repeats every \(\pi\) radians, and has vertical asymptotes at \(\dfrac{\pi}{2}+k\pi\) and \(x\)-intercepts at \(k\pi\) for \(k\in \mathbb{N}\).

[Image: Graph of tan]

Optical interference fringes due to a glass plate

Monochromatic light of intensity \(I_0\) propagates in air before impinging on a glass plate (see figure below). If a screen is placed beyond the plate then a pattern is observed including alternate light and dark regions. These are interference fringes.

[Image: Light hitting a glass plate at angle alpha, some light is internally reflected in the glass, some goes out at angle alph]

The intensity \(I\) of the light wave transmitted through the plate is given by

\[I=\dfrac{I_0|t|^4}{1+|r|^4-2|r|^2\cos\theta},\]

where \(t\) and \(r\) are the complex transmission and reflection coefficients. The phase angle \(\theta\) is the sum of

1. a phase proportional to the incidence angle \(\alpha\), and
2. a fixed phase lag due to multiple reflections.

The problem is to establish the form of the intensity pattern (i.e. the minima and maxima characteristics of interference fringes due to the plate), and deduce the shape and position \(\theta\) of the fringes captured by a screen beyond the plate.

Solution

The intensity of the optical wave outgoing from the glass plate is given by

\[I=\dfrac{I_0|t|^4}{1+|r|^4-2|r|^2\cos\theta} \tag{1}\]

The light intensity depends solely on the variable \(\theta\) as shown in equation \((1)\), and the objective is to find the values \(\theta\) that will minimize and maximize \(I\). The angle \(\theta\) is introduced in equation \((1)\) through the function \(\cos\theta\) in the denominator. We consider first the maxima of \(I\).

Light intensity maxima

\(I\) is maximum when the denominator is minimum. This condition is obtained when the factor \(2|r|\cos\theta\) is maximum due to the minus sign in the denominator. The maxima of \(2|r|\cos\theta\) occur when \(\cos\theta = +1.\) Values of \(\cos\theta = +1\) correspond to \(\theta = 2n\pi\) where \(n =\ldots -2, -1, 0, 1, 2,\ldots\) and \(\theta\) is measured in radians. Setting \(\cos\theta = +1\) in equation \((1)\) gives the intensity maxima

\[I_{{\max}}=\dfrac{I_0|t|^4}{1+|r|^4-2|r|^2}.\]

Since the denominator can be identified as the square of \((1-|r|^2)\), the final result for maximum intensity can be written as

\[I_{\max}=\dfrac{I_0|t|^4}{(1-|r|^2)^2} \tag{2}\]

Light intensity minima

\(I\) is minimum when the denominator in \((1)\) is maximum. As a result of the minus sign in the denominator, this condition is obtained when the factor \(2|r|\cos\theta\) is minimum. The minima of \(2|r|\cos\theta\) occur when \(\cos \theta = -1.\) Values of \(\cos \theta = -1\) correspond to \(\theta = \pi (2n+1)\) where \(n =\ldots-2, -1, 0, 1, 2,\ldots\). Setting \(\cos\theta = -1\) in equation \((1)\) gives an expression for the intensity minima

\[I_{\min}=\dfrac{I_0|t|^4}{1+|r|^4+2|r|^2}.\]

Since the denominator can be recognised as the square of \((1 + |r|^2),\) the final result for minimum intensity can be written as

\[I_{\min}=\dfrac{I_0|t|^4}{(1+|r|^2)^2} \tag{3}\]

Interpretation

The interference fringes for intensity maxima or minima occur at constant angle \(\theta\) and therefore describe concentric rings of alternating light and shadow as sketched in the figure below. From the centre to the periphery of the concentric ring system, the fringes occur in the following order

1. a fringe of maximum light at the centre (bright dot for \(\theta = 0\)),
2. a circular fringe of minimum light at angle \(\theta=\pi\),
3. a circular fringe of maximum light at \(2\pi\) etc.

[Image: Circular interference pattern]

Exercise

Without using a calculator write down the following values

1. \(\sin 120^{\circ}= \)
2. \(\sin 150^{\circ}= \)
3. \(\cos 120^{\circ}= \)
4. \(\cos 150^{\circ}= \)
5. \(\tan 120^{\circ}= \)
6. \(\tan 150^{\circ}= \)
Feedback: 1. \( \sin 120^\circ=\sin (180-60)=\sin 60^\circ=\dfrac{\sqrt{3}}{2} \) 2. \( \sin 150^\circ=\sin (180-30)=\sin 30^\circ=\dfrac{1}{2} \) 3. \(\cos 120^\circ=-\cos 60=-\dfrac{1}{2} \) 4. \(\cos 150^\circ=-\cos 30^\circ=-\dfrac{\sqrt{3}}{2} \) 5. \( \tan120^\circ=\dfrac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3} \) 6. \( \tan150^\circ=\dfrac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}} \)

Exercise

Using the projection definition write down the values of \( \cos(270^\circ) \) and \( \sin(270^\circ)\).

1. \( \cos(270^\circ)= \)
2. \( \sin(270^\circ)=\)
Feedback: 1. \(OP\) has zero projection along the positive \(x-\)axis. 2. \(OP\) is directed along the negative axis.

Exercise

Withouth using a calculator write down the following values

1. \(\sin(300^{\circ})= \)
2. \(\sin(-60^{\circ})= \)
3. \(\cos(330^{\circ})= \)
4. \(\cos(-30^{\circ})= \)
Feedback: 1. \(\sin(300^\circ)=-\sin(60^\circ) =-{\sqrt{3}}/{2} \) 2. \( \sin (-60^\circ)=-\sin(60^\circ)=-{\sqrt{3}}/{2} \) 3. \(\cos(330^\circ)=\cos(30^\circ)={\sqrt{3}}/{2} \) 4. \(\cos(-30^\circ)=\cos(30^\circ)={\sqrt{3}}/{2} \)

Exercise

Using the fact that \(\tan\theta \equiv\sin\theta /\cos\theta\) calculate the values of \(\tan 0,\) \(\tan\pi,\) \(\tan 2\pi\).

1. \(\tan 0=\)
2. \(\tan \pi=\)
3. \(\tan 2\pi=\)
Feedback: 1. \(\tan 0=\dfrac{\sin 0}{\cos 0}=\dfrac{0}{1}=0\) 2. \( \tan\pi=\dfrac{\sin\pi}{\cos\pi}=\dfrac{0}{-1}=0 \) 3. \(\tan 2\pi=\dfrac{\sin 2\pi}{\cos 2\pi}=\dfrac{0}{1}=0\)

Q: Exercise

In which quadrants is \( \tan \theta \) positive?
- Quadrant 1: \(0 \leq \theta < \dfrac{\pi}{2} \) (\( \quad\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{+\text{ve}}{+\text{ve}} =+\text{ve} \))
- Quadrant 2: \(\dfrac{\pi}{2} < \theta \leq \pi \) (\( \tan \theta=\dfrac{\sin\theta}{\cos\theta}= \dfrac{+\text{ve}}{-\text{ve}}=-\text{ve} \))
- Quadrant 3: \( \pi \leq \theta < \dfrac{3\pi}{2} \) (\( \tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\text{ve}}{-\text{ve}} =+\text{ve}\))
- Quadrant 4: \( \dfrac{3\pi}{2} < \theta \leq 2\pi \) (\(\tan\theta=\dfrac{\sin\theta}{\cos\theta} =\dfrac{-\text{ve}}{+\text{ve}}=-\text{ve} \))
A: Quadrant 1: \(0 \leq \theta < \dfrac{\pi}{2} \); Quadrant 3: \( \pi \leq \theta < \dfrac{3\pi}{2} \)

Exercise

Write down all values of \(\theta\) for which \(\tan \theta\) is not defined.

where \(n\in\mathbb{Z}\).

Your answer should be an expression involving \(n\).
Feedback: \(\tan\theta\) is not defined when \(\cos\theta=0\), i.e. when \(\theta = \dfrac{\pi}{2} + n\pi\)

Q: Exercise

Which trigonometric functions are positive in each of the 4 quadrants?
1. First quadrant \(\left( 0 \leq \theta \leq \dfrac{\pi}{2} \right)\): [...]
2. Second quadrant \( \left(\dfrac{\pi}{2} \leq \theta \leq \pi \right)\): [...]
3. Third quadrant \(\left( \pi \leq \theta \leq \dfrac{3\pi}{2} \right)\) : [...]
4. Fourth quadrant \( \left(\dfrac{3\pi}{2} \leq \theta \leq 2\pi \right)\): [...]
A: all; sin; tan; cos

Exercise

Convert the following angles from degrees to radians

1. \([value]^{\circ}=\) \(\text{rad}\)
2. \([value]^{\circ}=\) \(\text{rad}\)
3. \([value]^{\circ}=\) \(\text{rad}\)
4. \([value]^{\circ}=\) \(\text{rad}\)
Feedback: Solution 1. \( [value] \text{ degrees} = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \) rad 2. \( [value] \text{ degrees } = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \) rad 3. \( [value] \text{ degrees} = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \) rad 4. \( [value] \text{ degrees} = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \) rad

Exercise

Convert the following angles in radians to degrees

1. \(\displaystyle{[value]} \text{ rad}=\) \(\text{degrees}\)
2. \(\displaystyle{[value]} \text{ rad}=\) \(\text{degrees}\)
3. \(\displaystyle{[value]} \text{ rad}=\) \(\text{degrees}\)
4. \(\displaystyle{[value]} \text{ rad}=\) \(\text{degrees}\)
Feedback: 1. \( \displaystyle{[value]} \text{ rad } = \frac{180}{\pi} \times [value] = [value] \) degrees 2. \( \displaystyle{[value]} \text{ rad } = \frac{180}{\pi} \times [value] = [value] \) degrees 3. \( \displaystyle{[value]} \text{ rad } = \frac{180}{\pi} \times [value] = [value] \) degrees 4. \( \displaystyle{[value]} \text{ rad } = \frac{180}{\pi} \times [value] = [value] \) degrees

Exercise

Find all solutions in the range \(-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}\) to the equation [value]. Give your answer as a set of values.

Solutions are .
Feedback: We are given the equation [value]. The equation factorises as: \[[value] = 0,\] so the solutions are those \(x\) such that \[\sin(x) = [value], [value].\] In the range \(-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}\) these are the values: \(x = \displaystyle {[value], [value]}\).

Exercise

Put your calculator into radian mode (using the DRG button if necessary) for this exercise: Compute the following sines and cosines by first converting the angle to radians

1. \(\sin([value]^{\circ})=\)
2. \(\sin([value]^{\circ})=\)
3. \(\cos([value]^{\circ})=\)
4. \(\cos([value]^{\circ})=\)
Feedback: Recall that the convert degrees to radians one must multiply by \(\pi/180\). 1. First, we find that \([value]^\circ\) is equal to \([value]\text{ rad}\). Using a calculator or a table, we get \[\sin([value]^\circ) = \sin\left([value]\right) = [value].\] 2. We find that \([value]^\circ\) is equal to \([value]\text{ rad}\). Then \[\sin([value]^\circ) = \sin\left([value]\right) = [value].\] 3. We find that \([value]^\circ\) is equal to \([value]\text{ rad}\). Then \[\sin([value]^\circ) = \sin\left([value]\right) = [value].\] 4. We find that \([value]^\circ\) is equal to \([value]\text{ rad}\). Then \[\sin([value]^\circ) = \sin\left([value]\right) = [value].\]

Exercise

Obtain the exact value of all 6 trigonometric functions (\(\sin\),\(\cos\),\(\tan\),\( \mathrm{cosec}\),\(\sec\),\(\mathrm{cotan}\)) of the angle \( \theta \) which is given by the counterclockwise angle from the x-axis to the vector \(P = ([value],[value])\) (see figure):

[Image: Vector P]

1. \(\sin \theta=\)
2. \(\cos\theta=\)
3. \(\tan \theta=\)
4. \(\csc \theta=\)
5. \(\sec \theta=\)
6. \(\cot \theta=\)
Feedback: The distance of the point \(P\) from the origin is \(r=\sqrt{([value])^2+[value]^2}=[value]\). Since \(P\) lies on a circle of radius \([value]\) rather than a circle of unit radius, we have: \[\begin{array}{lll} \sin\theta = \dfrac{y}{r} = \displaystyle{[value]} &&{\rm cosec}\;\theta=\dfrac{1}{\sin \theta} = \displaystyle{[value]}\\ \\ \cos\theta=\dfrac{x}{r} = \displaystyle{[value]}&&\sec\theta=\dfrac{1}{\cos \theta} = \displaystyle{[value]}\\ \\ \tan\theta=\dfrac{y}{x} = \displaystyle{[value]} &&\cot\theta=\dfrac{1}{\tan \theta} = \displaystyle{[value]}\end{array}\]

Exercise

Obtain all values of \(\theta \) between \(-\pi\) and \( \pi\) such that: (give your answers in radians as a set)

1. \(\sin(\theta) = \displaystyle{[value]}\) gives  \(\theta \in \) .
2. \(\sin(\theta) = \displaystyle{[value]}\) gives  \(\theta \in \) .
3. \(\cos(\theta) = \displaystyle{[value]}\) gives  \(\theta \in \) .
4. \(\cos(\theta) = \displaystyle{[value]}\) gives  \(\theta \in \) .
5. \(\tan(\theta) = \displaystyle{[value]}\) gives  \(\theta \in \) .
6. \(\tan(\theta) = \displaystyle{[value]}\) gives  \(\theta \in \) .
Feedback: 1. Recall that \(\sin(\pi - x) = \sin x \), so that \(\theta \in [value]\) 2. \(\theta \in [value]\) 3. Recall that \(\cos(-x) = \cos x\), so that \(\theta \in [value]\) 4. \(\theta \in [value]\) 5. Recall that \(\tan x\) has period \(\pi\) radians, so that \(\theta \in [value]\) 6. \(\theta \in [value]\)

Exercise

1. What is the area \(A\) of a sector formed by a central angle \(\theta \) radians in a circle of radius \( r \)?
\(A =\)
(Hint: By proportionality the ratio of the area of the sector to the total area of the circle equals the ratio of \( \theta \) to the total angle at the centre of the circle.)
The shaded area \(S\) is the difference between a sector of radius \(R\) and a sector of radius \(r\), both of angle \(\theta\). What is the value of the shaded area \(S\) if \(\theta\) is measured (b) in radians, (c) in degrees?
[Image: Angular sector of angle theta]
2. \(S =\)
3. \(S =\)
Feedback: 1. Using the hint, \(\dfrac{\theta}{2\pi}=\dfrac{A}{\pi r^2}\) and from here we obtain \(A=\dfrac{\pi r^2\theta}{2\pi}=\dfrac{r^2\theta}{2}\) 2. With \(\theta\) in radians the shaded area is \(S=\dfrac{R^2\theta}{2}-\dfrac{r^2\theta}{2}=\dfrac{\theta}{2} (R^2-r^2)\) 3. If \(\theta\) is in degrees, then the proportion of the circle is \(\dfrac{\theta}{360^\circ}\) and we multiply this by the area of the the larger circle minus the smaller circle to give \(S=\dfrac{\pi\theta^\circ}{360^\circ}(R^2-r^2)\)

Exercise

Convert the following angles from degrees to radians:

1. [value]\(^{\circ} =\) \(\text{rad}\)
2. [value]\(^{\circ} =\) \(\text{rad}\)
3. [value]\(^{\circ} =\) \(\text{rad}\)
4. [value]\(^{\circ} =\) \(\text{rad}\)
Feedback: 1. \( [value] \text{ degrees} = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \text{ rad} \) 2. \( [value] \text{ degrees } = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \text{ rad} \) 3. \( [value] \text{ degrees} = \dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \text{ rad} \) 4. \( [value] \text{ degrees} =\dfrac{\pi}{180} \times \displaystyle{[value]} = \displaystyle{[value]} \text{ rad} \)