PYQ Vault
Ebooks

SPI Mathematics · Trigonometric Ratios and Identities

Trigonometric Ratios of an Acute Angle

In a right triangle, each trigonometric ratio of an acute angle is one side divided by another, named by where the sides sit relative to that angle.

All Trigonometric Ratios and Identities notes

Why this matters

Six questions in the bank, five EASY and one MODERATE: name the ratio, find every ratio from one given ratio, or turn a fraction of sines and cosines into tangents. Every later page in this chapter is built on these six ratios.

Concept 1 of 3: The six ratios of a right triangle

Stand at the acute angle you care about. The longest side, across from the right angle, is the hypotenuse. The side across from your angle is the opposite side, and the third side, touching your angle, is the adjacent side. Each ratio is one of these sides over another.

Definition

For an acute angle AA of a right triangle:

  • sin⁡A=oppositehypotenuse\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}, cos⁡A=adjacenthypotenuse\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}, tan⁡A=oppositeadjacent\tan A = \dfrac{\text{opposite}}{\text{adjacent}}.
  • The other three are their reciprocals: cosec⁡A=1sin⁡A\operatorname{cosec} A = \dfrac{1}{\sin A}, sec⁡A=1cos⁡A\sec A = \dfrac{1}{\cos A}, cot⁡A=1tan⁡A\cot A = \dfrac{1}{\tan A}.
  • A missing side comes from Pythagoras: hypotenuse2=opposite2+adjacent2\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2.
  • The ratios depend only on the angle, not on how big the triangle is.

The six trigonometric ratios

sin⁡A=opphyp, cos⁡A=adjhyp, tan⁡A=oppadj, cosec⁡A=hypopp, sec⁡A=hypadj, cot⁡A=adjopp\sin A = \frac{\text{opp}}{\text{hyp}},\ \cos A = \frac{\text{adj}}{\text{hyp}},\ \tan A = \frac{\text{opp}}{\text{adj}},\ \operatorname{cosec} A = \frac{\text{hyp}}{\text{opp}},\ \sec A = \frac{\text{hyp}}{\text{adj}},\ \cot A = \frac{\text{adj}}{\text{opp}}
  • oppthe side opposite the angle
  • adjthe side next to the angle (not the hypotenuse)
  • hypthe hypotenuse, across from the right angle

Worked example

Triangle ABC has a right angle at B, with AB = 5 cm and BC = 12 cm. Find sin⁡A\sin A, cos⁡A\cos A and tan⁡A\tan A.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Trigonometric Ratios and Identities · Trigonometric Ratios of an Acute Angle

Ratios of sides of a right triangle with respect to its acute angles are known as

Opposite and adjacent change with the angle

In a right triangle ABC with the right angle at B, side BC is opposite A but adjacent to C. Name the sides again every time the question moves to the other acute angle.

Reading sin A as sin times A

sin⁡A\sin A is one number, the value of a ratio for the angle A. You cannot cancel the A in sin⁡AA\frac{\sin A}{A} or split sin⁡(A+B)\sin(A + B) into sin⁡A+sin⁡B\sin A + \sin B.
Drill 1 more on the six ratios of a right triangle

Concept 2 of 3: Finding every ratio from one given ratio

One ratio gives you two sides of the triangle. Draw it, find the third side by Pythagoras, and you can read off all six ratios. The reciprocal and quotient relations give some of them without a drawing at all.

Definition

  • Read the given ratio as two sides: sin⁡θ=513\sin\theta = \frac{5}{13} means opposite 5, hypotenuse 13.
  • Find the third side: adjacent =132−52=12= \sqrt{13^2 - 5^2} = 12.
  • Quotient relations: tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta} and cot⁡θ=cos⁡θsin⁡θ\cot\theta = \dfrac{\cos\theta}{\sin\theta}.
  • Reciprocal pairs: sin with cosec, cos with sec, tan with cot. Each pair multiplies to 1.

Reciprocal and quotient relations

sin⁡θcosec⁡θ=1,cos⁡θsec⁡θ=1,tan⁡θcot⁡θ=1,tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\sin\theta \operatorname{cosec}\theta = 1,\quad \cos\theta \sec\theta = 1,\quad \tan\theta \cot\theta = 1,\quad \tan\theta = \frac{\sin\theta}{\cos\theta},\quad \cot\theta = \frac{\cos\theta}{\sin\theta}

Worked example

If sin⁡θ=513\sin\theta = \frac{5}{13}, find cos⁡θ\cos\theta, tan⁡θ\tan\theta and sec⁡θ\sec\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Trigonometric Ratios and Identities · Trigonometric Ratios of an Acute Angle

If cosec θ=10\text{cosec}\,\theta = \sqrt{10} then sec⁡θ\sec\theta = ?

Pairing sec with sin

The names cross over: sec is the reciprocal of cos, and cosec is the reciprocal of sin. If cos⁡θ=35\cos\theta = \frac{3}{5}, then sec⁡θ=53\sec\theta = \frac{5}{3}.

Forgetting the square root

With hypotenuse 13 and one side 5, the third side is 169−25=12\sqrt{169 - 25} = 12, not 144. Take the root before you write the ratio.
Drill 1 more on finding every ratio from one given ratio

Concept 3 of 3: Dividing by cos to turn a fraction into tan

When you know tan⁡θ\tan\theta but the question is a fraction of sines and cosines, you do not need the sides at all. Divide every term, top and bottom, by cos⁡θ\cos\theta: each sin⁡θ\sin\theta becomes tan⁡θ\tan\theta and each cos⁡θ\cos\theta becomes 1.

Definition

  • Works when every term on top and bottom has exactly one sin⁡θ\sin\theta or one cos⁡θ\cos\theta, as in asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ\dfrac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta}.
  • Dividing by cos⁡θ\cos\theta gives atan⁡θ+bctan⁡θ+d\dfrac{a\tan\theta + b}{c\tan\theta + d}.
  • If you are given cot⁡θ\cot\theta instead, divide by sin⁡θ\sin\theta: each cos⁡θ\cos\theta becomes cot⁡θ\cot\theta.

A sin and cos fraction in terms of tan

asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ=atan⁡θ+bctan⁡θ+d\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}

Worked example

If tan⁡θ=2\tan\theta = 2, find 3sin⁡θ+cos⁡θsin⁡θ+2cos⁡θ\dfrac{3\sin\theta + \cos\theta}{\sin\theta + 2\cos\theta}.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 3 · Trigonometric Ratios and Identities · Trigonometric Ratios of an Acute Angle

If 2 tan A = 3, then the value of 4sin⁡A+3cos⁡A4sin⁡A−3cos⁡A\frac{4\sin A + 3\cos A}{4\sin A - 3\cos A} is

Dividing only the top

Divide the numerator and the denominator by cos⁡θ\cos\theta. Dividing only one of them changes the value of the fraction.

A plain number breaks the trick

In sin⁡θ+1cos⁡θ\dfrac{\sin\theta + 1}{\cos\theta} the 1 has no sin or cos, so dividing by cos⁡θ\cos\theta leaves 1cos⁡θ\frac{1}{\cos\theta} behind. Then you need the actual values of sin⁡θ\sin\theta and cos⁡θ\cos\theta from the sides.
Drill 1 more on dividing by cos to turn a fraction into tan

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The six ratios of a right triangle

    The six trigonometric ratios

    sin⁡A=opphyp, cos⁡A=adjhyp, tan⁡A=oppadj, cosec⁡A=hypopp, sec⁡A=hypadj, cot⁡A=adjopp\sin A = \frac{\text{opp}}{\text{hyp}},\ \cos A = \frac{\text{adj}}{\text{hyp}},\ \tan A = \frac{\text{opp}}{\text{adj}},\ \operatorname{cosec} A = \frac{\text{hyp}}{\text{opp}},\ \sec A = \frac{\text{hyp}}{\text{adj}},\ \cot A = \frac{\text{adj}}{\text{opp}}
  • Finding every ratio from one given ratio

    Reciprocal and quotient relations

    sin⁡θcosec⁡θ=1,cos⁡θsec⁡θ=1,tan⁡θcot⁡θ=1,tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\sin\theta \operatorname{cosec}\theta = 1,\quad \cos\theta \sec\theta = 1,\quad \tan\theta \cot\theta = 1,\quad \tan\theta = \frac{\sin\theta}{\cos\theta},\quad \cot\theta = \frac{\cos\theta}{\sin\theta}
  • Dividing by cos to turn a fraction into tan

    A sin and cos fraction in terms of tan

    asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ=atan⁡θ+bctan⁡θ+d\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}

Watch out for (6)

Test yourself on a real paper

Sit a past SPI paper, timed and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Sit an SPI past paperDrill all 6 questions