OMPT drill — Trigonometry

Trigonometric identities

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

Lesson

An identity is an equation that is true for every angle: a rewriting rule rather than a puzzle to solve. The one that carries this whole topic is Pythagorean in origin: the unit-circle point $(cos x, sin x)$ sits at distance one from the origin, so $sin^{2} x + cos^{2} x = 1$ , always. That single line lets you trade sine for cosine and back whenever one of them is inconvenient, which turns out to be the move behind most identity questions on the test.

The skill being examined is not memorisation: the OMPT uses a short list, and the double-angle formulas plus the Pythagorean identity do nearly all the work. The skill is recognition: staring at $\frac{1 - c o s ^{2} x}{c o s x}$ and seeing that the top is $sin^{2} x$ wearing a disguise. That pattern-spotting gets fast the same way factoring did, through reps, and the payoff shows up again in trig equations and in calculus, where the right rewrite routinely converts an impossible-looking expression into arithmetic.

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.