Progressive Trigonometry practice questions

Use Clevolab for progressive practice in trigonometry. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.

Maths Progressive Trigonometry

About this topic

Clevolab treats trigonometry as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.

The current Progressive set covers triangle problems, angle reasoning, trigonometric ratios, and standard trigonometric methods. 143 reviewed questions currently published for this page.

What you can practise

  • Sine, cosine, and tangent
  • Right-angled triangle methods
  • Sine rule and cosine rule
  • Angle problems
  • Practical geometry links

Real sample questions from the current set

These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.

Level 1

Sample question

Which ratio is $\tan\theta$?

  • Ahypotenuse over adjacentAnswer option
  • Bopposite over adjacentCorrect answer
  • Cadjacent over hypotenuseAnswer option
  • Dhypotenuse over oppositeAnswer option

Why this answer is right

Tangent measures the steepness as opposite over adjacent. This comes from SOH CAH TOA, where TOA means tangent equals opposite over adjacent.

From SOH CAH TOA, TOA states the tangent relationship. $$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}.$$ It compares rise (opposite) to run (adjacent) in a right triangle.

Level 3

Sample question

Express $y=2\cos\!\left(x+\tfrac{\pi}{2}\right)$ in terms of $\sin x$.

  • A$2\sin x$Answer option
  • B$-2\cos x$Answer option
  • C$2\cos x$Answer option
  • D$-2\sin x$Correct answer

Why this answer is right

Use the shift identity $\cos\!\left(x+\tfrac{\pi}{2}\right)=-\sin x$. Thus $y=2\cdot(-\sin x)=-2\sin x$.

The angle addition identity gives $$\cos\!\left(x+\tfrac{\pi}{2}\right)=\cos x\cos\tfrac{\pi}{2}-\sin x\sin\tfrac{\pi}{2}=0-\sin x=-\sin x.$$ Multiplying by $2$ yields $$y=2\cos\!\left(x+\tfrac{\pi}{2}\right)=-2\sin x.$$

Level 5

Sample question

Solve for all real $x$: $\cos 2x=\sin x$.

  • A$x=\tfrac{\pi}{6}+2k\pi$ or $x=\tfrac{5\pi}{6}+2k\pi$Answer option
  • B$x=k\pi$Answer option
  • C$x=\tfrac{\pi}{6}+2k\pi$, $x=\tfrac{5\pi}{6}+2k\pi$, or $x=\tfrac{3\pi}{2}+2k\pi$Correct answer
  • D$x=\tfrac{\pi}{2}+k\pi$Answer option

Why this answer is right

Rewrite $\cos 2x=1-2\sin^2 x$. Solve $1-2\sin^2 x=\sin x$, i.e. $2y^2+y-1=0$ with $y=\sin x$. Then $y=\tfrac{1}{2}$ or $y=-1$, so $x=\tfrac{\pi}{6},\tfrac{5\pi}{6}$ or $x=\tfrac{3\pi}{2}$ modulo $2\pi$.

Use the double-angle identity $\cos 2x=1-2\sin^2 x$ and set $y=\sin x$.

How this page fits into Clevolab

Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.

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