### 2 Engineering Example 5

#### 2.1 Amplitude modulation

Introduction

Amplitude Modulation (the AM in AM radio) is a method of sending electromagnetic signals of a certain frequency (signal frequency) at another frequency (carrier frequency) which may be better for transmission. Modulation can be represented by the multiplication of the carrier and modulating signals. To demodulate the signal the carrier frequency must be removed from the modulated signal.

Problem in words

1. A single frequency of 200 Hz (message signal) is amplitude modulated with a carrier frequency of 2 MHz. Show that the modulated signal can be represented by the sum of two frequencies at $2×1{0}^{6}±200$ Hz
2. Show that the modulated signal can be demodulated by using a locally generated carrier and applying a low-pass filter.

Mathematical statement of problem

1. Express the message signal as $m=acos\left({\omega }_{m}t\right)$ and the carrier as $c=bcos\left({u}_{c}t\right)$ .

Assume that the modulation gives the product $mc=abcos\left({u}_{c}t\right)cos\left({\omega }_{m}t\right)$ .

Use trigonometric identities to show that

$mc=abcos\left({\omega }_{c}t\right)cos\left({u}_{m}t\right)={k}_{1}cos\left(\left({\omega }_{c}-{u}_{m}\right)t\right)+{k}_{2}cos\left(\left({\omega }_{c}+{u}_{m}\right)t\right)$

where ${k}_{1}$ and ${k}_{2}$ are constants.

Then substitute ${\omega }_{c}=2×1{0}^{6}×2\pi$ and ${\omega }_{m}=200×2\pi$ to calculate the two resulting frequencies.

2. Use trigonometric identities to show that multiplying the modulated signal by $bcos\left({u}_{c}t\right)$ results in the lowest frequency component of the output having a frequency equal to the original message signal.

Mathematical analysis

1. The message signal has a frequency of ${f}_{m}=200$ Hz so ${\omega }_{m}=2\pi {f}_{c}=2\pi ×200=400\pi$ radians per second.

The carrier signal has a frequency of ${f}_{c}=2×1{0}^{6}$ Hz.

Hence ${\omega }_{c}=2\pi {f}_{c}=2\pi ×2×1{0}^{6}=4×1{0}^{6}\pi$ radians per second.

So $mc=abcos\left(4×1{0}^{6}\pi t\right)cos\left(400\pi t\right)$ .

Key Point 13 includes the identity:

$\phantom{\rule{2em}{0ex}}cos\left(A+B\right)+cos\left(A-B\right)\equiv 2cos\left(A\right)cos\left(B\right)$

Rearranging gives the identity:

$\phantom{\rule{2em}{0ex}}cos\left(A\right)cos\left(B\right)\equiv \frac{1}{2}\left(cos\left(A+B\right)+cos\left(A-B\right)\right)\left(1\right)$

Using (1) with $A=4×1{0}^{6}\pi t$ and $B=400\pi t$ gives

$\begin{array}{rcll}mc& =& ab\left(cos\left(4×1{0}^{6}\pi t\right)cos\left(400\pi t\right)\right)& \text{}\\ & =& ab\left(cos\left(4×1{0}^{6}\pi t+400\pi t\right)+cos\left(4×1{0}^{6}\pi t-400\pi t\right)\right)& \text{}\\ & =& ab\left(cos\left(4000400\pi t\right)+cos\left(3999600\pi t\right)\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

So the modulated signal is the sum of two waves with angular frequency of 4000400 $\pi$ and 3999600 $\pi$ radians per second corresponding to frequencies of 4000400 $\pi ∕\left(2\pi \right)$ and 39996000 $\pi ∕\left(2\pi \right)$ , that is 2000200 Hz and 1999800 Hz i.e. $2×1{0}^{6}±200$ Hz.

2. Taking identity (1) and multiplying through by $cos\left(A\right)$ gives

$\phantom{\rule{2em}{0ex}}cos\left(A\right)cos\left(A\right)cos\left(B\right)\equiv \frac{1}{2}cos\left(A\right)\left(cos\left(A+B\right)+cos\left(A-B\right)\right)$

so

$\phantom{\rule{2em}{0ex}}cos\left(A\right)cos\left(A\right)cos\left(B\right)\equiv \frac{1}{2}\left(cos\left(A\right)cos\left(A+B\right)+cos\left(A\right)cos\left(A-B\right)\right)\left(2\right)$

Identity (1) can be applied to both expressions in the right-hand side of (2). In the first expression, using $A+B$ instead of ‘ $B$ ’, gives

$cos\left(A\right)cos\left(A+B\right)\equiv \frac{1}{2}\left(cos\left(A+A+B\right)+cos\left(A-A-B\right)\right)\equiv \frac{1}{2}\left(cos\left(2A+B\right)+cos\left(B\right)\right)$
where we have used $cos\left(-B\right)\equiv cos\left(B\right).$

Similarly, in the second expression, using $A-B$ instead of ‘ $B$ ’, gives

$\phantom{\rule{2em}{0ex}}cos\left(A\right)cos\left(A-B\right)\equiv \frac{1}{2}\left(cos\left(2A-B\right)+cos\left(B\right)\right)$

Together these give:

$\begin{array}{rcll}cos\left(A\right)cos\left(A\right)cos\left(B\right)& \equiv & \frac{1}{2}\left(cos\left(2A+B\right)+cos\left(B\right)+cos\left(2A-B\right)+cos\left(B\right)\right)& \text{}\\ & & & \text{}\\ & \equiv & cos\left(B\right)+\frac{1}{2}\left(cos\left(2A+B\right)+cos\left(2A-B\right)\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

With $A=4×1{0}^{6}\pi t$ and $B=400\pi t$ and substituting for the given frequencies, the modulated signal multiplied by the original carrier signal gives

$\phantom{\rule{2em}{0ex}}a{b}^{2}cos\left(4×1{0}^{6}\pi t\right)cos\left(4×1{0}^{6}\pi t\right)cos\left(400\pi t\right)=$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}a{b}^{2}cos\left(2\pi ×200t\right)+\frac{1}{2}a{b}^{2}\left(cos\left(2×4×1{0}^{6}\pi t+400\pi t\right)+cos\left(2×4×1{0}^{6}\pi t-400\pi t\right)\right)$

The last two terms have frequencies of $4×1{0}^{6}±200$ Hz which are sufficiently high that a low-pass filter would remove them and leave only the term

$a{b}^{2}cos\left(2\pi ×200t\right)$

which is the original message signal multiplied by a constant term.

Interpretation

Amplitude modulation of a single frequency message signal $\left({f}_{m}\right)$ with a single frequency carrier signal $\left({f}_{c}\right)$ can be shown to be equal to the sum of two cosines with frequencies ${f}_{c}±{f}_{m}$ . Multiplying the modulated signal by a locally generated carrier signal and applying a low-pass filter can reproduce the frequency, ${f}_{m}$ , of the message signal.

This is known as double side band amplitude modulation .

##### Example 2

Obtain expressions for $cos\theta$ in terms of the sine function and for $sin\theta$ in terms of the cosine function.

##### Solution

Using (9) with $A=\theta ,\phantom{\rule{1em}{0ex}}B=\frac{\pi }{2}$ we obtain

$\phantom{\rule{2em}{0ex}}cos\left(\theta -\frac{\pi }{2}\right)\equiv cos\theta \phantom{\rule{1em}{0ex}}cos\left(\frac{\pi }{2}\right)+sin\theta \phantom{\rule{1em}{0ex}}sin\left(\frac{\pi }{2}\right)\equiv cos\theta \phantom{\rule{1em}{0ex}}\left(0\right)+sin\theta \phantom{\rule{1em}{0ex}}\left(1\right)$

i.e. $sin\theta \equiv cos\left(\theta -\frac{\pi }{2}\right)\equiv cos\left(\frac{\pi }{2}-\theta \right)$

This result explains why the graph of $sin\theta$ has exactly the same shape as the graph of $cos\theta$ but it is shifted to the right by $\frac{\pi }{2}$ . (See Figure 29 on page 28). A similar calculation using (6) yields the result

$cos\theta \equiv sin\left(\theta +\frac{\pi }{2}\right).$

#### 2.2 Double angle formulae

If we put $B=A$ in the identity given in (6) we obtain Key Point 15:

##### Key Point 15

$sin2A\equiv sinAcosA+cosAsinA$ so $sin2A\equiv 2sinAcosA$ (12)

Substitute $B=A$ in identity (7) in Key Point 13 on page 38 to obtain an identity for $cos2A$ . Using ${sin}^{2}A+{cos}^{2}A\equiv 1$ obtain two alternative forms of the identity.

Using (7) with $B\equiv A$

$\begin{array}{rcll}cos\left(2A\right)& \equiv & \left(cosA\right)\left(cosA\right)-\left(sinA\right)\left(sinA\right)& \text{}\\ \therefore \phantom{\rule{2em}{0ex}}cos\left(2A\right)& \equiv & {cos}^{2}A-{sin}^{2}A\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

(13)

Substituting for ${sin}^{2}A$ in (13) we obtain

$\begin{array}{rcll}cos2A& \equiv & {cos}^{2}A-\left(1-{cos}^{2}A\right)& \text{}\\ & \equiv & 2{cos}^{2}A-1\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

(14)

Alternatively substituting for ${cos}^{2}A$ in (13)

$\begin{array}{rcll}cos2A& \equiv & \left(1-{sin}^{2}A\right)-{sin}^{2}A& \text{}\\ cos2A& \equiv & 1-2{sin}^{2}A\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

(15)

Use (14) and (15) to obtain, respectively, ${cos}^{2}A$ and ${sin}^{2}A$ in terms of $cos2A$ .

From (14) ${cos}^{2}A\equiv \frac{1}{2}\left(1+cos2A\right)$ . From (15) ${sin}^{2}A\equiv \frac{1}{2}\left(1-cos2A\right)$ .

Use (12) and (13) to obtain an identity for $tan2A$ in terms of $tanA$ .

$\phantom{\rule{2em}{0ex}}tan2A\equiv \frac{sin2A}{cos2A}\equiv \frac{2sinAcosA}{{cos}^{2}A-{sin}^{2}A}$

Dividing numerator and denominator by ${cos}^{2}A$ we obtain

$\phantom{\rule{2em}{0ex}}tan2A\equiv \frac{2\frac{sinA}{cosA}}{1-\frac{{sin}^{2}A}{{cos}^{2}A}}\equiv \frac{2tanA}{1-{tan}^{2}A}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(16\right)$

#### 2.3 Half-angle formulae

If we replace $A$ by $\frac{A}{2}$ and, consequently $2A$ by $A$ , in (12) we obtain

$\phantom{\rule{2em}{0ex}}sinA\equiv 2sin\left(\frac{A}{2}\right)cos\left(\frac{A}{2}\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(17\right)$

Similarly from (13)

$\phantom{\rule{2em}{0ex}}cosA\equiv 2{cos}^{2}\left(\frac{A}{2}\right)-1.\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(18\right)$

These are examples of half-angle formulae . We can obtain a half-angle formula for $tanA$ using (16). Replacing $A$ by $\frac{A}{2}$ and $2A$ by $A$ in (16) we obtain

$\phantom{\rule{2em}{0ex}}tanA\equiv \frac{2tan\left(\frac{A}{2}\right)}{1-{tan}^{2}\left(\frac{A}{2}\right)}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(19\right)$

Other formulae, useful for integration when trigonometric functions are present, can be obtained using (17), (18) and (19) shown in the Key Point 16.

##### Key Point 16

If $t=tan\left(\frac{A}{2}\right)$ then

$sinA=\frac{2t}{1+{t}^{2}}\left(20\right)$
$cosA=\frac{1-{t}^{2}}{1+{t}^{2}}\left(21\right)$
$tanA=\frac{2t}{1-{t}^{2}}\left(22\right)$

#### 2.4 Sum of two sines and sum of two cosines

Finally, in this Section, we obtain results that are widely used in areas of science and engineering such as vibration theory, wave theory and electric circuit theory.

$\begin{array}{rcll}sin\left(A+B\right)& \equiv & sinAcosB+cosAsinB& \text{}\\ sin\left(A-B\right)& \equiv & sinAcosB-cosAsinB\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

$\phantom{\rule{2em}{0ex}}sin\left(A+B\right)+sin\left(A-B\right)\equiv 2sinAcosB\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(23\right)$

Subtracting the identities produces

$\phantom{\rule{2em}{0ex}}sin\left(A+B\right)-sin\left(A-B\right)\equiv 2cosAsinB\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(24\right)$

It is now convenient to let $C=A+B$ and $D=A-B$  so that

$\phantom{\rule{2em}{0ex}}A=\frac{C+D}{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\frac{C-D}{2}$

Hence (23) becomes

$\phantom{\rule{2em}{0ex}}sinC+sinD\equiv 2sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(25\right)$

Similarly (24) becomes

$\phantom{\rule{2em}{0ex}}sinC-sinD\equiv 2cos\left(\frac{C+D}{2}\right)sin\left(\frac{C-D}{2}\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(26\right)$

Use (7) and (10) to obtain results for the sum and difference of two cosines.

$\begin{array}{rcll}cos\left(A+B\right)\equiv cosAcosB& -& sinAsinB\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}cos\left(A-B\right)\equiv cosAcosB+sinAsinB& \text{}\\ \therefore \phantom{\rule{2em}{0ex}}cos\left(A+B\right)+cos\left(A-B\right)& \equiv & 2cosAcosB& \text{}\\ cos\left(A+B\right)-cos\left(A-B\right)& \equiv & -2sinAsinB& \text{}\end{array}$

Hence with $C=A+B$  and  $D=A-B$

$cosC+cosD\equiv 2cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(27\right)$

$\phantom{\rule{2em}{0ex}}cosC-cosD\equiv -2sin\left(\frac{C+D}{2}\right)sin\left(\frac{C-D}{2}\right)\left(28\right)$

#### 2.5 Summary

In this Section we have covered a large number of trigonometric identities. The most important of them and probably the ones most worth memorising are given in the following Key Point.

##### Key Point 17

$\begin{array}{rcll}{cos}^{2}\theta +{sin}^{2}\theta & \equiv & 1& \text{}\\ sin2\theta & \equiv & 2sin\theta cos\theta & \text{}\\ cos2\theta & \equiv & {cos}^{2}\theta -{sin}^{2}\theta & \text{}\\ & \equiv & 2{cos}^{2}\theta -1& \text{}\\ & \equiv & 1-2{sin}^{2}\theta & \text{}\\ sin\left(A±B\right)& \equiv & sinAcosB±cosAsinB& \text{}\\ cos\left(A±B\right)& \equiv & cosAcosB∓sinAsinB& \text{}\end{array}$

A projectile is fired from the ground with an initial speed $u\text{m}{\text{s}}^{-1}$ at an angle of elevation ${\alpha }^{∘}$ . If air resistance is neglected, the vertical height, $y$ m, is related to the horizontal distance, $x$ m, by the equation

$\phantom{\rule{2em}{0ex}}y=xtan\alpha -\frac{g{x}^{2}{sec}^{2}\alpha }{2{u}^{2}}$ where $g\text{m}{\text{s}}^{-2}$ is the gravitational constant.

[This equation is derived in HELM booklet  34 Modelling Motion pages 16-17.]

1. Confirm that $y=0$ when $x=0$ :

When $y=0$ , the left-hand side of the equation is zero. Since $x$ appears in both of the terms on the right-hand side, when $x=0$ , the right-hand side is zero.

2. Find an expression for the value of $x$ other than $x=0$ at which $y=0$ and state how this value is related to the maximum range of the projectile:

When $y=0$ , the equation can be written $\phantom{\rule{1em}{0ex}}\frac{g{x}^{2}{sec}^{2}\alpha }{2{u}^{2}}-xtan\alpha =0$

If $x=0$ is excluded from consideration, we can divide through by $x$ and rearrange to give

$\phantom{\rule{2em}{0ex}}\frac{gx{sec}^{2}\alpha }{2{u}^{2}}=tan\alpha$

To make $x$ the subject of the equation we need to multiply both sides by $\frac{2{u}^{2}}{g{sec}^{2}\alpha }$ .

Given that $1∕{sec}^{2}\alpha \equiv {cos}^{2}\alpha ,\phantom{\rule{1em}{0ex}}tan\alpha \equiv sin\alpha ∕cos\alpha$ and $sin2\alpha \equiv 2sin\alpha cos\alpha$ , this results in

$\phantom{\rule{2em}{0ex}}x=\frac{2{u}^{2}sin\alpha cos\alpha }{g}=\frac{{u}^{2}sin2\alpha }{g}$

This represents the maximum range.

3. Find the value of $x$ for which the value of $y$ would be a maximum and thereby obtain an expression for the maximum height:

If air resistance is neglected, we can assume that the parabolic path of the projectile is symmetrical about its highest point. So the highest point will occur at half the maximum range i.e. where

$\phantom{\rule{2em}{0ex}}x=\frac{{u}^{2}sin2\alpha }{2g}$

Substituting this expression for $x$ in the equation for $y$ gives

$\phantom{\rule{2em}{0ex}}y=\left(\frac{{u}^{2}sin2\alpha }{2g}\right)tan\alpha -{\left(\frac{{u}^{2}sin2\alpha }{2g}\right)}^{2}\frac{g{sec}^{2}\alpha }{2{u}^{2}}$

Using the same trigonometric identities as before,

$\phantom{\rule{2em}{0ex}}y=\frac{{u}^{2}{sin}^{2}\alpha }{g}-\frac{{u}^{2}{sin}^{2}\alpha }{2g}=\frac{{u}^{2}{sin}^{2}\alpha }{2g}$ This represents the maximum height.

4. Assuming $u=20\text{m}{\text{s}}^{-1}$ , $\alpha =6{0}^{∘}$ and $g=10\text{m}{\text{s}}^{-2}$ , find the maximum value of the range and the horizontal distances travelled when the height is 10 m:

Substitution of $u=20,\phantom{\rule{1em}{0ex}}\alpha =60,\phantom{\rule{1em}{0ex}}g=10$ and $y=10$ in the original equation gives a quadratic for $x$ :

$\phantom{\rule{2em}{0ex}}10=1.732x-0.05{x}^{2}$ or $0.05{x}^{2}-1.732x+10=0$

Solution of this quadratic yields $x=7.33$ or $x=27.32$ as the two horizontal ranges at which $y=10$ . These values are illustrated in the diagram below which shows the complete trajectory of the projectile.

##### Exercises
1. Show that $sintsect\equiv tant$ .
2. Show that $\left(1+sint\right)\left(1+sin\left(-t\right)\right)\equiv {cos}^{2}t.$
3. Show that $\frac{1}{tan\theta +cot\theta }\equiv \frac{1}{2}sin2\theta$ .
4. Show that ${sin}^{2}\left(A+B\right)-{sin}^{2}\left(A-B\right)\equiv sin2Asin2B$ .

(Hint: the left-hand side is the difference of two squared quantities.)

5. Show that $\frac{sin4\theta +sin2\theta }{cos4\theta +cos2\theta }\equiv tan3\theta .$
6. Show that ${cos}^{4}A-{sin}^{4}A\equiv cos2A$
7. Express each of the following as the sum (or difference) of 2 sines (or cosines)
1. $sin5xcos2x$
2. $8cos6xcos4x$
3. $\frac{1}{3}sin\frac{1}{2}xcos\frac{3}{2}x$
8. Express
1. $sin3\theta$ in terms of $cos\theta$ .
2. $cos3\theta$ in terms of $cos\theta$ .
9. By writing $cos4x$ as $cos2\left(2x\right)$ , or otherwise, express $cos4x$ in terms of $cosx$ .
10. Show that $tan2t\equiv \frac{2tant}{2-{sec}^{2}t}$ .
11. Show that $\frac{cos10t-cos12t}{sin10t+sin12t}\equiv tant$ .
12. Show that the area of an isosceles triangle with equal sides of length $x$ is $\frac{{x}^{2}}{2}sin\theta$

where $\theta$ is the angle between the two equal sides. Hint: use the following diagram:

1. $sint.sect\equiv sint.\frac{1}{cost}\equiv \frac{sint}{cost}\equiv tant$ .
2. $\left(1+sint\right)\left(1+sin\left(-t\right)\right)\equiv \left(1+sint\right)\left(1-sint\right)\equiv 1-{sin}^{2}t\equiv {cos}^{2}t$
3. $\frac{1}{tan\theta +cos\theta }\equiv \frac{1}{\frac{sin\theta }{cos\theta }+\frac{cos\theta }{sin\theta }}\equiv \frac{1}{\frac{{sin}^{2}\theta +{cos}^{2}\theta }{sin\theta cos\theta }}\equiv \frac{sin\theta cos\theta }{{sin}^{2}\theta +{cos}^{2}\theta }\equiv sin\theta cos\theta \equiv \frac{1}{2}sin2\theta$
4. Using the hint and the identity ${x}^{2}-{y}^{2}\equiv \left(x-y\right)\left(x+y\right)$ we have

$\phantom{\rule{2em}{0ex}}{sin}^{2}\left(A+B\right)-{sin}^{2}\left(A-B\right)\equiv \left(sin\left(A+B\right)-sin\left(A-B\right)\right)\left(sin\left(A+B\right)+sin\left(A-B\right)\right)$

The first bracket gives

$\phantom{\rule{2em}{0ex}}sinAcosB+cosAsinB-\left(sinAcosB-cosAsinB\right)\equiv 2cosAsinB$

Similarly the second bracket gives $2sinAcosB.$

Multiplying we obtain   $\left(2cosAsinA\right)\left(2cosBsinB\right)\equiv sin2A.sin2B$

5. $\frac{sin4\theta +sin2\theta }{cos4\theta +cos2\theta }\equiv \frac{2sin3\theta cos\theta }{2cos3\theta cos\theta }\equiv \frac{sin3\theta }{cos3\theta }\equiv tan3\theta$
6. $\begin{array}{rcll}{cos}^{4}A-{sin}^{4}A\equiv {\left(cosA\right)}^{4}-{\left(sinA\right)}^{4}& \equiv & {\left({cos}^{2}A\right)}^{2}-{\left({sin}^{2}A\right)}^{2}& \text{}\\ & \equiv & \left({cos}^{2}A-{sin}^{2}A\right)\left({cos}^{2}A+{sin}^{2}A\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & \equiv & {cos}^{2}A-{sin}^{2}A\equiv cos2A& \text{}\end{array}$
7. (a) Using $sinA+sinB\equiv 2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)$

Clearly here $\frac{A+B}{2}=5x$ $\frac{A-B}{2}=2x$ giving $A=7x$ $B=3x$

$\therefore \phantom{\rule{2em}{0ex}}sin5xcos2x\equiv \frac{1}{2}\left(sin7x+sin3x\right)$

(b) Using $cosA+cosB\equiv 2cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)$ .

With $\phantom{\rule{1em}{0ex}}\frac{A+B}{2}=6x$ $\frac{A-B}{2}=4x$ giving $A=10x\phantom{\rule{1em}{0ex}}B=2x$

$\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{1em}{0ex}}8cos6xcos4x\equiv 4\left(cos6x+cos2x\right)$

(c) $\frac{1}{3}sin\left(\frac{1}{2}x\right)cos\left(\frac{3x}{2}\right)\equiv \frac{1}{6}\left(sin2x-sinx\right)$

8. $\begin{array}{rcll}\text{(a)}\phantom{\rule{2em}{0ex}}sin3\theta & \equiv & sin\left(2\theta +\theta \right)=sin2\theta cos\theta +cos2\theta sin\theta & \text{}\\ & \equiv & 2sin\theta {cos}^{2}\theta +\left({cos}^{2}\theta -{sin}^{2}\theta \right)sin\theta & \text{}\\ & \equiv & 3sin\theta {cos}^{2}\theta -{sin}^{3}\theta & \text{}\\ & \equiv & 3sin\theta \left(1-{sin}^{2}\theta \right)-{sin}^{3}\theta \equiv 3sin\theta -4{sin}^{3}\theta \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}& \text{}\end{array}$

$\begin{array}{rcll}\text{(b)}\phantom{\rule{2em}{0ex}}cos3\theta & \equiv & cos\left(2\theta +\theta \right)\equiv cos2\theta cos\theta -sin2\theta sin\theta & \text{}\\ & \equiv & \left({cos}^{2}\theta -{sin}^{2}\theta \right)cos\theta -2sin\theta cos\theta sin\theta & \text{}\\ & \equiv & {cos}^{3}\theta -3{sin}^{2}\theta cos\theta & \text{}\\ & \equiv & {cos}^{3}\theta -3\left(1-{cos}^{2}\theta \right)cos\theta & \text{}\\ & \equiv & 4{cos}^{3}\theta -3cos\theta \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
9.    $\begin{array}{rcll}cos4x=cos2\left(2x\right)& \equiv & 2{cos}^{2}\left(2x\right)-1& \text{}\\ & \equiv & 2{\left(cos2x\right)}^{2}-1& \text{}\\ & \equiv & 2{\left(2{cos}^{2}x-1\right)}^{2}-1& \text{}\\ & \equiv & 2\left(4{cos}^{4}x-4{cos}^{2}x+1\right)-1\equiv 8{cos}^{4}x-8{cos}^{2}x+1.\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
10.    $tan2t\equiv \frac{2tant}{1-{tan}^{2}t}\equiv \frac{2tant}{1-\left({sec}^{2}t-1\right)}\equiv \frac{2tant}{2-{sec}^{2}t}$
11.    $cos10t-cos12t\equiv 2sin11tsint\phantom{\rule{2em}{0ex}}sin10t+sin12t\equiv 2sin11tcos\left(-t\right)$

$\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}\frac{cos10t-cos12t}{sin10t+sin12t}\equiv \frac{sint}{cos\left(-t\right)}\equiv \frac{sint}{cost}\equiv tant$

12.   The right-angled triangle $ACD$ has area    $\frac{1}{2}\left(CD\right)\left(AD\right)$ $\begin{array}{rcll}\text{But}\phantom{\rule{2em}{0ex}}sin\left(\frac{\theta }{2}\right)=\frac{CD}{x}\phantom{\rule{2em}{0ex}}& \therefore & \phantom{\rule{1em}{0ex}}CD=xsin\left(\frac{\theta }{2}\right)& \text{}\\ & & & \text{}\\ cos\left(\frac{\theta }{2}\right)=\frac{AD}{x}\phantom{\rule{2em}{0ex}}& \therefore & \phantom{\rule{1em}{0ex}}AD=xcos\left(\frac{\theta }{2}\right)& \text{}\end{array}$