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Introduction to Trigonometry
Introduction
Trigonometric Ratios
Practice: Ratios in a Right Triangle
Trigonometric Ratios of Some Specific Angles
Practice: Ratios of Special Angles
Trigonometric Identities
Practice: Trigonometric Identities
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Glossary

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Introduction to TrigonometryTrigonometric Ratios

Reading time: ~50 min

Every scene in Section 8.1 came down to a right triangle, so let us study one closely: triangle ABC, right-angled at B. We keep our eye on the angle at A, \angle CAB, or simply "angle A".

The three sides get names from the point of view of angle A. The side BC faces angle A: it is the side opposite to angle A. The side AC faces the right angle and is the longest: it is the hypotenuse. The side AB is one of the arms of angle A: it is the side adjacent to angle A.

Fig. 8.4 is live. Pick a name, then tap the side of triangle ABC that it belongs to.

From now on every figure keeps the same colours: the opposite side is blue, the adjacent side is green, the hypotenuse is purple, and the angle we are looking from is marked in orange.

Change the Angle, Change the Names

Here is the catch: "opposite" and "adjacent" depend on which angle you stand at. Move to angle C of the same triangle (Fig. 8.5) and name the sides again.

The hypotenuse never moves — it always faces the right angle. But the side that was opposite to A is adjacent to C, and the other way round. So always say which angle you mean before you name a side.

In earlier classes you met the idea of a ratio. Divide one side of a right triangle by another and you get a ratio; six of these ratios are so useful that they have names of their own.

Sine, Cosine, Tangent

The trigonometric ratios of angle A in the right triangle ABC (Fig. 8.4) are defined as follows.

The sine of angle A:

\sin A = \dfrac{\text{side opposite to } A}{\text{hypotenuse}} = \dfrac{BC}{AC}

The cosine of angle A:

\cos A = \dfrac{\text{side adjacent to } A}{\text{hypotenuse}} = \dfrac{AB}{AC}

The tangent of angle A:

\tan A = \dfrac{\text{side opposite to } A}{\text{side adjacent to } A} = \dfrac{BC}{AB}

Build them yourself. A ratio card is selected — tap the side that goes on top, then the side that goes underneath. Tap a card to switch to it.

Turn each of these three fractions upside down and you get the other three ratios: the cosecant, secant and cotangent of angle A.

\operatorname{cosec} A = \dfrac{1}{\sin A} = \dfrac{AC}{BC}

\sec A = \dfrac{1}{\cos A} = \dfrac{AC}{AB}

\cot A = \dfrac{1}{\tan A} = \dfrac{AB}{BC}

In the triangle above, that makes cosec A = , sec A = and cot A = .

Six Ratios, Three Pairs

We shorten the six names to sin A, cos A, tan A, cosec A, sec A and cot A. The last three are the reciprocals of the first three, so they come in pairs: sin and cosec, cos and sec, tan and cot.

There is one more link worth spotting. Divide sin A by cos A and the hypotenuse cancels:

\dfrac{\sin A}{\cos A} = \dfrac{BC / AC}{AB / AC} = \dfrac{BC}{AB} = \tan A

So \tan A = \dfrac{\sin A}{\cos A}, and turning it upside down, cot A = .

SOH-CAH-TOA

A well-known way to remember the three main ratios:

Sine = Opposite ÷ Hypotenuse, Cosine = Adjacent ÷ Hypotenuse, Tangent = Opposite ÷ Adjacent.

The other three need no extra memory work: they are the same fractions turned upside down.

Now write the ratios for angle C of the same triangle (Fig. 8.5) — remember that opposite and adjacent swap places:

sin C = , cos C = and tan C = .

So a trigonometric ratio of an acute angle in a right triangle ties that angle to the lengths of the sides around it.

From Half-Chord to Sine

The sine, in the form we use it today, first appears in the Aryabhatiyam, written by Aryabhata around C.E. 500. He worked with half of a chord of a circle and called it ardha-jya, "half-chord", a name that soon shrank to jya or jiva.

Then the word travelled. Arabic scholars who translated the Aryabhatiyam kept jiva as it was. When those Arabic texts were later put into Latin, the word was rendered sinus, meaning a curve or a fold — and sinus became the English sine. The English astronomy professor Edmund Gunter (1581–1626) was the first to write the short form "sin".

The cosine came much later, out of the need to find the sine of the complementary angle. Aryabhata called it kotijya; the name cosinus is Gunter's; and in 1674 the English mathematician Sir Jonas Moore first wrote "cos".

Put the journey of the word "sine" in order, earliest at the top.

Arabic translators keep the word jiva
Edmund Gunter writes the short form "sin"
Aryabhata writes ardha-jya, the half-chord
Latin translators turn it into sinus
The word shrinks to jya or jiva

A word born in India around C.E. 500 now sits on every scientific calculator in the world.

sin A Is One Symbol

Riya writes "sin A" in her notebook, and her friend Aman asks, "So that means sin times A?" What should Riya tell him?

Yes — sin is a number and A is an angle, so sin A is their product
sin A is one symbol: the sine of the angle A, a single number
sin A means the angle A divided by sin

Note

sin A is shorthand for "the sine of angle A". The letters "sin" on their own, separated from A, mean nothing, so sin A is not a product of "sin" and A. The same goes for cos A, tan A and the other three ratios.

Does the Size of the Triangle Matter?

Take triangle ABC again, right-angled at B. Choose a point P on the hypotenuse AC and drop the perpendicular PM to AB — or choose a point Q on AC extended beyond C and drop QN to AB extended. Now three right triangles share angle A: PAM, CAB and QAN. Do they give angle A three different sines?

Fig. 8.6 is live. Drag the point along the line AC and complete all three missions. Watch the side lengths change — and watch the ratios.

The lengths changed every time, but the three ratios did not budge. Here is why. Triangles PAM and CAB share angle A and both have a right angle, so by the similarity criterion they are similar. In similar triangles the corresponding sides are :

\dfrac{AM}{AB} = \dfrac{AP}{AC} = \dfrac{MP}{BC}

Rearranging these equal fractions gives \dfrac{MP}{AP} = \dfrac{BC}{AC}, and the right-hand side is exactly .

In the same way,

\dfrac{AM}{AP} = \dfrac{AB}{AC} = \cos A, and \dfrac{MP}{AM} = \dfrac{BC}{AB} = \tan A

and triangle QAN behaves just the same.

The values of the trigonometric ratios of an angle do not depend on the lengths of the sides of the triangle — only on the angle.

Tip

Think of a ratio as the triangle's shape, not its size. Double every side and a ratio like BC/AC doubles on top and on the bottom, so its value does not change. That is why 30° has exactly one sine, whichever right triangle you measure it in.

Note

We write \sin^2 A for (\sin A)^2, \cos^2 A for (\cos A)^2, and so on. But \operatorname{cosec} A = (\sin A)^{-1} is not the same as \sin^{-1} A — that symbol means "sine inverse A", an idea for higher classes. We will also often use the Greek letter \theta (theta) for an angle.

One Ratio Is Enough

Know one trigonometric ratio of an angle, and you can find all six. Suppose that in a right triangle ABC, \sin A = \frac{1}{3}. That says \frac{BC}{AC} = \frac{1}{3}: the sides BC and AC are in the ratio 1 : 3. So we can take BC = k and AC = 3k, where k is any positive number (Fig. 8.7).

The third side comes from the Pythagoras theorem:

AB^2 = AC^2 - BC^2 = (3k)^2 - k^2 = 8k^2

So AB = \pm 2\sqrt{2}\,k. Which one do we keep?

So AB = 2\sqrt{2}\,k, and now cos A = AB/AC = and tan A = BC/AB = .

Tip

The hypotenuse is the longest side of a right triangle, and sin A and cos A each divide a shorter side by it. So for an acute angle A, sin A and cos A are always less than 1.

Example 1 : Given \tan A = \frac{4}{3}, find the other trigonometric ratios of the angle A.

Solution : Draw a right triangle ABC, right-angled at B (Fig. 8.8). Since \tan A = \frac{BC}{AB} = \frac{4}{3}, take BC = 4k and AB = 3k for some positive number k.

By the Pythagoras theorem, AC^2 = AB^2 + BC^2 = (3k)^2 + (4k)^2 = 25k^2, so AC = .

Now read every ratio straight off the triangle:

\sin A = \frac{BC}{AC} = \frac{4k}{5k} = and \cos A = \frac{AB}{AC} = \frac{3k}{5k} =

and their partners: cot A = , cosec A = and sec A = .

Example 2 : If ∠B and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.

Solution : Take two right triangles ABC and PQR, right-angled at C and R, in which sin B = sin Q (Fig. 8.9).

In each triangle, the sine compares the opposite side with the hypotenuse:

\sin B = \dfrac{AC}{AB} and \sin Q = \dfrac{PR}{PQ}, so \dfrac{AC}{AB} = \dfrac{PR}{PQ}.

Therefore \dfrac{AC}{PR} = \dfrac{AB}{PQ} = k, say.   (1)

So AC = k·PR and AB = k·PQ. By the Pythagoras theorem, BC = \sqrt{AB^2 - AC^2} and QR = \sqrt{PQ^2 - PR^2}. So

\dfrac{BC}{QR} = \dfrac{\sqrt{k^2 PQ^2 - k^2 PR^2}}{\sqrt{PQ^2 - PR^2}}

= \dfrac{k\sqrt{PQ^2 - PR^2}}{\sqrt{PQ^2 - PR^2}} = k   (2)

From (1) and (2), \dfrac{AC}{PR} = \dfrac{AB}{PQ} = \dfrac{BC}{QR} — all three pairs of sides are in the same ratio. So by Theorem (the SSS similarity criterion), \triangle ACB \sim \triangle PRQ, and therefore ∠B = ∠Q.

Example 3 : Consider △ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ABC = θ (Fig. 8.10). Determine the values of (i) \cos^2\theta + \sin^2\theta and (ii) \cos^2\theta - \sin^2\theta.

Solution : First the missing side. A difference of squares keeps the numbers small:

AC = \sqrt{AB^2 - BC^2} = \sqrt{(29 - 21)(29 + 21)}

= \sqrt{8 \times 50} = \sqrt{400} = units

So \sin\theta = \frac{AC}{AB} = and \cos\theta = \frac{BC}{AB} = .

(i) \cos^2\theta + \sin^2\theta = \dfrac{21^2 + 20^2}{29^2} = \dfrac{441 + 400}{841} =

(ii) \cos^2\theta - \sin^2\theta = \dfrac{(21 + 20)(21 - 20)}{29^2} =

Did (i) come out as exactly 1 by luck? Keep that question in mind — Section 8.4 answers it.

Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

Solution : tan A = BC/AB = 1 means BC = AB (Fig. 8.11). Let AB = BC = k, where k is a positive number.

AC = \sqrt{AB^2 + BC^2} = \sqrt{k^2 + k^2} =

So \sin A = \frac{BC}{AC} = and \cos A = \frac{AB}{AC} = .

Therefore 2\sin A\cos A = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1, which is what we had to verify.

A Side Hidden in a Difference

Example 5 : In △OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (Fig. 8.12). Determine the values of sin Q and cos Q.

Fig. 8.12 is live. Before any algebra, drag PQ until the hypotenuse OQ is exactly 1 cm longer than PQ.

PQ = ${pq} cm

Solution : In △OPQ, by the Pythagoras theorem, OQ^2 = OP^2 + PQ^2.

Since OQ = 1 + PQ, this becomes (1 + PQ)^2 = OP^2 + PQ^2

i.e., 1 + 2PQ + PQ^2 = 49 + PQ^2

i.e., 1 + 2PQ = 49, because the PQ^2 terms cancel.

So PQ = cm and OQ = 1 + PQ = cm — the very triangle the slider found.

Hence \sin Q = \frac{OP}{OQ} = and \cos Q = \frac{PQ}{OQ} = .

Speed Round: Read the Ratio

Each triangle has its sides marked and one angle θ. Find the opposite side, the adjacent side and the hypotenuse as seen from θ, then divide.

Get five right to move on.

Sina