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10th class > Introduction to Trigonometry > Trigonometric ratios

Trigonometric ratios

Since all of these triangles are similar, we know that their sides are proportional. In particular, the following ratios are the same for all of these triangles:

OppositeHypotenuse AdjacentHypotenuse OppositeAdjacent

Let’s try to summarise this: we picked a certain value for α, and got lots of similar, right-angled triangles. All of these triangles have the same ratios of sides. Since α was our only variable, there must be some relationship between α and those ratios.

These relationships are the Trigonometric ratios – and there are three of them:

The three Trigonometric ratios are relationships between the angles and the ratios of sides in a right-angles triangle. They each have a name, as well as a 3-letter abbreviation:

  • Sine: sinα=OppositeHypotenuse
  • Cosine: cosα=AdjacentHypotenuse
  • Tangent: tanα=OppositeAdjacent

Easy way to remember name and sides: SOH-CAH-TOA

Let's expand on this idea. We have seen:

OppositeHypotenuse AdjacentHypotenuse OppositeAdjacent

and given names for these ratios.

What about the inverse of these ratios?

HypotenuseOpposite HypotenuseAdjacent AdjacentOpposite

Let's give these ratios also a name.

The inverse of the above ratios also each have a name, as well as a 3-letter abbreviation:

  • Cosecant: cosecα=HypotenuseOpposite
  • Secant: secα=HypotenuseAdjacent
  • Cotangent: cotα=AdjacentOpposite

These are the 6 trigonometric ratios of an acute angle. The angle α is related to the sides of the triangle in these 6 different ways only.

Now, when we consider β

sinβ= cosβ= tanβ=

Let's consider the triangle ABC. Two sides are given, AB = 3k and BC = k. From trigonometric ratios we can find the value of .

Considering angle A i.e. angle α we get: sin A = k3k = . To find the values of cos A and tan A we do not have the value for AC.

We know that ABC is a triangle. And we know the two sides, AB and BC.

So, if we have to find the third side AC, we can use the theorem.

Solving for AC

AC2 = AB2 - BC2 = 3k2 -k2 = 8k2= 2 2k2

AC = +- 2k

AC = 2k (since length can't be a negative value)

We now have all the sides of the triangle.

cos A = ,

tan A =

The first use of the idea of sine in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500. Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation sin.

Aryabhata C.E. 476 – 550.

The origin of the terms cosine and tangent was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation cos.

Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’. sin A is not the product of ‘sin’ and A. ‘sin’ separated from A has no meaning.

Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.

Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see adjacent figure),

how will the trigonometric ratios of ∠A in △ PAM differ from those of ∠A in △ CAB or from those of ∠A in △ QAN ?

To answer this, first look at these triangles. Is △ PAM similar to △ CAB? From Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the triangles PAM and CAB are similar.

Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.

So, we have AMAB = =

From re-arranging the last two ratios, we find MPAP = BCAC = sin . Similarly, AMAP = = cosA, MPAM = = tan A and so on.

This shows that the trigonometric ratios of angle A in △ PAM do not differ from those of angle A in △ CAB.

In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in △ QAN also.

From our observations, it is now clear that:

The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the .

Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of sinA2, cosA2, etc., respectively. But cosec A = sinA1sin1 A (it is called sine inverse A). sin1 A has a different meaning, which will be discussed in higher classes.

Similar conventions hold for the other trigonometric ratios as well. Sometimes, the Greek letter θ (theta) is also used to denote an angle.

We have defined six trigonometric ratios of an acute angle. If we know any one of the ratios, can we obtain the other ratios? Let us see.

If in a right triangle ABC, sin A = 13. But, we know that sin A = BCAC BCAC = ,

i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio : .

Let's change the side lengths and try it again! Say,

AB2 + BC2 = AC2 i.e. 3k2 + 4k2 = 25k2 =

Therefore, AC = ± k

So we get AC = k

Now,

cosA = ABBC = 3k4k =

Similarly, you can obtain the other trigonometric ratios of the angle A.

Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than (or, in particular, equal to 1).

Let us consider some examples.

From the figure tan A =

Apply Pythagoras theorem.

AC = with sin A = and cos A =

Example 1

1. Given tan A = 43 , find the other trigonometric ratios of the angle A.

Solution : Let us first draw a right △ ABC.

Now, we know that tan A = =

Therefore, if BC = k, then AB = k, where k is a positive number.

Now, by using the Pythagoras Theorem, we have:

AC2 = AB2 + BC2 = 4k2 + 3k2 = 25k2

So, AC =

Now, we can write all the trigonometric ratios using their definitions.

sin A = BCAC = 4k5k =

cosA = ABAC = 3k5k =

Therefore, cotA = 1tanA =

cosecA = 1sinA =

secA = 1cosA =

Example 2

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

Solution :

Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).

We have sin B = ACAB and sin Q = PRPQ

Then ACAB =

Therefore, ACPR = ABPQ = k , say (1)

Now, using Pythagoras theorem:

BC = AB2AC2 and QR = PQ2PR2

So, BCQR = AB2AC2PQ2PR2 = kPQ2kPR2PQ2PR2

= kPQ2PR2PQ2PR2 = (2)

From (1) and (2), we have:

ACPR = ABPQ = BCQR

Then, by using Theorem 6.4, Δ ACB ~ Δ PRQ and therefore, ∠ B = ∠ .

Ex 3

3 : Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ. Determine the values of :

(i) cos2θ+sin2θ,

(ii) cos2θsin2θ

Solution :

In Δ ACB, we have:

AC = AB2BC2 = 292212 = 29+212921 = 850 = 400= units

So, sin θ = ACAB = while cos θ = BCAB =

(i) cos2θ+sin2θ = 21292 + 20292 = 202+212292 = 400+441841=

(ii) cos2θsin2θ = 21292 - 20292 = 212021+20292 =

Ex 4

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

Solution : In ΔABC, tan A = BCAB = 1 i.e., BC =

Let AB = BC = k, where k is a positive number.

Now, AC = AB2+BC2 = k2+k2 =

Therefore, sin A = BCAC = and cos A = ABAC =

So, 2 sin A cos A = 2 1212 = which is the required value.

Ex 5

5 : In Δ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. Determine the values of sin Q and cos Q.

Solution : In Δ OPQ, we have:

OQ2 = OP2 +

1+PQ2 = OP2 + PQ2

1 + PQ2 + 2PQ = OP2 + PQ2

+ 2PQ =

PQ = cm and OQ = 1 + PQ = cm

So, sin Q = and cos Q =