When do I convert \cos^{2}x to \sin x (or the other way)?

When an equation mixes a squared ratio with a first-power ratio, use sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1sin2x+cos2x=1 to write everything in one ratio. That turns it into a quadratic you can factorise.

Trigonometric Functions Practice

Q: How do I know which quadrants give the answers?

The sign of the trig ratio decides them. Sine is positive in the first and second quadrants; cosine is positive in the first and fourth; tangent is positive in the first and third. Find the basic angle from the positive value, then place the answers in the quadrants that match the sign.

Q: Which form of \cos 2\theta should I use?

All three are correct: cos⁡2θ−sin⁡2θ\cos^{2}\theta-\sin^{2}\thetacos2θ−sin2θ, 1−2sin⁡2θ1-2\sin^{2}\theta1−2sin2θ, and 2cos⁡2θ−12\cos^{2}\theta-12cos2θ−1. Pick the one that matches the ratio you already know, if you know sin⁡θ\sin\thetasinθ, use 1−2sin⁡2θ1-2\sin^{2}\theta1−2sin2θ.

Q: Does a wrong final answer cost me every mark?

No. Because marking is analytic, a correct basic angle and correct quadrant reasoning still earn marks even if you miss one solution. Always show the basic angle before listing the answers in the range.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from reading exact ratios off a right triangle to using the double-angle formulae. Give yourself roughly four to seven minutes per question and work on paper first, writing every line the way you would in the real exam, quote the identity or rule you are using, find the basic angle, then decide the correct quadrants before you list every answer in the given range.

Resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, a correct basic angle still earns method marks even when you miss one of the answers in the range, so always show the basic angle and the quadrant reasoning clearly.

Six practice questions

Q1[2 marks]

Given that $\theta$ is acute and $\sin\theta=\dfrac{3}{5}$ , find the exact values of $\cos\theta$ and $\tan\theta$ .

Show worked solution

Sketch a right triangle. Since $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{3}{5}$ , the opposite side is $3$ and the hypotenuse is $5$ .

Find the adjacent side by Pythagoras:

\text{adjacent}=\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}=4

Because $\theta$ is acute, all ratios are positive:

\cos\theta=\frac{4}{5},\qquad \tan\theta=\frac{3}{4}

Answer

$\cos\theta=\dfrac{4}{5}$ and $\tan\theta=\dfrac{3}{4}$ . Check with the identity: $\sin^{2}\theta+\cos^{2}\theta=\dfrac{9}{25}+\dfrac{16}{25}=1$ .

Q2[3 marks]

Given that $\cos A=-\dfrac{5}{13}$ and $A$ is obtuse $(90^{\circ}<A<180^{\circ})$ , find the exact values of $\sin A$ and $\tan A$ .

Show worked solution

Use the identity $\sin^{2}A+\cos^{2}A=1$ to find $\sin A$ :

\sin^{2}A=1-\left(-\tfrac{5}{13}\right)^{2}=1-\frac{25}{169}=\frac{144}{169}

\sin A=\pm\frac{12}{13}

An obtuse angle lies in the second quadrant, where sine is positive, so we take the positive value. Then $\tan A=\dfrac{\sin A}{\cos A}$ :

\sin A=\frac{12}{13},\qquad \tan A=\frac{12/13}{-5/13}=-\frac{12}{5}

Answer

$\sin A=\dfrac{12}{13}$ and $\tan A=-\dfrac{12}{5}$ . The quadrant is what fixes the sign: in the second quadrant sine is positive but tangent is negative.

Q3[3 marks]

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

Show worked solution

First isolate $\sin x$ :

2\sin x-1=0 \;\Rightarrow\; \sin x=\frac{1}{2}

The basic (reference) angle is $\sin^{-1}\tfrac{1}{2}=30^{\circ}$ . Sine is positive in the first and second quadrants, so:

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

Answer

$x=30^{\circ}$ or $x=150^{\circ}$ . Check: $\sin 30^{\circ}=\sin 150^{\circ}=\tfrac{1}{2}$ .

Deciding the quadrants from the sign of $\sin x$ is what gives you both answers.

Q4[4 marks]

Prove the identity $\dfrac{1-\cos^{2}\theta}{\sin\theta\cos\theta}=\tan\theta$ .

Show worked solution

Work on the left-hand side and aim to reach the right-hand side. Start from the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta=1$ , which rearranges to $1-\cos^{2}\theta=\sin^{2}\theta$ .

Replace the numerator:

\text{LHS}=\frac{1-\cos^{2}\theta}{\sin\theta\cos\theta}=\frac{\sin^{2}\theta}{\sin\theta\cos\theta}

Cancel one factor of $\sin\theta$ , then use $\dfrac{\sin\theta}{\cos\theta}=\tan\theta$ :

=\frac{\sin\theta}{\cos\theta}=\tan\theta=\text{RHS}

Answer

Proved. The key move is recognising $1-\cos^{2}\theta=\sin^{2}\theta$ ; after that, cancelling one $\sin\theta$ leaves $\tan\theta$ .

Always transform one side only and finish by writing $=\text{RHS}$ .

Q5[5 marks]

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

Show worked solution

The equation mixes $\cos^{2}x$ and $\sin x$ , so convert everything to $\sin x$ using $\cos^{2}x=1-\sin^{2}x$ :

2(1-\sin^{2}x)+\sin x-1=0

2-2\sin^{2}x+\sin x-1=0 \;\Rightarrow\; 2\sin^{2}x-\sin x-1=0

This is a quadratic in $\sin x$ . Factorise:

(2\sin x+1)(\sin x-1)=0 \;\Rightarrow\; \sin x=-\frac{1}{2}\ \text{ or }\ \sin x=1

For $\sin x=1$ : $x=90^{\circ}$ . For $\sin x=-\tfrac{1}{2}$ , the basic angle is $30^{\circ}$ and sine is negative in the third and fourth quadrants:

x=180^{\circ}+30^{\circ}=210^{\circ} \quad\text{or}\quad x=360^{\circ}-30^{\circ}=330^{\circ}

Answer

$x=90^{\circ},\ 210^{\circ},\ 330^{\circ}$ . Check $x=210^{\circ}$ : $2\cos^{2}210^{\circ}+\sin 210^{\circ}-1=2(\tfrac{3}{4})-\tfrac{1}{2}-1=0$ .

Converting to a single ratio before factorising is the method mark.

Q6[6 marks]

Given that $\sin\theta=\dfrac{4}{5}$ where $\theta$ is acute, find the exact values of (a) $\sin 2\theta$ , (b) $\cos 2\theta$ , (c) $\tan 2\theta$ .

Show worked solution

First find $\cos\theta$ . From a $3$ – $4$ – $5$ triangle (opposite $4$ , hypotenuse $5$ , adjacent $3$ ), and since $\theta$ is acute, $\cos\theta=\dfrac{3}{5}$ .

(a) Use $\sin 2\theta=2\sin\theta\cos\theta$ :

\sin 2\theta=2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)=\frac{24}{25}

(b) Use $\cos 2\theta=1-2\sin^{2}\theta$ :

\cos 2\theta=1-2\left(\frac{4}{5}\right)^{2}=1-\frac{32}{25}=-\frac{7}{25}

\tan 2\theta=\frac{24/25}{-7/25}=-\frac{24}{7}

Answer

(a) $\dfrac{24}{25}$ , (b) $-\dfrac{7}{25}$ , (c) $-\dfrac{24}{7}$ . Check: $\sin^{2}2\theta+\cos^{2}2\theta=\dfrac{576}{625}+\dfrac{49}{625}=1$ .

A negative $\cos 2\theta$ tells you $2\theta$ is obtuse even though $\theta$ is acute.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for a correct basic angle, the right quadrants chosen, a clean use of an identity, and every answer inside the stated range.

Ratio mark: did you use $\sin^{2}\theta+\cos^{2}\theta=1$ or a right triangle correctly to find the missing ratio?
Quadrant mark: did the sign of the ratio decide the correct quadrants, and did you fix the sign of the final answer accordingly?
Basic-angle mark: did you find the reference angle first, then build the answers as $180^{\circ}\pm$ or $360^{\circ}-$ the basic angle?
Identity mark: in a proof, did you transform one side only and reach the other, quoting each identity you used?
If your basic angle is right but you dropped one answer in the range, give yourself the method marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, choosing the wrong quadrant, forgetting the second solution in the range, or mixing up which double-angle form of $\cos 2\theta$ to use.

Because our teachers are experienced, you work with someone who explains the why behind each identity. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How do I know which quadrants give the answers?

The sign of the trig ratio decides them. Sine is positive in the first and second quadrants; cosine is positive in the first and fourth; tangent is positive in the first and third.

Find the basic angle from the positive value, then place the answers in the quadrants that match the sign.

When do I convert

\cos^{2}x

\sin x

(or the other way)?

When an equation mixes a squared ratio with a first-power ratio, use $\sin^{2}x+\cos^{2}x=1$ to write everything in one ratio. That turns it into a quadratic you can factorise.

Which form of

\cos 2\theta

should I use?

All three are correct: $\cos^{2}\theta-\sin^{2}\theta$ , $1-2\sin^{2}\theta$ , and $2\cos^{2}\theta-1$ . Pick the one that matches the ratio you already know, if you know $\sin\theta$ , use $1-2\sin^{2}\theta$ .

Does a wrong final answer cost me every mark?

No. Because marking is analytic, a correct basic angle and correct quadrant reasoning still earn marks even if you miss one solution.

Always show the basic angle before listing the answers in the range.

Source:SRC-DSKP-EN

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

Trigonometric Functions, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

Ready to get started?