In this chapter, we intend to study an important branch in mathematics called "Trigonometry". The word 'Trigonometry' is derived from the two Greek words.

i) trigonon and ii) metron. The word trigonon means a triangle and the word metron means a measure. Hence, trigonometry means the science of measuring triangles. In detail we say that it is the branch of mathematics which deals with the measurement of the sides and the angles of a triangle and the problems allied with angles. In this chapter we confine for right angled triangle only.

Angle: We know that an angle is considered as the figure obtained by rotating a given ray about its end point.

Generally we denote the angle by "θ" (Theeta)

Measure of an angle: The measure of an angle is an amount of rotation from the initial side to the terminal side. We know that in right angled triangle, one angle is right angle and remaining two angles should be acute.

Trigonometric Ratios: The most important task of trigonometry is to find the measuring sides and angles of a triangle when some of its sides and angles are given. This problem is should by using some ratios of the sides of a triangle with respect to its acute angles. These ratios of acute angles are called 'trigonometric ratios' of angles. Let us now define various trigonometric ratios.

Consider the right angled triangle ABC, right angle at B. A and C are acute angles. Let angle at A be θ. Then we can say that angle at C is (90 − θ). Now we shall write the ratios with respect to θ.

We define the following six trigonometric ratios:

Relations between trigonometric ratios: The trigonometric ratios in θ, cos θ and tan θ  of an angle θ are very closely connected by a relation. If any one of them is known, the other two can be easily calculated.

Similarly sin θ . cosec θ = 1 and cos θ . sec θ = 1

EXAMPLES

1. In ΔABC, right angle at B, if AB = 12, AC = 13 and BC = 5 find all trigonometric (six) ratios of angle A.

Sol: Here opposite side to A = BC = 5

          Adjacent side to A = AB = 12

          Hypotenuse = AC = 13

2. In ΔABC, right angle at B, if AB = 4 and BC = 3, find all six trigonometric ratios of ∠A.

Sol: Here we have AB = 4, BC = 3 and ∠B = 90° By Pythagoras theorem

          AC2 = AB2 + BC2

                  = 42 + 32

                  = 16 + 9

                  = 25

3. In a ΔABC right angle at B, if AB = 12 and BC = 5, find

(i) sin A and tan A (ii) cos C and cot C.

Sol: Here AB = 12, BC = 5 and ∠B = 90°

By Pythagoras theorem, AC² = AB² + BC²

               = 12² + 5²

               = 144 + 25

               = 169

4. If cos θ = , find the other five trigonometric ratios.
S0l:

We have cos θ = 

So AB = 8, AC = 17 and right angle at B

       By Pythagoras theorem, AC² = AB² + BC²

             17² = 8² + BC²

           ⇒  BC² = 189 − 64

5. In ΔPQR right angle at R, in which PQ = 29, QR = 21 and ∠PQR = θ. Then find the values of i) cos² θ + sin² θ ii) cos² θ − sin² θ

Sol: Here we have PQ = 29, QR = 21 and R = 90°

6. In a Δ ABC right angle at C, If tan A =  find the value of sin A.cos B + cos A.sin B.

⇒ BC = 1 k units, AC =  k units

By Pythagoras theorem AB² = BC² + CA²

          = (1k)² + ( k)²

          = k² + 3k²

          = 4k²

          = 2K

Trigonometric ratios of some specific angles

Trigonometric ratios of 45°:

Let us take an isosceles right angled triangle ABC right angle at 'B'

Let AB = BC = a

By Pythagoras theorem AC² = AB² + BC²

          = a² + a²

          = 2a²

Trigonometric Ratios of 30° and 60°:

Let us take ABC is an equilateral triangle AD r BC.

Hence AB = BC = AC = a (say)

Since ABC is an equilateral triangle, each angle

∠BAC = ∠B = ∠C = 60°

We know that AD bisects ∠A

∴ ∠BAD = ∠DAC =

 =  30°

In ΔABD, AB = a, BD =  and ∠BAD = 30°, ∠ADB = 90°, ∠B = 60°

Trigonometric ratios of 0° and 90°:

Let ∠XAY = θ be an acute angle and Let 'P' be a point on its terminal side AY. Draw perpendicular PM from P on AX.

In Δ AMP

It is evident from ΔAMP that as θ becomes smaller and smaller, line segment PM also becomes smaller and smaller and finally when θ becomes 0°; the point 'P' will coincide with M. Consequently, we have

PM = 0 and AP = AM

From ΔAMP, it is evident that as θ increase, line segment AM becomes smaller and smaller and finally when θ becomes 90° the point M will coincide with A. Consequently, we have AM = 0, AP = PM

Now, Let us see the values of trigonometric ratios of all the above discussed angles in the form of a table

7. Evaluate the following.

(i) cos 60° . cos 30° − sin 60°. sin 30°

(ii) cosec2 30°. sin2 45° − sec2 60°

        = 2 − 4 = −2

8. Find the value of 'x'  each case

(i) tan 3x = sin 45° . cos 45° + sin 30°

(ii) sin 2x = sin 60°. cos 30° − cos60° . sin 30°

Sol: (i) tan 3x = sin 45° . cos 45° + sin 30°

∴ tan 3x = tan 45°

⇒ 3x = 45°

⇒ x = 15°

(ii) sin 2x = sin 60°. cos 30° − cos 60°. sin 30°

    sin 2x = sin 30°

  ⇒  2x = 30°

  ⇒   x = 15°

9. If sin(A + B) = 1 and cos(A − B) = , 0° < A + B ≤ 90°, A > B then find A and B.

Sol: We have sin(A + B) = 1 = sin 90°

           ∴ A + B = 90° ....... (1)

           cos (A − B) =

 = cos 30°

           ∴ A − B = 30° ........ (2)

          By adding (1) & (2) we get

         (A + B) + (A − B) = 90° + 30°

        ⇒  2A = 120°

       ⇒  A = 60°

        Substitute A = 60° in (1), we get

        60° + B = 90°

        => B = 90° − 60° = 30°

       ∴  A = 60°, B = 30°

10. In ΔPQR right angle at Q, PQ = 3 cm and PR = 6 cm. Determine ∠QPR and ∠PRQ.

Sol: Given PQ = 3 cm, PR = 6 cm

       

         sin R = sin 30°

        ⇒  R = 30º

         ∴ PRQ = 30° ⇒ QPR = 60°

Trigonometric ratios of complementary Angles:

We know that two angles are said to be complementary, if their sum is equal to 90°. Consider a right angle triangle ABC with right angle at B.

Since ∠B = 90°, ∠A + ∠C = 90°

Hence ∠A and ∠C are said to be complementary. Let us assume that ∠A = x, then for ∠x, BC is opposite side and AB is adjacent side.

Since ∠A + ∠C = 90°

      ∠C = 90° − ∠A

              = 90° − x

for (90 − x)°, AB is opposite side and BC is adjacent side.

Now compare the ratios of angles x and (90 − x) from the above values of different trigonometric terms.

    

11. 

Sol: cosec A = sec (90 − A)°

 cosec 55° = cosec (90° − 35°) = sec 35

12. Evaluate cos (40º  − θ)  − sin (50º  + θ) + 

 

Sol: cos (40° − θ) − sin  (50° + θ) + 

         = sin [90° − (40° −θ)] − sin(50° + θ) + 

        =  sin (50°+ θ) − sin (50° + θ) +  

        = 0 + 1 = 1

Trigonometric Identities:

We know that an identity is that mathematical equation which is true for all the values of the variables in the equation.

For example (a − b)2 = a2 − 2ab + b2 is an identity

In the same way, an identity equation having trigonometric ratios of an angle is called trigonometric identity. And is true for all the values of the angles involved in it.

Hence, we will derive a trigonometric identity and remaining would be based on that.

Consider a right angle triangle ABC with right angle is at B, So By Pythagoras theorem

We have AB² + BC² = AC² ...........(1)

Dividing each term by AC², we get

(i.e) cos² A + sin² A = 1

We have given an equation having a variable 'A' and the equation is true for all values of A. Hence the equation is a trigonometric identity

cos² A + sin² A = 1

Divide this identity by cos² A, we get

1 + tan² A = sec² A

Again divide cos² A + sin² A = 1 by sin² A, we get

⇒ cot² A + 1 = cosec² A (or) 1 + cot² A = cosec² A

13. Prove that

Sol:   

14. Without using trigonometric tables, evaluate the following.

Sol: We have

= 1 + cos² θ + sin² θ

= 1 + 1

= 2

Writer: T.S.V.S. Suryanarayana Murthy