# Definition of Trigonometric Functions & Basic Formulae

In a right angled triangle ABC, ∠CAB = A and ∠BCA = 90° = π/2. AC is the base, BC the altitude and AB is the hypotenuse. We refer to the base as the adjacent side and to the altitude as the opposite side. There are six trigonometric ratios, also called trigonometric functions or circular functions. With reference to angle A, the six ratios are:

$\large \frac{BC}{AB} = \frac{opposite \; side}{hypotenuse}$  ;

is called sine of A , and written as sinA

$\large \frac{AC}{AB} = \frac{adjacent \; side}{hypotenuse}$  ;

is called the cosine of A , and written as cosA

$\large \frac{BC}{AC} = \frac{opposite \; side}{adjacent \; side}$  ;

is called the tangent of A , and written as tanA

Obviously, $\large tanA = \frac{sinA}{cosA}$

The reciprocals of sine, cosine and tangent are called the cosecant, secant and cotangent of A respectively. We write these as cosecA , secA , cot A respectively.

Since the hypotenuse is the greatest side in a right angle triangle, sin A and cos A can never be greater than unity and cosecA and sec A can never be less than unity.

Hence ,  | sinA | ≤ 1, |cos A| ≤ 1, |cosec A| ≥ 1, |sec A | ≥1 , while tan A and cot A may have any numerical value lying between – ∞ to +∞

##### Notes:

♦ Above mentioned method relating trigonometric functions to angles and sides of a triangle is called geometric definition of trigonometric functions.

♦ All the six trigonometric functions have got a very important property in common that is periodicity.

♦ Remember that the trigonometrical ratios are real numbers and remain same so long as the angles are real.

### Basic Formulae:

sin2A + cos2A = 1

⇒ sin2A = 1 – cos2A

or , cos2A = 1 – sin2A

1 + tan2A = sec2A

=> sec2A – tan2A = 1

cot2A + 1 = cosec2A

⇒ cosec2A – cot2A = 1

$\large tanA = \frac{sinA}{cosA}$

It is possible to express a trigonometrical ratio in terms of any one of the other ratios:

$\large |sin\theta| = \frac{1}{\sqrt{1 + cot^2 \theta}}$

$\large |cos\theta| = \frac{cot\theta}{\sqrt{1 + cot^2 \theta}}$

$\large tan\theta = \frac{1}{cot\theta}$

$\large cosec\theta = \sqrt{1 + cot^2 \theta}$

$\large |sec\theta| = \frac{\sqrt{1 + cot^2 \theta}}{cot\theta}$

i.e. all trigonometrical functions have been expressed in terms of cotθ . This is left as an exercise for you to derive these results. Just as a hint for you, express the denominator of fraction that defines cotθ as unity (i.e. base as unity) and form a right-angled triangle to express the sides and proceed.

Illustration : Express tanθ in terms of cosθ.

Solution:

By definition ;

$\large cos\theta = \frac{OA}{OB} = \frac{x}{1}$ ; where OB is taken as unity and OA = x

Hence in ΔAOB , OA = x ,

OB = 1 , By definition,

$\large tan\theta = \frac{AB}{OA} = \frac{\sqrt{1-x^2}}{x} = \frac{\sqrt{1-cos^2 \theta}}{cos\theta}$

Illustration : If sinθ + sin2θ = 1 , then prove that

cos12θ + 3 cos10θ + 3 cos8θ + cos6θ – 1 = 0.

Solution:

Given that sinθ = 1 – sin2θ = cos2θ

L.H.S = cos6θ (cos2θ + 1)3 – 1

= sin3θ(1 + sinθ)3 – 1

= (sinθ + sin2θ)3 – 1 = 0.

Illustration : Prove that :

(tanθ + cotθ)2 = sec2 θ + cosec2 θ = sec2 θ cosec2 θ.

Solution:

(tanθ + cotθ)2 = tan2 θ + cot2 θ + 2

= sec2 θ – 1 + cosec2 θ – 1 + 2

= sec2 θ + cosec2 θ

$\large = \frac{1}{cos^2 \theta} + \frac{1}{sin^2 \theta}$

$\large = \frac{1}{cos^2 \theta \; sin^2 \theta}$

= sec2 θ cosec2 θ

Exercise:

(i) If sin x + cos x = m and sec x + cosec x = n prove that n(m2 – 1) = 2 m.

(ii) If x sin3θ + y cos3θ = sinθ and x sinθ – y cosθ  = 0, prove that x2 + y2 = 1

(iii) Show that $\large \frac{sin^2 \theta}{1- cot\theta} + \frac{cos^2 \theta}{1- tan\theta} = 1 + sin\theta cos\theta$

(iv) If  a sec α + btan α = d and b sec α + a tan α = c , prove that a2 + c2 = b2 + d2

(v) If  a cos3θ + 3a cosθ sin2θ = x and a sin3θ + 3a cos2θ sinθ = y,

Prove that : (x + y)2/3 + (x – y)2/3 = 2 a2/3