Year 11 Maths Advanced: Trigonometric Identities and Equations

A right-angled triangle has an acute angle of \(30^\circ\). Its opposite side is half the length of its hypotenuse, so \(\sin30^\circ=\frac12\).

Now reverse that ratio. What is the hypotenuse divided by the opposite side? It is \(2\).

But what happens if you divide the hypotenuse by the adjacent side instead, then gradually increase the angle towards \(90^\circ\)? The adjacent side gets shorter and shorter, so this ratio grows without bound.

These three ways of comparing the sides lead to three additional trigonometric functions: secant, cosecant and cotangent.

The NSW 2024 Year 11 Mathematics Advanced syllabus develops these functions further, moving from right-angled triangles to the unit circle, reciprocal and complementary angle identities, and equations involving trigonometric ratios. The important skill is knowing which relationship applies, and exactly when it is valid.

01Secant, cosecant and cotangent in a right-angled triangle

Start with a right-angled triangle containing an acute angle \(\theta\).

Relative to this angle, the three sides are:

• Opposite: the side directly across from \(\theta\).
• Adjacent: the side next to \(\theta\), excluding the hypotenuse.
• Hypotenuse: the longest side, opposite the right angle.

You already know the three primary trigonometric ratios:

\[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \]

\[ \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \]

\[ \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \]

The additional three ratios reverse some of these comparisons.

The three new trigonometric functions

Secant, written \(\sec\theta\), is the ratio of the hypotenuse to the adjacent side.

\[ \boxed{\sec\theta= \frac{\text{hypotenuse}}{\text{adjacent}}} \]

Cosecant, written \(\cosec\theta\), is the ratio of the hypotenuse to the opposite side.

\[ \boxed{\cosec\theta= \frac{\text{hypotenuse}}{\text{opposite}}} \]

Cotangent, written \(\cot\theta\), is the ratio of the adjacent side to the opposite side.

\[ \boxed{\cot\theta= \frac{\text{adjacent}}{\text{opposite}}} \]

Notice that each is the reciprocal of one of the original three ratios.

For an acute angle, all six trigonometric ratios are positive because all three side lengths are positive.

Exact values at \(30^\circ\), \(45^\circ\) and \(60^\circ\)

You can find the exact values using two familiar special triangles.

The first is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle, with side lengths in the ratio:

\[ 1:\sqrt3:2 \]

The second is a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle, with side lengths:

\[ 1:1:\sqrt2 \]

Consider the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.

For the \(30^\circ\) angle:

• Opposite = \(1\)
• Adjacent = \(\sqrt3\)
• Hypotenuse = \(2\)

Therefore:

\[ \sec30^\circ=\frac{2}{\sqrt3} =\frac{2\sqrt3}{3} \]

\[ \cosec30^\circ=\frac21=2 \]

\[ \cot30^\circ=\frac{\sqrt3}{1} =\sqrt3 \]

For \(45^\circ\), the two shorter sides are equal, so secant and cosecant have the same value.

The complete set of exact values is:

A useful exam habit: Keep exact values in surd form. For example, write \(\frac{2\sqrt3}{3}\), not \(1.155\). Unless a question specifically requests a decimal approximation, exact values are generally preferred.

What happens at \(0^\circ\) and \(90^\circ\)?

The definitions above use an acute angle in a non-degenerate right-angled triangle. At \(0^\circ\) and \(90^\circ\), the triangle becomes degenerate.

However, we can investigate what happens as the angle approaches these values.

Imagine keeping the hypotenuse at length \(1\).

As \(\theta\) approaches \(0^\circ\):

• The opposite side approaches \(0\).
• The adjacent side approaches \(1\).
• The hypotenuse remains \(1\).

Consequently:

\[ \sec\theta\to1 \]

\[ \cosec\theta\to+\infty \]

\[ \cot\theta\to+\infty \]

As \(\theta\) approaches \(90^\circ\):

• The opposite side approaches \(1\).
• The adjacent side approaches \(0\).
• The hypotenuse remains \(1\).

Therefore:

\[ \sec\theta\to+\infty \]

\[ \cosec\theta\to1 \]

\[ \cot\theta\to0 \]

These limits explain the behaviour, but an infinite limit does not mean that the function has an infinite value at the endpoint.

Using the circle definitions developed below, the exact endpoint values are:

The distinction is important.

At \(0^\circ\), the opposite side is zero, so cosecant and cotangent have a zero denominator.

At \(90^\circ\), the adjacent side is zero, so secant is undefined. However, cotangent is zero because its numerator is zero and its denominator is non-zero.

This is one reason it is important to understand the original side ratios rather than simply memorising reciprocal formulas.

02Extending the definitions using the circle

Right-angled triangles are useful, but they only describe acute angles naturally.

What happens when an angle is \(120^\circ\), \(225^\circ\), or even \(-45^\circ\)?

To handle angles of any magnitude, we use a circle centred at the origin.

Consider a circle of radius \(r\), centred at \(O=(0,0)\). A point \(P=(x,y)\) lies on its circumference.

Measure the angle \(\theta\) anticlockwise from the positive \(x\)-axis to the radius \(OP\).

The radius, horizontal coordinate and vertical coordinate form a right-angled triangle.

By Pythagoras:

\[ r^2=x^2+y^2 \]

The original trigonometric definitions become:

\[ \cos\theta=\frac{x}{r} \]

\[ \sin\theta=\frac{y}{r} \]

\[ \tan\theta=\frac{y}{x} \]

The three additional functions can now be defined in terms of the circle:

\[ \boxed{\sec\theta=\frac{r}{x}} \]

\[ \boxed{\cosec\theta=\frac{r}{y}} \]

\[ \boxed{\cot\theta=\frac{x}{y}} \]

These definitions extend secant, cosecant and cotangent to angles of any magnitude, provided the relevant denominator is non-zero.

The key difference is that \(x\) and \(y\) can now be negative.

For example, in the second quadrant:

• \(x\) is negative.
• \(y\) is positive.
• \(r\) remains positive.

Therefore, secant and cotangent are negative, while cosecant is positive.

This gives us a way to determine the signs of the trigonometric ratios without drawing a new right-angled triangle for every angle.

Why the reciprocal identities work

From the circle definitions:

\[ \cos\theta=\frac{x}{r} \]

and

\[ \sec\theta=\frac{r}{x} \]

Multiplying them gives:

\[ \cos\theta\sec\theta=1 \]

Therefore:

\[ \boxed{\sec\theta=\frac{1}{\cos\theta}} \]

provided \(\cos\theta\ne0\).

The same reasoning establishes the other reciprocal identities:

\[ \boxed{\cosec\theta=\frac{1}{\sin\theta}} \]

provided \(\sin\theta\ne0\), and

\[ \boxed{\cot\theta=\frac{1}{\tan\theta}} \]

provided \(\tan\theta\) exists and is non-zero.

These restrictions matter. We cannot divide by zero, even if an algebraic expression looks as though it should simplify.

Conditions for reciprocal and quotient identities

The syllabus expects you to identify the angles excluded from these identities.

Let \(k\) be any integer.

For radians, replace \(180^\circ\) with \(\pi\) and \(90^\circ\) with \(\frac{\pi}{2}\).

For example:

\[ \sec x=\frac1{\cos x}, \qquad x\ne\frac{\pi}{2}+k\pi \]

\[ \cosec x=\frac1{\sin x}, \qquad x\ne k\pi \]

An important distinction involving cotangent

You might notice something interesting in the table.

The identity

\[ \cot\theta=\frac{\cos\theta}{\sin\theta} \]

is valid at \(90^\circ\), because:

\[ \cot90^\circ=\frac01=0 \]

However, the identity

\[ \cot\theta=\frac1{\tan\theta} \]

is not valid at \(90^\circ\), because \(\tan90^\circ\) is undefined.

Both expressions describe cotangent when the reciprocal of tangent exists, but the second expression has a narrower domain.

This is a common source of mistakes when simplifying expressions.

03Exact values for multiples of \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\)

The NSW Mathematics Advanced course also requires you to work with exact trigonometric values in radians.

You already know the special angles \(30^\circ\), \(45^\circ\), and \(60^\circ\).

In radians, these are:

\[ 30^\circ=\frac{\pi}{6} \]

\[ 45^\circ=\frac{\pi}{4} \]

\[ 60^\circ=\frac{\pi}{3} \]

The other exact values follow from the same special triangles, together with the signs of the trigonometric functions in each quadrant.

Multiples of \(\frac{\pi}{6}\)

The following table gives the exact values for the integer multiples of \(\frac{\pi}{6}\) from \(0\) to \(2\pi\).

Multiples of \(\frac{\pi}{4}\)

The equivalent table for multiples of \(\frac{\pi}{4}\) is:

What about angles outside these tables?

The tables cover one complete revolution, but trigonometric functions repeat.

Secant and cosecant have period \(2\pi\), while cotangent has period \(\pi\).

For example:

\[ \sec\left(\frac{13\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right) \]

because:

\[ \frac{13\pi}{6}-2\pi=\frac{\pi}{6} \]

Therefore:

\[ \sec\left(\frac{13\pi}{6}\right) =\frac{2\sqrt3}{3} \]

For cotangent:

\[ \cot\left(\frac{7\pi}{6}\right) = \cot\left(\frac{\pi}{6}\right) =\sqrt3 \]

because the angles differ by \(\pi\).

The useful strategy is to reduce an unfamiliar angle to one of the familiar angles, then use the appropriate sign and exact value.

04Complementary angle identities

Two acute angles are complementary if their sum is \(90^\circ\).

For example, \(30^\circ\) and \(60^\circ\) are complementary.

The important observation is that the opposite side of one angle becomes the adjacent side of the other.

Consider a right-angled triangle with acute angles \(\theta\) and \(90^\circ-\theta\).

For the angle \(\theta\):

\[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \]

For the complementary angle, the original adjacent side becomes the opposite side.

Therefore:

\[ \boxed{\sin(90^\circ-\theta)=\cos\theta} \]

Similarly, the original opposite side becomes adjacent, giving:

\[ \boxed{\cos(90^\circ-\theta)=\sin\theta} \]

The same reasoning produces four more identities.

The six complementary angle identities

\[ \boxed{\sin(90^\circ-\theta)=\cos\theta} \]

\[ \boxed{\cos(90^\circ-\theta)=\sin\theta} \]

\[ \boxed{\tan(90^\circ-\theta)=\cot\theta} \]

\[ \boxed{\cot(90^\circ-\theta)=\tan\theta} \]

\[ \boxed{\sec(90^\circ-\theta)=\cosec\theta} \]

\[ \boxed{\cosec(90^\circ-\theta)=\sec\theta} \]

The identities can also be established for angles of any magnitude using the circle definition. Reflecting a point on the unit circle across the line \(y=x\) exchanges its coordinates, which is exactly the relationship between the sine and cosine of complementary angles.

When are complementary angle identities undefined?

The first two identities involve only sine and cosine, which are defined for every real angle.

The remaining identities require restrictions.

Let \(k\) be any integer.

For radians, the equivalent identities use \(\frac{\pi}{2}-x\), with the corresponding restrictions:

• Exclude \(x=k\pi\) for the first and third identities.
• Exclude \(x=\frac{\pi}{2}+k\pi\) for the second and fourth identities.

05Evaluating trigonometric expressions for angles of any magnitude

An angle does not have to be between \(0^\circ\) and \(90^\circ\).

It might be negative, larger than \(360^\circ\), or expressed in radians.

The unit circle handles all these cases.

There are two ideas to use:

1. Periodicity: reduce the angle to an equivalent angle.
2. Quadrant signs: determine whether the ratio is positive or negative.

Periodicity

The sine, cosine, secant and cosecant functions repeat every \(360^\circ\), or \(2\pi\) radians.

The tangent and cotangent functions repeat every \(180^\circ\), or \(\pi\) radians.

For example:

\[ \sin390^\circ=\sin30^\circ \]

because:

\[ 390^\circ-360^\circ=30^\circ \]

Similarly:

\[ \tan225^\circ=\tan45^\circ=1 \]

because tangent repeats every \(180^\circ\).

Signs in each quadrant

The sign of a trigonometric ratio depends on the quadrant.

For example, consider \(150^\circ\).

Its reference angle is:

\[ 180^\circ-150^\circ=30^\circ \]

It lies in quadrant II, where cosine and secant are negative.

Therefore:

\[ \sec150^\circ =-\sec30^\circ =-\frac{2\sqrt3}{3} \]

The reference angle gives the magnitude of the ratio, while the quadrant determines its sign.

06Pythagorean trigonometric identities

One of the most useful relationships in trigonometry comes directly from Pythagoras.

For a point \((x,y)\) on a circle of radius \(r\):

\[ x^2+y^2=r^2 \]

Recall that:

\[ \cos\theta=\frac{x}{r} \]

and

\[ \sin\theta=\frac{y}{r} \]

Divide the Pythagorean equation by \(r^2\):

\[ \frac{x^2}{r^2}+\frac{y^2}{r^2}=1 \]

Substituting the trigonometric definitions gives:

\[ \boxed{\cos^2\theta+\sin^2\theta=1} \]

This is the fundamental Pythagorean trigonometric identity.

The identity is valid for every real angle.

Deriving the secant identity

Start with:

\[ \cos^2\theta+\sin^2\theta=1 \]

Divide every term by \(\cos^2\theta\), provided \(\cos\theta\ne0\).

\[ 1+\frac{\sin^2\theta}{\cos^2\theta} =\frac{1}{\cos^2\theta} \]

Using the definitions of tangent and secant:

\[ \boxed{1+\tan^2\theta=\sec^2\theta} \]

This identity is valid when:

\[ \cos\theta\ne0 \]

Deriving the cosecant identity

Return to the original Pythagorean identity.

Divide every term by \(\sin^2\theta\), provided \(\sin\theta\ne0\).

\[ \frac{\cos^2\theta}{\sin^2\theta} +1 =\frac{1}{\sin^2\theta} \]

Using cotangent and cosecant:

\[ \boxed{1+\cot^2\theta=\cosec^2\theta} \]

This identity is valid when:

\[ \sin\theta\ne0 \]

The three Pythagorean identities

You should be able to recognise and use all three:

\[ \boxed{\sin^2x+\cos^2x=1} \]

\[ \boxed{1+\tan^2x=\sec^2x} \]

\[ \boxed{1+\cot^2x=\cosec^2x} \]

Remember that \(\sin^2x\) means \((\sin x)^2\), not \(\sin(x^2)\).

07Proving and simplifying further trigonometric identities

A trigonometric identity is an equality that holds for every angle in the common domain of both sides.

For example:

\[ 1+\tan^2x=\sec^2x \]

is an identity, not an equation that is true only at certain angles.

To prove an identity, you need to show that one side can be transformed into the other using valid trigonometric relationships.

A reliable method for proving identities

When a question asks you to prove an identity, follow this approach:

1. Start with the more complicated side.

Usually, one side contains more functions, fractions or powers.

2. Replace secant, cosecant, tangent and cotangent with sine and cosine.

Use:

\[ \sec x=\frac1{\cos x} \]

\[ \cosec x=\frac1{\sin x} \]

\[ \tan x=\frac{\sin x}{\cos x} \]

\[ \cot x=\frac{\cos x}{\sin x} \]

3. Look for a Pythagorean identity.

The most common substitutions are:

\[ 1-\sin^2x=\cos^2x \]

and

\[ 1-\cos^2x=\sin^2x \]

4. Simplify carefully.

Factorise, find common denominators and cancel common factors only when they are non-zero.

5. Finish with the required expression.

You should not assume the two sides are equal and manipulate them simultaneously without showing why.

Common identity transformations

Why domain restrictions still matter

Suppose you simplify an expression by cancelling \(\sin x\).

That cancellation is only valid if \(\sin x\ne0\).

Similarly, dividing by \(\cos x\) requires \(\cos x\ne0\).

An identity must hold wherever the original expression is defined. You cannot introduce new values by cancelling a factor that was zero in the original expression.

This is particularly important in longer exam proofs, where a correct algebraic-looking simplification can accidentally change the domain.

08Solving trigonometric equations on a restricted domain

An identity is true for every angle in its domain.

A trigonometric equation is different. It asks you to find the particular angles that satisfy a condition.

For example:

\[ \sin\theta=\frac12 \]

does not hold for every angle.

On the domain \(0^\circ\leq\theta<360^\circ\), it has two solutions:

\[ \theta=30^\circ,\quad150^\circ \]

The domain is important because trigonometric functions repeat.

A general strategy for solving trigonometric equations

When solving a trigonometric equation:

1. Isolate the trigonometric function where possible.
2. Use reciprocal, quotient or Pythagorean identities if needed.
3. Find the reference angle.
4. Identify the quadrants where the required sign occurs.
5. List every solution inside the specified domain.
6. Check that each solution is valid in the original equation.

Do not stop after finding the first angle. Most trigonometric equations have multiple solutions over a complete revolution.

Solving equations in radians

The same method applies when angles are measured in radians.

The main difference is that you use radian reference angles and express the solutions using multiples of \(\pi\).

For example, if:

\[ \cos x=-\frac12 \]

on:

\[ 0\leq x<2\pi \]

then:

\[ \boxed{x=\frac{2\pi}{3},\ \frac{4\pi}{3}} \]

Be careful not to mix degrees and radians in the same calculation.

09Solving trigonometric equations that reduce to quadratics

Some trigonometric equations initially contain squared functions, or a combination of sine and cosine.

The trick is to use an identity to reduce the equation to a quadratic in one trigonometric function.

This is similar to solving an ordinary quadratic equation, except that you must then convert the resulting values back into angles.

What if the quadratic is in tangent or cotangent?

The same approach works if you substitute a different trigonometric function.

For example:

\[ 2\tan^2x-3\tan x+1=0 \]

can be treated as an ordinary quadratic by temporarily writing:

\[ u=\tan x \]

Then:

\[ 2u^2-3u+1=0 \]

Factorising:

\[ (2u-1)(u-1)=0 \]

Therefore:

\[ \tan x=\frac12 \]

or:

\[ \tan x=1 \]

You then find the angles in the required domain.

The important point is that the algebraic solutions are values of a trigonometric function, not the final angles.

A common mistake: Losing solutions when dividing

Suppose an equation contains:

\[ \sin x\cos x=0 \]

It is tempting to divide both sides by \(\cos x\).

However, this would remove every solution where \(\cos x=0\).

Similarly, dividing by \(\sin x\) could remove valid solutions where \(\sin x=0\).

Before dividing by a trigonometric expression, check whether it could equal zero. If it can, investigate that case separately.

This small habit can prevent you from losing valid solutions in a multi-step exam question.

10Common mistakes to avoid

The following errors appear frequently when working with trigonometric identities and equations.

11Questions and solutions

The following questions progress from straightforward applications to more demanding HSC-style problems. The final question combines reciprocal identities, the unit circle, restricted domains and the number of solutions to an equation.

Question 1

Evaluate the following expression exactly:

\[ \sec60^\circ+\cosec30^\circ-\cot45^\circ \]

Solution 1

The exact value is:

\[ \boxed{3} \]

Using the special-angle values:

\[ \sec60^\circ=2 \]

\[ \cosec30^\circ=2 \]

\[ \cot45^\circ=1 \]

Substituting:

\[ =2+2-1 \]

Therefore:

\[ \boxed{3} \]

The important skill here is recognising the exact values immediately rather than converting them into decimals.

Question 2

Prove the identity:

\[ \frac{\sec x-\cos x} {\cosec x-\sin x} =\tan^3x \]

State the restrictions on \(x\), where \(x\) is measured in radians.

Solution 2

The identity is valid for:

\[ \boxed{x\ne\frac{n\pi}{2},\quad n\in\mathbb Z} \]

To prove it, rewrite the left-hand side entirely in terms of sine and cosine.

Step 1: Expand the numerator using the reciprocal identity.

\[ \sec x-\cos x =\frac1{\cos x}-\cos x \]

Combining the fractions:

\[ =\frac{1-\cos^2x}{\cos x} \]

Using the Pythagorean identity:

\[ =\frac{\sin^2x}{\cos x} \]

Step 2: Simplify the denominator.

\[ \cosec x-\sin x =\frac1{\sin x}-\sin x \]

Combining the fractions:

\[ =\frac{1-\sin^2x}{\sin x} \]

Using the Pythagorean identity:

\[ =\frac{\cos^2x}{\sin x} \]

Step 3: Substitute into the original expression.

\[ \frac{\sec x-\cos x} {\cosec x-\sin x} = \frac{\frac{\sin^2x}{\cos x}} {\frac{\cos^2x}{\sin x}} \]

Dividing by a fraction is equivalent to multiplying by its reciprocal:

\[ = \frac{\sin^2x}{\cos x} \times \frac{\sin x}{\cos^2x} \]

Therefore:

\[ =\frac{\sin^3x}{\cos^3x} \]

Step 4: Recognise the tangent ratio.

\[ =\left(\frac{\sin x}{\cos x}\right)^3 \]

Hence:

\[ \boxed{\tan^3x} \]

The identity is proved.

The restriction is important because the original expression contains both secant and cosecant, so neither sine nor cosine can equal zero.

Question 3

Solve the equation:

\[ 2\sin^2x+\sin x\cos x=1 \]

for:

\[ 0\leq x<2\pi \]

Give your answers exactly in radians.

Solution 3

There are four solutions:

\[ \boxed{ \begin{aligned} x&=\alpha\\ x&=\pi+\alpha\\ x&=\pi-\beta\\ x&=2\pi-\beta \end{aligned} } \]

where:

\[ \alpha=\tan^{-1} \left(\frac{\sqrt5-1}{2}\right) \]

and:

\[ \beta=\tan^{-1} \left(\frac{\sqrt5+1}{2}\right) \]

To obtain these solutions, first reduce the equation to a quadratic in tangent.

Step 1: Use the Pythagorean identity.

\[ \sin^2x+\cos^2x=1 \]

Rewrite the right-hand side of the original equation:

\[ 2\sin^2x+\sin x\cos x =\sin^2x+\cos^2x \]

Rearranging:

\[ \sin^2x+\sin x\cos x-\cos^2x=0 \]

Step 2: Check whether \(\cos x=0\).

If \(\cos x=0\), then \(\sin^2x=1\).

The original equation becomes:

\[ 2=1 \]

which is impossible.

Therefore, \(\cos x\ne0\), so dividing by \(\cos^2x\) is valid.

Step 3: Divide through by \(\cos^2x\).

\[ \frac{\sin^2x}{\cos^2x} + \frac{\sin x\cos x}{\cos^2x} -1=0 \]

Therefore:

\[ \tan^2x+\tan x-1=0 \]

Step 4: Solve the quadratic.

Using the quadratic formula:

\[ \tan x= \frac{-1\pm\sqrt{1+4}}{2} \]

Hence:

\[ \tan x=\frac{-1+\sqrt5}{2} \]

or:

\[ \tan x=\frac{-1-\sqrt5}{2} \]

Step 5: Find the angles.

For the positive tangent value, define:

\[ \alpha=\tan^{-1} \left(\frac{\sqrt5-1}{2}\right) \]

Tangent is positive in quadrants I and III.

Therefore:

\[ x=\alpha,\quad\pi+\alpha \]

For the negative tangent value, define:

\[ \beta=\tan^{-1} \left(\frac{\sqrt5+1}{2}\right) \]

Tangent is negative in quadrants II and IV.

Therefore:

\[ x=\pi-\beta,\quad2\pi-\beta \]

All four solutions lie in the required interval.

The key step was checking \(\cos x=0\) before dividing. Without that check, you would not know whether the algebra had removed a possible solution.

Question 4

Solve:

\[ \sec\theta+\tan\theta=2 \]

for:

\[ 0\leq\theta<2\pi \]

Give your answer exactly, using an inverse trigonometric function if necessary.

Solution 4

There is exactly one solution:

\[ \boxed{ \theta=\tan^{-1}\left(\frac34\right) } \]

The key is to use the Pythagorean identity to obtain a second equation.

Step 1: Recall the secant identity.

\[ \sec^2\theta-\tan^2\theta=1 \]

Factorising:

\[ (\sec\theta+\tan\theta) (\sec\theta-\tan\theta)=1 \]

Step 2: Substitute the information from the question.

We know:

\[ \sec\theta+\tan\theta=2 \]

Therefore:

\[ 2(\sec\theta-\tan\theta)=1 \]

Hence:

\[ \sec\theta-\tan\theta=\frac12 \]

Step 3: Solve the two simultaneous equations.

We have:

\[ \sec\theta+\tan\theta=2 \]

\[ \sec\theta-\tan\theta=\frac12 \]

Adding:

\[ 2\sec\theta=\frac52 \]

Therefore:

\[ \sec\theta=\frac54 \]

Subtracting the second equation from the first:

\[ 2\tan\theta=\frac32 \]

Therefore:

\[ \tan\theta=\frac34 \]

Step 4: Determine the quadrant.

Since:

\[ \sec\theta=\frac54 \]

we have:

\[ \cos\theta=\frac45 \]

Also:

\[ \tan\theta=\frac34 \]

Since tangent and cosine are both positive, sine must also be positive.

Therefore, \(\theta\) lies in quadrant I.

Step 5: Find the angle.

\[ \tan\theta=\frac34 \]

Therefore:

\[ \boxed{ \theta=\tan^{-1}\left(\frac34\right) } \]

This is the only solution in the specified interval.

A common mistake would be to assume that a trigonometric equation must have two solutions over a complete revolution. The signs of both sine and cosine restrict this equation to quadrant I.

Question 5

Consider the function:

\[ F(\theta)= \frac{\sec\theta+\cosec\theta} {\tan\theta+\cot\theta} \]

where:

\[ 0\leq\theta<2\pi \]

and the original expression is defined.

(a) Simplify \(F(\theta)\) using trigonometric identities, without changing its domain.

(b) Determine the exact range of \(F\).

(c) For each real number \(k\), determine the number of solutions to:

\[ F(\theta)=k \]

within the specified domain.

Explain carefully why some values that appear to be possible from the simplified expression are not actually attained.

Solution 5

The exact range is:

\[ \boxed{ [-\sqrt2,\sqrt2]\setminus\{-1,1\} } \]

The number of solutions is:

• One solution when \(k=\sqrt2\) or \(k=-\sqrt2\).
• Two solutions when \(-\sqrt2<k<\sqrt2\), excluding \(k=\pm1\).
• No solutions for all other values of \(k\).

The interesting part is that the original expression is not defined at four angles, even though the simplified expression would appear to allow them.

Part (a): Simplify the expression.

Step 1: Replace each trigonometric function using sine and cosine.

\[ F(\theta)= \frac{ \frac1{\cos\theta}+\frac1{\sin\theta} }{ \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} } \]

The original expression requires:

\[ \sin\theta\ne0 \]

and:

\[ \cos\theta\ne0 \]

Step 2: Combine the fractions in the numerator.

\[ \frac1{\cos\theta}+\frac1{\sin\theta} = \frac{\sin\theta+\cos\theta} {\sin\theta\cos\theta} \]

Step 3: Combine the fractions in the denominator.

\[ \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \]

Taking a common denominator:

\[ = \frac{\sin^2\theta+\cos^2\theta} {\sin\theta\cos\theta} \]

Using the Pythagorean identity:

\[ = \frac1{\sin\theta\cos\theta} \]

Step 4: Divide the two expressions.

\[ F(\theta)= \frac{ \frac{\sin\theta+\cos\theta} {\sin\theta\cos\theta} }{ \frac1{\sin\theta\cos\theta} } \]

The common denominator cancels:

\[ \boxed{F(\theta)=\sin\theta+\cos\theta} \]

However, the original domain must be retained:

\[ \theta\ne0,\frac{\pi}{2},\pi,\frac{3\pi}{2} \]

within \(0\leq\theta<2\pi\).

Part (b): Determine the range.

Step 1: Rewrite the simplified expression.

Recall the complementary angle relationship:

\[ \sin\left(\theta+\frac{\pi}{4}\right) = \sin\theta\cos\frac{\pi}{4} + \cos\theta\sin\frac{\pi}{4} \]

Since both \(\sin\frac{\pi}{4}\) and \(\cos\frac{\pi}{4}\) equal \(\frac{\sqrt2}{2}\):

\[ \sin\left(\theta+\frac{\pi}{4}\right) = \frac{\sin\theta+\cos\theta}{\sqrt2} \]

Therefore:

\[ F(\theta)= \sqrt2\sin\left(\theta+\frac{\pi}{4}\right) \]

Step 2: Determine the maximum and minimum.

Since:

\[ -1\leq\sin x\leq1 \]

we obtain:

\[ -\sqrt2\leq F(\theta)\leq\sqrt2 \]

However, we must now consider the original domain.

When:

\[ F(\theta)=1 \]

we need:

\[ \sin\theta+\cos\theta=1 \]

Within one complete revolution, the only solutions are:

\[ \theta=0,\quad\frac{\pi}{2} \]

Both are excluded from the original domain.

Similarly, when:

\[ F(\theta)=-1 \]

the only solutions are:

\[ \theta=\pi,\quad\frac{3\pi}{2} \]

These are also excluded.

Therefore, although the simplified expression can equal \(1\) and \(-1\), the original function cannot.

The exact range is:

\[ \boxed{ [-\sqrt2,\sqrt2]\setminus\{-1,1\} } \]

Part (c): Determine the number of solutions for each value of \(k\).

We need to solve:

\[ \sin\theta+\cos\theta=k \]

subject to the original domain.

Using the circle interpretation, let:

\[ x=\cos\theta,\qquad y=\sin\theta \]

The circle satisfies:

\[ x^2+y^2=1 \]

The equation becomes:

\[ x+y=k \]

This represents a straight line intersecting the unit circle.

There are three important cases.

Case 1: \(k=\sqrt2\) or \(k=-\sqrt2\).

The line is tangent to the circle.

It touches the circumference at exactly one point.

Therefore, there is one solution for each value.

Case 2: \(-\sqrt2<k<\sqrt2\).

The line passes through the circle at two points.

Normally, this gives two solutions.

However, when \(k=1\), the intersections are:

\[ (1,0),\quad(0,1) \]

These correspond to:

\[ \theta=0,\quad\frac{\pi}{2} \]

Both are excluded.

Likewise, when \(k=-1\), the intersections are:

\[ (-1,0),\quad(0,-1) \]

These correspond to:

\[ \theta=\pi,\quad\frac{3\pi}{2} \]

Again, both are excluded.

Therefore, \(k=\pm1\) gives no solutions, despite lying inside the interval \([-\sqrt2,\sqrt2]\).

Case 3: \(k<-\sqrt2\) or \(k>\sqrt2\).

The line does not intersect the unit circle.

Therefore, there are no solutions.

The complete answer is:

This question highlights an important principle: simplifying a trigonometric expression does not automatically preserve its domain.

The simplified function may appear to have a value at an angle where the original expression was undefined. Those missing values can change both the range and the number of solutions.

The next useful step is connecting these identities to the graphs of sine, cosine and tangent. The unit circle explains which angles produce a particular value, while the graphs make periodicity, restricted domains and the number of solutions much easier to visualise. Together, these ideas provide the foundation for more advanced trigonometric equations in Mathematics Advanced.