OMPT drill — Trigonometry

OMPT trigonometric identities practice: lesson and 10 problems

The Pythagorean identity and the double-angle formulas do nearly all the work. The skill is spotting which rewrite turns the question into arithmetic. Keep a unit circle sketch next to you for the first pass, then redo the set without it.

Tested inOMPT-B OMPT-D

OMPT trigonometric identities: the lesson

What the OMPT tests in trigonometric identities

An identity holds for every angle, so it is a rewriting rule rather than a puzzle to solve. The one that carries this topic comes from the unit circle: the point $(cos x, sin x)$ sits at distance one from the origin, so $sin^{2} x + cos^{2} x = 1$ for every $x$ . That line lets you trade sine for cosine and back whenever one of them is in the way. Add the double-angle and addition formulas and you have the entire working set the OMPT draws on.

The chapter sits on the OMPT-B and the OMPT-D only. On the OMPT-B the questions are short: one ratio and a quadrant given, another ratio asked for; a fraction to simplify down to a single function; or the true identity to pick out of four candidates. On the OMPT-D the identity is rarely the question itself. It is the step that turns $sin^{2} x$ into something you can integrate, or $sin 2 x = cos x$ into something you can factor. The students who struggle here do not lack formulas. They fail to recognise $1 - cos^{2} x$ as $sin^{2} x$ in disguise, and that recognition only comes from reps.

Notation and answer format the OMPT expects

Answers go into a box, and the platform wants an exact form unless the question says to round.

Type fractions as fractions: $- 3/4$ , not $- 0.75$ . Roots are typed with sqrt, so $\frac{6 - 2}{4}$ becomes (sqrt(6)-sqrt(2))/4.
Settle the sign of a square root from the quadrant before you type. The box takes one number, not $\pm$ .
A simplification question wants the shortest form: $tan x$ rather than $\frac{s i n x}{c o s x}$ .
Angles are in radians unless the question shows a degree sign, and $π$ is typed as pi. An identity with a fraction in it only holds where the denominator is non-zero, and the OMPT-D sometimes asks you to say so.

Worked example 1: from one ratio to all of them

Given $sin x = \frac{3}{5}$ with $x$ in the second quadrant, find $cos x$ and $tan x$ .
Pythagorean identity: $cos^{2} x = 1 - (\frac{3}{5})^{2} = 1 - \frac{9}{25} = \frac{16}{25}$ .
So $cos x = \pm \frac{4}{5}$ , and here the quadrant information earns its place in the question. Second quadrant means cosine is negative: $cos x = - \frac{4}{5}$ .
Then $tan x = \frac{s i n x}{c o s x} = \frac{3/5}{- 4/5} = - \frac{3}{4}$ .
The quadrant step is not decoration. Answering $+ \frac{4}{5}$ is the wrong answer I see most, and the question was built to catch it.

Worked example 2: simplifying with the identity

Simplify $\frac{s i n ^{2} x}{1 - c o s x}$ , for values of $x$ where $cos x \neq = 1$ .
Spot the disguise: $sin^{2} x = 1 - cos^{2} x$ , and that is a difference of squares.
Factor it: $1 - cos^{2} x = (1 - cos x) (1 + cos x)$ .
Cancel the common factor with the denominator: $\frac{( 1 - c o s x ) ( 1 + c o s x )}{1 - c o s x} = 1 + cos x$ .
Check with a landmark angle: at $x = \frac{π}{2}$ the original is $\frac{1}{1 - 0} = 1$ and $1 + cos \frac{π}{2} = 1$ . This is difference-of-squares factoring in trigonometric clothes, which is how most identity questions are built.

The full lesson lays out the working set with the rule for choosing from it, reaches the double angles straight from a tangent, works an addition-formula question of the OMPT-B kind, and finishes with the two OMPT-D shapes: squaring a sum to reach a product, and collapsing a mixed expression to a single cosine. It closes with the slips that cost the most marks.

How each OMPT variant tests trigonometric identities

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Not tested
The OMPT-A has no trigonometry chapter, so no identities appear; spend the time on exponents and roots, where the same rewriting instinct is tested with powers instead of sines.
OMPT-B
64 questions
Testedroughly 2 to 3 of 64 questions
Official chapter: Trigonometry, 15% of the paper.
Short numeric or multiple-choice items: one ratio and a quadrant given, another ratio or a double angle asked for; a fraction to simplify to a single function; the true identity to pick out of four.
OMPT-C
30 questions
Not tested
The OMPT-C geometry chapter uses right-triangle ratios but never asks for an identity; drill the geometry topic and the exact values instead.
OMPT-D
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Trigonometry, 15% of the paper.
Rarely a standalone item. The identity is the hinge inside an integration or equation question: rewrite a squared sine as a cosine of a double angle, or a sum as a product, then carry on.
OMPT-E
40 questions
Not tested
The OMPT-E replaces trigonometry with statistics and probability, so no identities appear; use the time on probability and random variables.
OMPT-F
21 questions
Not tested
The OMPT-F has six chapters and trigonometry is not one of them; its differentiation questions use polynomials, exponentials and logarithms only.
OMPT-G
26 questions
Not tested
The OMPT-G trigonometry chapter stays with functions, exact values and equations for physics problems; identities beyond the Pythagorean one are not asked, so drill trig equations and angles instead.

OMPT trigonometric identities practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Numeric answer

Given $sin θ = \frac{2}{3}$ , find $cos 2 θ$ as a decimal rounded to two decimals.

Show the worked solution

Answer: 0.11

I pick the double-angle form that only needs sine: $cos 2 θ = 1 - 2 sin^{2} θ$ . With $sin^{2} θ = \frac{4}{9}$ , that gives $1 - \frac{8}{9} = \frac{1}{9} \approx 0.11$ . Notice the quadrant of $θ$ is irrelevant here because sine only appears squared.

Problem 2Multiple choice

Simplify $\frac{1 - cos ^{2} x}{sin x cos x}$ for values where it is defined.

A $tan x$
B $cot x$
C $sin x$
D $sec x$

Show the worked solution

Answer: A — $tan x$

The Pythagorean identity turns the numerator into $sin^{2} x$ . Then $\frac{sin ^{2} x}{sin x cos x} = \frac{sin x}{cos x} = tan x$ .

Problem 3Multiple choice

Which of the following is a valid identity for all $x$ ?

A $sin 2 x = 2 sin x$
B $sin 2 x = 2 sin x cos x$
C $sin 2 x = sin^{2} x + cos^{2} x$
D $sin 2 x = 2 cos^{2} x - 1$

Show the worked solution

Answer: B — $sin 2 x = 2 sin x cos x$

The double-angle formula for sine is $sin 2 x = 2 sin x cos x$ . The first option treats sine as linear, which it is not; the third equals 1 for every $x$ ; the fourth is actually the formula for $cos 2 x$ .

Problem 4Numeric answer

Use an addition formula to find the exact value of $cos 7 5^{\circ}$ , then give it as a decimal rounded to two decimals.

Show the worked solution

Answer: 0.26

I write $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ and expand: $cos (4 5^{\circ} + 3 0^{\circ}) = cos 4 5^{\circ} cos 3 0^{\circ} - sin 4 5^{\circ} sin 3 0^{\circ} = \frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} = \frac{6 - 2}{4} \approx 0.26$ .

Problem 5Spot the error

A student derives a formula for $sin^{2} x$ . Step 1: start from $cos 2 x$ . Step 2: substitute $cos 2 x = 1 - sin^{2} x$ . Step 3: rearrange to $sin^{2} x = 1 - cos 2 x$ . Which step contains the error?

Show the worked solution

Answer: Step 2

The double-angle identity is $cos 2 x = 1 - 2 sin^{2} x$ ; Step 2 dropped the factor 2. With the correct identity, rearranging gives $sin^{2} x = \frac{1 - c o s 2 x}{2}$ . Step 3 rearranges the (wrong) equation correctly, so the fault lies in Step 2 alone.

Problem 6Numeric answer

Given $tan x = 2$ , evaluate $\frac{sin x + cos x}{sin x - cos x}$ .

Show the worked solution

Answer: 3

Dividing numerator and denominator by $cos x$ turns everything into tangents: $\frac{tan x + 1}{tan x - 1} = \frac{2 + 1}{2 - 1} = 3$ . No need to find $sin x$ or $cos x$ individually.

Problem 7Numeric answer

Given $sin x = 0.6$ with $x$ in the second quadrant, find $sin 2 x$ .

Show the worked solution

Answer: -0.96

I need $cos x$ first. From $cos^{2} x = 1 - 0.36 = 0.64$ I get $cos x = \pm 0.8$ , and the second quadrant makes it $- 0.8$ . Then $sin 2 x = 2 sin x cos x = 2 \cdot 0.6 \cdot (- 0.8) = - 0.96$ . Check: $x$ is between $9 0^{\circ}$ and $18 0^{\circ}$ , so $2 x$ is between $18 0^{\circ}$ and $36 0^{\circ}$ , where sine is negative. The sign agrees.

Problem 8Multiple choice

Simplify $\frac{sin 2 x}{1 + cos 2 x}$ for values of $x$ where it is defined.

A $tan x$
B $cot x$
C $2 tan x$
D $sin x$

Show the worked solution

Answer: A — $tan x$

I pick the cosine double-angle form that swallows the 1: $1 + cos 2 x = 1 + (2 cos^{2} x - 1) = 2 cos^{2} x$ . The numerator is $2 sin x cos x$ . Dividing, $\frac{2 sin x cos x}{2 cos ^{2} x} = \frac{sin x}{cos x} = tan x$ , valid wherever $cos x \neq = 0$ . Check at $x = 4 5^{\circ}$ : $\frac{sin 9 0 ^{\circ}}{1 + cos 9 0 ^{\circ}} = \frac{1}{1} = 1 = tan 4 5^{\circ}$ .

Problem 9Numeric answer

Given $sin x - cos x = 0.5$ , find $sin 2 x$ .

Show the worked solution

Answer: 0.75

Squaring both sides gives $sin^{2} x - 2 sin x cos x + cos^{2} x = 0.25$ . The two squares add to 1, so $1 - 2 sin x cos x = 0.25$ , which means $2 sin x cos x = 0.75$ . That left side is $sin 2 x$ , so $sin 2 x = 0.75$ . Check that the data is possible: $sin x - cos x$ ranges between $- 2$ and $2$ , and 0.5 sits inside, and $0.75$ is a legal value for a sine.

Problem 10Numeric answer

Let $f (x) = 2 sin^{2} x - 4 cos^{2} x$ . Rewrite $f$ in the form $a + b cos 2 x$ , then give the maximum value of $f$ .

Show the worked solution

Answer: 2

I remove one square with the Pythagorean identity: $2 sin^{2} x = 2 - 2 cos^{2} x$ , so $f (x) = 2 - 6 cos^{2} x$ . Then the backwards double angle, $2 cos^{2} x = 1 + cos 2 x$ , turns $6 cos^{2} x$ into $3 + 3 cos 2 x$ , giving $f (x) = - 1 - 3 cos 2 x$ . The cosine runs from $- 1$ to $1$ , so $f$ runs from $- 4$ to $2$ , and the maximum is 2. Check at $x = 9 0^{\circ}$ : $2 \cdot 1 - 4 \cdot 0 = 2$ , and $cos 18 0^{\circ} = - 1$ gives $- 1 + 3 = 2$ from the rewritten form.

OMPT trigonometric identities: questions students ask

Are trigonometric identities on the OMPT-A?

No. The OMPT-A has eight chapters and none of them is trigonometry, so identities, double angles and addition formulas do not appear. If your programme asks for the OMPT-A, skip this topic and put the hours into algebra, functions and differentiation. Identities are tested on the OMPT-B and used as a working step on the OMPT-D.

Which trig identities do I need to know for the OMPT-B?

A short list: the Pythagorean identity and its two rearrangements, the tangent definition, the three forms of the cosine double angle, the sine double angle, and the addition formulas for sine and cosine. Everything else on the paper, including half angles and the tangent forms of the double angles, is derived from those in two lines, and the lesson shows each derivation.

How do I find cos x when I am given sin x on the OMPT?

Use the Pythagorean identity: cosine squared equals one minus sine squared. Take the square root, then pick the sign from the quadrant the question names. Second and third quadrant give a negative cosine, first and fourth a positive one. The OMPT builds these questions around 3, 4, 5 and 5, 12, 13 triangles, so the root is always clean, and the mark is lost on the sign rather than the arithmetic.

Do I need a calculator for trig identities on the OMPT?

No, and none is provided on the maths variants unless a question supplies one. Identity questions are built from exact fractions and the special angles, so the answers are values like 7/25 or (sqrt(6) minus sqrt(2))/4, typed exactly. A decimal is only expected when the question says to round, and even then the exact form comes first in your working.

What is the difference between a trig identity and a trig equation on the OMPT?

An identity is true for every angle, so the task is to rewrite or to prove, never to solve. An equation is true only for some angles, and the task is to find them, usually on a given interval. The two meet on the OMPT-B and OMPT-D: an equation such as sin 2x = cos x is solved by first applying the double-angle identity and then factoring, so the identity is the first step of the equation.

How do I check whether an identity is true on the OMPT?

Substitute a landmark angle into both sides. Pick one where every function is easy, such as 30 degrees, 45 degrees or 90 degrees, and compare the numbers. A single mismatch proves the identity false. This takes ten seconds and catches the usual fakes, such as sin 2x = 2 sin x, which fails at 90 degrees because the left side is 0 and the right side is 2.

Test hubs that cover trigonometric identities:OMPT-B OMPT-D All 28 topics

OMPT trigonometric identities practice: lesson and 10 problems

OMPT trigonometric identities: the lesson

What the OMPT tests in trigonometric identities

Notation and answer format the OMPT expects

Worked example 1: from one ratio to all of them

Worked example 2: simplifying with the identity

How each OMPT variant tests trigonometric identities

OMPT trigonometric identities practice problems

OMPT trigonometric identities: questions students ask

Related OMPT topics and test hubs