Trigonometric proof and periodic modelling

Trigonometry has two complementary uses. In a proof, known identities are combined to show that a statement is true for every value for which it is defined. In a model, a sine or cosine function approximates a repeating real phenomenon and its parameters acquire practical meanings.

The central periodic model is

$$\boxed{y=M+A\cos\bigl(\omega(t-t_0)\bigr)},$$

or an equivalent sine form. Here $M$ is the midline, $|A|$ is the amplitude, $\omega$ is the angular frequency, $t_0$ is a phase shift, and the period is

$$\boxed{T=\frac{2\pi}{|\omega|}}.$$

Prerequisites

You should be able to:

• use exact trigonometric values;
• use the Pythagorean and reciprocal identities;
• expand compound and double angles;
• interpret trigonometric graphs and transformations;
• solve trigonometric equations;
• work in radians.

Proving trigonometric identities

An identity such as

$$\frac{1-\cos^2x}{\sin x}\equiv\sin x$$

is true throughout the common domain of its two sides. The left side is undefined when $\sin x=0$, so the displayed identity applies only where $\sin x\ne0$. This is different from an equation, which asks for particular values of $x$.

A reliable proof strategy

1. Start with the more complicated side.
2. Replace expressions using known identities.
3. Factorise, combine fractions or change everything to sine and cosine.
4. Work towards the other side without assuming it.
5. Record any restrictions caused by denominators.

You may manipulate both sides independently until each reaches the same expression. Do not begin with the claimed identity and use $\Leftrightarrow$ unless every step is known to be reversible. That approach can conceal circular reasoning or lost domain restrictions.

Worked example 1: choose the useful identity

Prove that

$$\frac{\sec^2x-1}{\tan x}\equiv\tan x$$

where both sides are defined.

The numerator suggests the identity $\sec^2x-1\equiv\tan^2x$. Therefore

$$\begin{aligned} \frac{\sec^2x-1}{\tan x} &\equiv\frac{\tan^2x}{\tan x}\\ &\equiv\tan x. \end{aligned}$$

Cancellation requires $\tan x\ne0$, which was already required by the original denominator. The original expression also requires $\cos x\ne0$.

Worked example 2: convert to sine and cosine

Prove that

$$\frac{\cos x}{1-\sin x}\equiv\sec x+\tan x$$

where defined.

The right side written over a common denominator is

$$\sec x+\tan x \equiv\frac1{\cos x}+\frac{\sin x}{\cos x} \equiv\frac{1+\sin x}{\cos x}.$$

Transform the left side by multiplying by a carefully chosen form of $1$:

$$\begin{aligned} \frac{\cos x}{1-\sin x} &\equiv\frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)}\\ &\equiv\frac{\cos x(1+\sin x)}{1-\sin^2x}\\ &\equiv\frac{\cos x(1+\sin x)}{\cos^2x}\\ &\equiv\frac{1+\sin x}{\cos x}\\ &\equiv\sec x+\tan x. \end{aligned}$$

This proof applies on the common domain. In fact, $1-\sin x\ne0$ and $\cos x\ne0$ together reduce to $\cos x\ne0$ for the identity as written.

Multiplying by a conjugate

The factor $1+\sin x$ is not an arbitrary trick. It creates

$$(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x,$$

which can cancel with the numerator. Look for $1\pm\sin x$ or $1\pm\cos x$ when the Pythagorean identity could help.

Worked example 3: compound-angle proof

Prove that

$$\sin(x+y)\sin(x-y)\equiv\sin^2x-\sin^2y.$$

Expand the left side:

$$\begin{aligned} \sin(x+y)\sin(x-y) &\equiv(\sin x\cos y+\cos x\sin y) (\sin x\cos y-\cos x\sin y)\\ &\equiv\sin^2x\cos^2y-\cos^2x\sin^2y. \end{aligned}$$

Now replace $\cos^2y$ by $1-\sin^2y$ and $\cos^2x$ by $1-\sin^2x$:

$$\begin{aligned} \sin^2x(1-\sin^2y)-(1-\sin^2x)\sin^2y &\equiv\sin^2x-\sin^2y. \end{aligned}$$

Hence the identity is proved.

Worked example 4: proof before solving

Show that

$$\frac{\sin2x}{1+\cos2x}\equiv\tan x$$

where the left side is defined. Hence solve

$$\frac{\sin2x}{1+\cos2x}=\sqrt3, \qquad 0<x<2\pi.$$

Using double-angle formulae,

$$\begin{aligned} \frac{\sin2x}{1+\cos2x} &\equiv\frac{2\sin x\cos x}{1+(2\cos^2x-1)}\\ &\equiv\frac{2\sin x\cos x}{2\cos^2x}\\ &\equiv\tan x. \end{aligned}$$

The original denominator is zero when $\cos x=0$, so those values are excluded. The equation becomes

$$\tan x=\sqrt3.$$

The reference angle is $\pi/3$, and tangent has period $\pi$. Thus

$$x=\frac\pi3+k\pi.$$

Within $0<x<2\pi$,

$$\boxed{x=\frac\pi3,\ \frac{4\pi}3.}$$

Neither value makes the original denominator zero.

Self-check 1

Prove that

$$(\sec x-\tan x)(\sec x+\tan x)\equiv1$$

where defined.

Answer

Use the difference of two squares:

$$\begin{aligned} (\sec x-\tan x)(\sec x+\tan x) &\equiv\sec^2x-\tan^2x\\ &\equiv1. \end{aligned}$$

The expressions require $\cos x\ne0$.

Reading the parameters of a periodic model

For

$$y=M+A\cos\bigl(\omega(t-t_0)\bigr),$$

the parameters control different features.

Parameter	Meaning	How to identify it
$M$	midline or mean level	$M=(y_{\max}+y_{\min})/2$
$	A	$
$\omega$	angular frequency	$
$T$	period	time for one complete cycle
$t_0$	phase shift	a time at which the cosine argument is $0$

If $A>0$, $t=t_0$ gives a maximum. If $A<0$, it gives a minimum. Since $-1\leq\cos\theta\leq1$, the unrestricted range is

$$M-|A|\leq y\leq M+|A|.$$

Always include units. If $t$ is measured in hours, $T$ and $t_0$ are in hours, while $\omega$ is in radians per hour.

Degrees or radians

In a model with period $T$, the coefficient is $2\pi/T$ when the angle is in radians, but $360/T$ when it is in degrees. A-level calculus and most mathematical models use radians unless stated otherwise.

Building a model from maximum, minimum and period

Worked example 5: daylight hours

At a location, daylight is modelled as varying sinusoidally from a minimum of $8$ hours to a maximum of $16$ hours. The period is $365$ days, and a maximum occurs at $t=172$, where $t$ is the day number after 1 January. Construct a cosine model.

First find the midline and amplitude:

$$M=\frac{16+8}{2}=12, \qquad A=\frac{16-8}{2}=4.$$

The angular frequency is

$$\omega=\frac{2\pi}{365}.$$

Cosine is convenient because its value is $1$ when its argument is $0$. Shifting the maximum to $t=172$ gives

$$\boxed{D(t)=12+4\cos\left(\frac{2\pi}{365}(t-172)\right).}$$

Checks:

• $D(172)=16$, the stated maximum;
• half a period later, $D(354.5)=8$;
• $D(t+365)=D(t)$.

The half-day value $354.5$ is not an error. The continuous model is a smooth approximation, even though day numbers are usually recorded as integers.

Worked example 6: starting at a midline

The depth of water in a harbour varies between $2.4$ m and $6.8$ m with period $12.4$ hours. At $t=0$, the depth is at its mean level and rising. Construct a sine model.

The midline and amplitude are

$$M=\frac{6.8+2.4}{2}=4.6, \qquad A=\frac{6.8-2.4}{2}=2.2.$$

Also

$$\omega=\frac{2\pi}{12.4}=\frac{5\pi}{31}.$$

Since $\sin0=0$ and sine initially increases, no phase shift is needed:

$$\boxed{d(t)=4.6+2.2\sin\left(\frac{5\pi t}{31}\right).}$$

To predict the first high tide, set the sine equal to $1$:

$$\frac{5\pi t}{31}=\frac\pi2.$$

Therefore

$$\boxed{t=3.1\text{ hours}.}$$

The quarter-period answer $12.4/4=3.1$ provides a quick independent check.

Self-check 2

A temperature varies between $11^\circ\text{C}$ and $23^\circ\text{C}$ with period $24$ hours. Its maximum occurs at 15:00. Let $t$ be hours after midnight. Write a cosine model.

Answer

$$M=\frac{23+11}{2}=17, \qquad A=\frac{23-11}{2}=6, \qquad \omega=\frac{2\pi}{24}=\frac\pi{12}.$$

Therefore one suitable model is

$$\boxed{T(t)=17+6\cos\left(\frac\pi{12}(t-15)\right).}$$

Determining a phase shift from an observation

One observed value may produce two possible phases because sine and cosine are not one-to-one. The direction of change or another observation is then needed to select the correct model.

Worked example 7: resolve phase ambiguity

A wheel has radius $5$ m and its centre is $6$ m above the ground. It turns once every $40$ s. A passenger’s height is modelled by

$$h(t)=6+5\sin\left(\frac{\pi t}{20}+\phi\right).$$

At $t=0$, the passenger is $8.5$ m above the ground and rising. Find $\phi$ in $0\leq\phi<2\pi$.

Substitute $t=0$:

$$8.5=6+5\sin\phi,$$

so

$$\sin\phi=\frac12.$$

This gives $\phi=\pi/6$ or $5\pi/6$. The passenger is rising, so the graph must have positive gradient. Since

$$h'(t)=\frac\pi4\cos\left(\frac{\pi t}{20}+\phi\right),$$

we need $\cos\phi>0$. This selects

$$\boxed{\phi=\frac\pi6.}$$

Without calculus, inspect the sine graph: it is rising at $\pi/6$ and falling at $5\pi/6$.

Fitting the form $a\cos x+b\sin x+c$

Some models are given as a linear combination of sine and cosine. Harmonic form converts

$$a\cos x+b\sin x$$

into a single shifted sinusoid with amplitude $R=\sqrt{a^2+b^2}$. This reveals its range and phase.

Worked example 8: interpret a fitted model

The temperature inside a greenhouse is modelled by

$$T(t)=18-3\cos\left(\frac{\pi t}{12}\right)+4\sin\left(\frac{\pi t}{12}\right),$$

where $t$ is measured in hours. Find the period and range.

The argument has coefficient $\omega=\pi/12$, so

$$T_{\text{period}}=\frac{2\pi}{\pi/12}=24\text{ hours}.$$

The oscillating part has amplitude

$$R=\sqrt{(-3)^2+4^2}=5.$$

It therefore lies between $-5$ and $5$. Adding the midline $18$ gives

$$\boxed{13\leq T(t)\leq23},$$

so the model predicts temperatures from $13^\circ\text{C}$ to $23^\circ\text{C}$.

Using and assessing a model

A formula is not a complete model. State the variables, units, domain and assumptions, then check predictions against the context.

Worked example 9: solve in context

For the daylight model

$$D(t)=12+4\cos\left(\frac{2\pi}{365}(t-172)\right),$$

find the modelled days on which there are $14$ hours of daylight, for $0\leq t<365$.

Set $D(t)=14$:

$$\cos\left(\frac{2\pi}{365}(t-172)\right)=\frac12.$$

Let

$$\theta=\frac{2\pi}{365}(t-172).$$

Around the maximum at $\theta=0$, the relevant solutions are $\theta=\pm\pi/3$. Hence

$$\frac{2\pi}{365}(t-172)=\pm\frac\pi3,$$

so

$$t-172=\pm\frac{365}{6}.$$

Therefore

$$\boxed{t\approx111.2\text{ or }232.8.}$$

These correspond approximately to one date in spring and one in late summer. Reporting exact-looking calendar dates would overstate the accuracy of this simplified model.

Assumptions and limitations

A sinusoidal model usually assumes that:

• the phenomenon repeats with a constant period;
• maxima and minima are equally spaced in time;
• the rise and fall have the smooth, symmetric shape of a sine curve;
• the midline and amplitude remain constant;
• random variation and unusual events can be ignored.

These assumptions can fail. Tides combine several astronomical cycles, daily temperature is affected by weather, and daylight variation is only approximately sinusoidal. A model may still be useful within a stated interval, but extrapolation across many cycles needs justification.

To assess a prediction, ask:

1. Is the input inside the stated domain?
2. Are the units consistent?
3. Is the output within the model’s theoretical range?
4. Is the answer sensible in context?
5. How closely does the model agree with observed data?

If observed values $y_i$ are available, the residuals

$$r_i=y_i-\widehat y_i$$

measure observation minus prediction. Small residuals scattered without a pattern support the model. A systematic pattern suggests that the period, phase, amplitude, midline, or even the sinusoidal form is inadequate.

Common modelling errors

• The amplitude is half the total range, not the full range.
• The period of $\sin(kt)$ or $\cos(kt)$ is $2\pi/|k|$, not $k$.
• A horizontal shift belongs inside the argument: $\cos(\omega(t-t_0))$.
• One height usually gives two possible phases. Use whether the quantity is rising or falling.
• Calculator angle mode must match the model.
• A mathematically valid solution may be outside the physical domain.

Self-check 3

A buoy’s height above the seabed is modelled by

$$h(t)=7+1.5\cos\left(\frac{\pi}{3}(t-1)\right), \qquad 0\leq t\leq12,$$

where $h$ is in metres and $t$ is in hours.

1. State the midline, amplitude and period.
2. Find the first maximum height and when it occurs.
3. Find the first time at which $h=6.25$ m.

Answer

1. The midline is $7$ m, the amplitude is $1.5$ m, and

   $$T=\frac{2\pi}{\pi/3}=6\text{ hours}.$$

2. The maximum is $7+1.5=8.5$ m. It first occurs when the cosine argument is $0$, so $t=1$ hour.

3. Solve

   $$7+1.5\cos\left(\frac\pi3(t-1)\right)=6.25.$$

   Then

   $$\cos\left(\frac\pi3(t-1)\right)=-\frac12.$$

   After the maximum at $t=1$, the first suitable angle is $2\pi/3$:

   $$\frac\pi3(t-1)=\frac{2\pi}{3}.$$

   Hence $t-1=2$, so

   $$\boxed{t=3\text{ hours}.}$$

Final checklist

For a proof:

• distinguish an identity from an equation;
• begin with known results and show each algebraic step;
• watch the domain when dividing or cancelling;
• check that the conclusion is exactly what was required.

For a model:

• define variables, units and domain;
• obtain $M$, $|A|$, $T$, $\omega$ and the phase from the information;
• select sine or cosine to make the phase simple;
• check an extreme, a midline crossing and one complete period;
• interpret solutions and discuss assumptions and limitations.

Next steps

• Practise rearranging expressions in trigonometric identities.
• Use harmonic form to analyse fitted sine and cosine combinations.
• Apply approximations in small-angle approximations.
• Connect periodic models with rates of change in differentiating standard functions.
• Compare this modelling cycle with functions in mathematical modelling.