# TRIGONOMETRIC FUNCTIONS

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: is called sine of A , and written as sinA is called the cosine of A , and written as cosA is called the tangent of A , and written as tanA
Obviously, .
The reciprocals of sine, cosine and tangent are called the cosecant, secant and cotangent of A respectively. We write these as cosec A, sec A, 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 cosec A and sec A can never be less than unity.

Notes:

• Above mentioned method relating trigonometric functions to angles and sides of a triangle is called geometric definition of trigonometric functions. One can define these functions in analytical ways without any reference to geometry, which is beyond the scope of this material. You just have to understand that the argument of these functions can also be a real number, not necessarily the angles only. You may refer to the chapter ‘Functions’ in Phase-II of RSM for detailed study of behaviour of these functions and their graphs.
• 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 : Trigonometric functions

• sin2A + cos2A = 1

=> sin2A = 1 – cos2A

or cos2A = 1 – sin2A

• 1 + tan2A = sec2A
• => sec2A – tan2A = 1
• cot2A + 1 = cosec2A

=> cosec2A – cot2A = 1

• • Fundamental inequalities: For 0 < A < π/2 , 0 < cosA < sinA/A < 1/cosA .
• It is possible to express a trigonometrical ratio in terms of any one of the other ratios:

e.g.  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 1. Express tanθ in terms of cosθ.

Solution:

By definition
cosθ = OA/OB = x/1 = x where OB is
taken as unity and OA = x
Hence in ΔAOB , OA = x ,
OB = 1 , By definition, Illustration 2. 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 3. 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  = sec2 θ cosec2 θ .

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