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Deduce exact trigonometric ratios from one ratio

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Deducing trigonometric ratios

Choose side lengths proportional to the given ratio in a right-angled triangle. Use the acute angle to name the sides, find the missing length and form the requested ratio

  1. 1Name the side roles Use the acute angle as the reference. The hypotenuse is opposite the right angle; the other sides are opposite or adjacent to the acute angle \(\displaystyle \begin{gathered}\sin\theta=\frac{O}{H}\\\cos\theta=\frac{A}{H}\\\tan\theta=\frac{O}{A}\end{gathered}\)
  2. 2Choose a proportional triangle A ratio fixes relative side lengths. Choose the numerator and denominator as convenient side lengths for a similar triangle \(\displaystyle \sin\theta=\frac{a}{b}\quad\Longrightarrow\quad O=a,\ H=b\)
  3. 3Use Pythagoras Subtract a known leg square from the hypotenuse square to find the other leg square. Add the two leg squares when the hypotenuse is unknown \(\displaystyle H^2=O^2+A^2\)
  4. 4Take a positive length A side length is positive. Take the positive square root, then form the requested ratio from the named sides \(\displaystyle \begin{gathered}A=\sqrt{H^2-O^2}\\\cos\theta=\frac{A}{H},\qquad\tan\theta=\frac{O}{A}\end{gathered}\)
Year 11, VCE Units 1 & 2, Chapter 9, 4 difficulty levels, No calculatorBuild a worksheet

Deducing trigonometric ratios

Name the sides relative to the acute angle, choose a proportional triangle and keep exact values throughout

  1. Name the side roles Use the acute angle as the reference. The hypotenuse is opposite the right angle; the other sides are opposite or adjacent to the acute angle \(\displaystyle \begin{gathered}\sin\theta=\frac{O}{H}\\\cos\theta=\frac{A}{H}\\\tan\theta=\frac{O}{A}\end{gathered}\)
  2. Choose a proportional triangle A ratio fixes relative side lengths. Choose the numerator and denominator as convenient side lengths for a similar triangle \(\displaystyle \sin\theta=\frac{a}{b}\quad\Longrightarrow\quad O=a,\ H=b\)
  3. Use Pythagoras Subtract a known leg square from the hypotenuse square to find the other leg square. Add the two leg squares when the hypotenuse is unknown \(\displaystyle H^2=O^2+A^2\)
  4. Take a positive length A side length is positive. Take the positive square root, then form the requested ratio from the named sides \(\displaystyle \begin{gathered}A=\sqrt{H^2-O^2}\\\cos\theta=\frac{A}{H},\qquad\tan\theta=\frac{O}{A}\end{gathered}\)

Worked example

An acute angle in a right-angled triangle has the stated sine ratio. Find its exact cosine and tangent ratios

Right-angled triangle with acute angle theta at the left vertex. A is the adjacent horizontal leg, O is the opposite vertical leg, and H is the sloping hypotenuse opposite the right angleAOHθ
Relative to θ: A is adjacent, O is opposite and H is the hypotenuse. Letters label lengths; no numerical scale is implied
\[\sin\theta=\frac{28}{53}\]
  1. \(\displaystyle \sin\theta=\frac{O}{H}=\frac{28}{53}\)Choose a similar triangle with these proportional side lengths
  2. \(\displaystyle O=28,\qquad H=53\)The sine ratio gives the opposite side and hypotenuse
  3. \(\displaystyle A^2=53^2-28^2\)Subtract the opposite-side square from the hypotenuse square
  4. \(\displaystyle A^2=2809-784=2025\)Evaluate the squares and subtract
  5. \(\displaystyle A=\sqrt{2025}=45\)Use the positive root for a side length
  6. \(\displaystyle \begin{gathered}\cos\theta=\frac{45}{53}\\\tan\theta=\frac{28}{45}\end{gathered}\)Form the requested exact ratios; both fractions are in lowest terms

Common mistakes

Opposite and adjacent sides swapped
✕\(\displaystyle \sin\theta=\frac{28}{53}\quad\Longrightarrow\quad A=28,\ H=53\)✓\(\displaystyle \sin\theta=\frac{28}{53}\quad\Longrightarrow\quad O=28,\ H=53\)
Sine identifies the opposite side and hypotenuse. Calling its numerator the adjacent side swaps the leg roles
Skipping the square root
✕\(\displaystyle A^2=49\quad\Longrightarrow\quad A=49\)✓\(\displaystyle A^2=49\quad\Longrightarrow\quad A=7\)
Pythagoras first gives a square. Take its positive square root to obtain the side length
Inverting a trigonometric ratio
✕\(\displaystyle \tan\theta=\frac{A}{O}\)✓\(\displaystyle \tan\theta=\frac{O}{A}\)
Tangent is opposite divided by adjacent, so reversing the fraction changes the ratio

Difficulty levels

introUse integer side lengths and one target ratio after finding the third side
standardFind both other ratios when the missing side contains a non-square surd
challengeUse rational given side lengths and simplify exact target ratios after cancellation
examChoose the exact-form ratio produced by the correct side roles and square-root step

Questions about this topic

Why is the positive square root used for a side length?+

A length is positive. Use the positive square root of its square

\[A>0\quad\Longrightarrow\quad\sqrt{A^2}=A\]
How do I choose the sine, cosine or tangent ratio?+

Name the opposite and adjacent legs relative to the acute angle. The hypotenuse is opposite the right angle

\[\begin{gathered}\sin\theta=\frac{O}{H}\\\cos\theta=\frac{A}{H}\\\tan\theta=\frac{O}{A}\end{gathered}\]
Why are exact surd coefficients entered separately?+

The labelled coefficient fields distinguish the rational term, radical coefficient, radicand and denominator. Keep the value exact

\[\frac{P\pm Q\sqrt R}{S}\]
Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 9: Circular functions

In this chapter

Convert angle measure and use the radian arc relationLocate extended angles and generate coterminal anglesDeduce exact trigonometric ratios from one ratioUse unit-circle coordinates to evaluate circular functionsEvaluate exact circular-function values by symmetry and periodicityKey points and features of the basic sine and cosine graphsLocate and bisect a circular-equation rootUse local polynomial approximations for circular functions