Why does the interval become 0^{\circ} to 720^{\circ} for \sin 2x?

You are solving for the angle 2x2x2x, not xxx. If xxx ranges over 0∘0^{\circ}0∘ to 360∘360^{\circ}360∘, then 2x2x2x ranges over twice that, 0∘0^{\circ}0∘ to 720∘720^{\circ}720∘. Find every 2x2x2x in the wider range, then divide by 222. Skipping this step is the usual reason answers go missing.

Trigonometric Functions Worked Examples (Medium)

Q: How do I know a trig equation can be treated as a quadratic?

Look for a squared ratio and its first power in the same equation, such as 2sin⁡2x+sin⁡x−1=02\sin^{2}x+\sin x-1=02sin2x+sinx−1=0. Substitute a single letter for the ratio, factorise or use the quadratic formula, then translate each root back and solve for the angle in the given range.

Q: When proving an identity, which side should I start from?

Start from the more complicated side, usually the one with fractions or higher powers, and simplify it towards the other. Combine fractions over a common denominator, expand, and use sin⁡2θ+cos⁡2θ=1\sin^{2}\theta+\cos^{2}\theta=1sin2θ+cos2θ=1. Do not move terms across the equals sign as if solving an equation.

Q: How many times can a line cut y=2\sin 2x between 0^{\circ} and 360^{\circ}?

The period is 180∘180^{\circ}180∘, so two full cycles fit the range. A horizontal line at a height inside the amplitude, such as y=1y=1y=1, crosses each cycle twice, giving four intersections, which matches the four solutions of 2sin⁡2x=12\sin 2x=12sin2x=1.

What these examples cover

These medium Trigonometric Functions examples move past single-step recall to the reasoning the middle of a paper expects. The first treats a trigonometric equation as a quadratic in $\sin x$ : factorise, split into two simple equations, then read every angle in the range.

The second is a proof, combine two fractions over a common denominator, expand carefully, and use the identity $\sin^{2}\theta+\cos^{2}\theta=1$ to collapse the numerator until the two sides match. The third is a graph-and-equation question with a multiple angle: because the angle is $2x$ , you must widen the working interval to $0^{\circ}$ to $720^{\circ}$ so no solution is lost, then divide back.

Numbers stay small and clean so the method, not the arithmetic, is the focus. Use the set honestly: cover the solution, attempt the question in full, and only then check line by line.

Where your answer differs, find the exact step where the two solutions part company, that single line is usually where the real learning is.

Worked examples

Work through all three. Attempt each fully before you read the matching solution, and watch how one habit, reduce the problem to a familiar sub-problem, then finish it carefully, runs through the quadratic equation, the identity proof and the multiple-angle graph question.

As you read a full solution, notice where the marks actually sit. In an equation the method marks come from a correct factorisation and from placing the basic angle in the right quadrants; in a proof they come from the honest algebra of combining fractions and naming the identity you use; in a graph question they come from stating amplitude and period correctly and from widening the interval before you list solutions.

A final answer with no visible method earns far less than tidy working that shows each of these steps. So write every substitution as its own line, and never erase a line just because it looks long, under analytic marking, that line is often what is being marked.

Q1[4 marks]

Solve $2\sin^{2}x+\sin x-1=0$ for $0^{\circ}\le x\le 360^{\circ}$ .

Show worked solution

The equation is a quadratic in $\sin x$ . Write $s=\sin x$ so the structure is clear:

2s^{2}+s-1=0

Factorise the quadratic. Two numbers multiplying to $2\times(-1)=-2$ and adding to $+1$ are $+2$ and $-1$ , which gives:

(2s-1)(s+1)=0

So $s=\dfrac{1}{2}$ or $s=-1$ ; that is, $\sin x=\dfrac{1}{2}$ or $\sin x=-1$ . Solve each separately.

For $\sin x=\dfrac{1}{2}$ : the basic angle is $30^{\circ}$ , and sine is positive in the first and second quadrants:

x=30^{\circ}\quad\text{or}\quad x=180^{\circ}-30^{\circ}=150^{\circ}

For $\sin x=-1$ : sine reaches its lowest value $-1$ once in a full turn, at the bottom of the curve:

x=270^{\circ}

Answer

$x=30^{\circ},\ 150^{\circ},\ 270^{\circ}$ . Check the factorisation by expanding: $(2s-1)(s+1)=2s^{2}+2s-s-1=2s^{2}+s-1$ , which is the original.

Then $\sin 30^{\circ}=\sin 150^{\circ}=\tfrac{1}{2}$ and $\sin 270^{\circ}=-1$ , so all three angles satisfy the equation.

Q2[4 marks]

Prove that $\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\operatorname{cosec}\theta$ .

Show worked solution

Work on the left-hand side (LHS) and combine the two fractions over the common denominator $\sin\theta(1+\cos\theta)$ :

\text{LHS}=\dfrac{\sin\theta\cdot\sin\theta+(1+\cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\dfrac{\sin^{2}\theta+(1+\cos\theta)^{2}}{\sin\theta(1+\cos\theta)}

Expand the square in the numerator, keeping $(1+\cos\theta)^{2}=1+2\cos\theta+\cos^{2}\theta$ :

=\dfrac{\sin^{2}\theta+1+2\cos\theta+\cos^{2}\theta}{\sin\theta(1+\cos\theta)}

Group $\sin^{2}\theta+\cos^{2}\theta$ and replace it with $1$ , using the Pythagorean identity:

Given in the exam

\sin^{2}\theta+\cos^{2}\theta=1

=\dfrac{1+1+2\cos\theta}{\sin\theta(1+\cos\theta)}=\dfrac{2+2\cos\theta}{\sin\theta(1+\cos\theta)}

Factor $2$ from the numerator, then cancel the common factor $(1+\cos\theta)$ :

=\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\dfrac{2}{\sin\theta}=2\operatorname{cosec}\theta=\text{RHS}

Answer

Since the left-hand side simplifies exactly to $2\operatorname{cosec}\theta$ , the identity is proven. The turning point is spotting that $\sin^{2}\theta+\cos^{2}\theta$ hides inside the expanded numerator; without that substitution the fraction never simplifies.

Q3[5 marks]

The graph of $y=2\sin 2x$ is drawn for $0^{\circ}\le x\le 360^{\circ}$ . (a) State its amplitude and period.

(b) Find the number of solutions of the equation $2\sin 2x=1$ in this interval, and solve it.

Show worked solution

(a) Compare with $y=a\sin bx+c$ : here $a=2$ , $b=2$ , $c=0$ . The amplitude is $|a|=2$ , and the period is $\dfrac{360^{\circ}}{b}=180^{\circ}$ .

The curve therefore completes two full cycles between $0^{\circ}$ and $360^{\circ}$ .

(b) Divide the equation by $2$ to isolate the sine:

2\sin 2x=1\ \Rightarrow\ \sin 2x=\dfrac{1}{2}

The multiple angle $2x$ is the key. As $x$ runs from $0^{\circ}$ to $360^{\circ}$ , the angle $2x$ runs from $0^{\circ}$ to $720^{\circ}$ , so look for every solution of $\sin 2x=\dfrac{1}{2}$ in that wider interval.

The basic angle is $30^{\circ}$ , and sine is positive in the first and second quadrants, so within one turn the solutions are $30^{\circ}$ and $150^{\circ}$ ; add $360^{\circ}$ to each for the second turn:

2x=30^{\circ},\ 150^{\circ},\ 390^{\circ},\ 510^{\circ}

Divide every value by $2$ :

x=15^{\circ},\ 75^{\circ},\ 195^{\circ},\ 255^{\circ}

Answer

Amplitude $2$ , period $180^{\circ}$ ; there are $4$ solutions: $x=15^{\circ},\ 75^{\circ},\ 195^{\circ},\ 255^{\circ}$ . Check the last one: $2x=510^{\circ}$ , and $\sin 510^{\circ}=\sin(510^{\circ}-360^{\circ})=\sin 150^{\circ}=\dfrac{1}{2}$ .

The count matches the graph: two cycles, each crossing the line $y=1$ twice.

Q4[3 marks]

Given that $\cos\theta=-\dfrac{4}{5}$ and $180^{\circ}<\theta<270^{\circ}$ , find the value of $\sin\theta$ and $\tan\theta$ .

Show worked solution

Since $180^{\circ}<\theta<270^{\circ}$ , $\theta$ lies in the third quadrant, where sine and cosine are both negative but tangent is positive. Use the Pythagorean identity to find $\sin\theta$ first.

\sin^{2}\theta=1-\cos^{2}\theta=1-\left(-\dfrac{4}{5}\right)^{2}=1-\dfrac{16}{25}=\dfrac{9}{25}

Take the square root and choose the negative sign, since sine is negative in the third quadrant:

\sin\theta=-\dfrac{3}{5}\quad\text{so}\quad\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{3}{5}}{-\frac{4}{5}}=\dfrac{3}{4}

Answer

$\sin\theta=-\dfrac{3}{5}$ and $\tan\theta=\dfrac{3}{4}$ . This is a 3–4–5 right triangle in the third quadrant, and $\tan\theta$ coming out positive matches the rule that tangent is positive when sine and cosine share the same sign.

Q5[4 marks]

Without using a calculator, find the exact value of $\sin 75^{\circ}$ , giving your answer in surd form.

Show worked solution

Write $75^{\circ}$ as a sum of two angles with known exact ratios, $45^{\circ}+30^{\circ}$ , and apply the addition formula:

\sin 75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin 45^{\circ}\cos 30^{\circ}+\cos 45^{\circ}\sin 30^{\circ}

Substitute the exact values $\sin45^{\circ}=\cos45^{\circ}=\dfrac{\sqrt{2}}{2}$ , $\cos30^{\circ}=\dfrac{\sqrt{3}}{2}$ and $\sin30^{\circ}=\dfrac{1}{2}$ :

\sin 75^{\circ}=\left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)+\left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{6}}{4}+\dfrac{\sqrt{2}}{4}

\sin 75^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}

Answer

$\sin 75^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$ . As a check, $\dfrac{\sqrt{6}+\sqrt{2}}{4}\approx\dfrac{2.449+1.414}{4}\approx0.966$ , which matches $\sin75^{\circ}$ from a calculator.

Q6[3 marks]

Given that $\sin A=\dfrac{5}{13}$ and $A$ is acute, find the value of $\cos 2A$ .

Show worked solution

Because $\sin A$ is already known, use the double angle formula written in terms of sine only, so there is no need to find $\cos A$ first:

\cos 2A=1-2\sin^{2}A

Substitute $\sin A=\dfrac{5}{13}$ :

\cos 2A=1-2\left(\dfrac{5}{13}\right)^{2}=1-2\left(\dfrac{25}{169}\right)=1-\dfrac{50}{169}=\dfrac{119}{169}

Answer

$\cos 2A=\dfrac{119}{169}$ . Since $\sin A\approx0.385$ , $A\approx22.6^{\circ}$ , so $2A\approx45.2^{\circ}$ ; $\cos45^{\circ}\approx0.707$ is close to $\dfrac{119}{169}\approx0.704$ , confirming the answer is reasonable.

Q7[4 marks]

Solve $\sqrt{3}\sin x=\cos x$ for $0^{\circ}\le x\le 360^{\circ}$ .

Show worked solution

Both sine and cosine appear to the first power only, and $\cos x\ne0$ at any solution here, so divide every term by $\cos x$ to reduce the equation to a single ratio:

\sqrt{3}\sin x=\cos x\ \Rightarrow\ \sqrt{3}\tan x=1\ \Rightarrow\ \tan x=\dfrac{1}{\sqrt{3}}

The basic angle is $30^{\circ}$ . Tangent is positive in the first and third quadrants, so:

x=30^{\circ}\quad\text{or}\quad x=180^{\circ}+30^{\circ}=210^{\circ}

Answer

$x=30^{\circ},\ 210^{\circ}$ . Check $x=210^{\circ}$ : $\sqrt{3}\sin210^{\circ}=\sqrt{3}\left(-\tfrac{1}{2}\right)\approx-0.866$ and $\cos210^{\circ}\approx-0.866$ , so both sides agree.

Q8[3 marks]

The graph of $y=3\cos x+2$ is drawn for $0^{\circ}\le x\le 360^{\circ}$ . State the maximum and minimum values of $y$ , and the corresponding values of $x$ .

Show worked solution

Compare with $y=a\cos x+c$ : here $a=3$ and $c=2$ . Because $\cos x$ itself only ever ranges from $-1$ to $1$ , the vertical shift $c$ simply moves that whole range up or down.

The maximum occurs when $\cos x=1$ , and the minimum when $\cos x=-1$ :

y_{\max}=3(1)+2=5\ \text{at}\ \cos x=1\ \Rightarrow\ x=0^{\circ},\ 360^{\circ}

y_{\min}=3(-1)+2=-1\ \text{at}\ \cos x=-1\ \Rightarrow\ x=180^{\circ}

Answer

Maximum $y=5$ at $x=0^{\circ}$ and $x=360^{\circ}$ ; minimum $y=-1$ at $x=180^{\circ}$ . The curve oscillates symmetrically about the midline $y=2$ , which matches the vertical shift $c=2$ .

Across all three, the winning move is the same: reshape the problem into something you already know how to finish. A trig equation becomes a quadratic; a stubborn fraction becomes a single term once the identity is used; a multiple-angle equation becomes an ordinary one once you widen the range and divide back.

Do the reshaping in clear lines and the marks follow. It also pays to check your answer count against the shape of the problem: a quadratic in a trig ratio gives up to two ratio values, and each value that lies strictly between the limits of sine or cosine contributes two angles per turn, so a single full-range equation often has three or four solutions rather than one.

When your list is shorter than that, look first for a solution you skipped near the edges of the range.

Key method points

These three examples rehearse the mid-paper skills that separate a steady scorer from a shaky one. Keep the following points in mind as you practise more.

When an equation contains $\sin^{2}x$ or $\cos^{2}x$ and the matching first power, treat it as a quadratic, substitute $s$ , factorise, then solve each simple equation in the given range.
To prove an identity, work one side only, put fractions over a common denominator, expand fully, and look for $\sin^{2}\theta+\cos^{2}\theta$ to replace with $1$ .
For a multiple angle such as $2x$ , first widen the interval (here to $0^{\circ}$ – $720^{\circ}$ ), find all solutions there, then divide back by the multiplier.
Reading amplitude $|a|$ and period $\dfrac{360^{\circ}}{b}$ tells you how many cycles fill the range, which is also the number of times a horizontal line can cut the curve.
A value of $-1$ or $+1$ for sine or cosine gives exactly one solution per turn; a value strictly between gives two per turn.
Keep every substitution and factorising line, with analytic marking a clear method line still earns method marks even if the final digit slips.

How a teacher helps

The commonest mid-paper losses here are quiet ones: forgetting to widen the range for a $2x$ equation, so half the answers vanish; or hunting for a shortcut in a proof instead of combining the fractions and trusting the identity. In a one-to-one lesson our teacher spots the missing solutions or the stalled proof immediately and shows the reliable route, so the habit forms correctly rather than by luck.

Because our teachers are experienced, you get someone who explains why the interval doubles, not just the steps. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same either way.

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Frequently asked questions

Why does the interval become

0^{\circ}

720^{\circ}

for

\sin 2x

You are solving for the angle $2x$ , not $x$ . If $x$ ranges over $0^{\circ}$ to $360^{\circ}$ , then $2x$ ranges over twice that, $0^{\circ}$ to $720^{\circ}$ .

Find every $2x$ in the wider range, then divide by $2$ . Skipping this step is the usual reason answers go missing.

How do I know a trig equation can be treated as a quadratic?

Look for a squared ratio and its first power in the same equation, such as $2\sin^{2}x+\sin x-1=0$ . Substitute a single letter for the ratio, factorise or use the quadratic formula, then translate each root back and solve for the angle in the given range.

When proving an identity, which side should I start from?

Start from the more complicated side, usually the one with fractions or higher powers, and simplify it towards the other. Combine fractions over a common denominator, expand, and use $\sin^{2}\theta+\cos^{2}\theta=1$ .

Do not move terms across the equals sign as if solving an equation.

How many times can a line cut

y=2\sin 2x

between

0^{\circ}

and

360^{\circ}

The period is $180^{\circ}$ , so two full cycles fit the range. A horizontal line at a height inside the amplitude, such as $y=1$ , crosses each cycle twice, giving four intersections, which matches the four solutions of $2\sin 2x=1$ .

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Trigonometric Functions, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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