Question

☛ Give the Revision Notes of Exercise :- 3

"" TRIGONOMETRIC FUCTIONS ""

Class 11th.

Answer 1

Hey mate....

here's ur answer....



We know that when sin θ = k, k has to be such that –1 ≤ k ≤ 1 

We can always find some α ∈ [–π/2, π/2]

As sin (-π)/2 = -1 & sin π/2 = 1, such that sin θ = k, i.e. α = sin-1k

i.e. sin θ = sin α, α ∈ [–π/2, π/2]

⇒ sin θ – sin α = 0

⇒ 2 sin {(θ – α)/2} cos {θ + α)/2} = 0 

from the above equation to be satisfied, either sin {(θ – α)/2) = 0

and consequently ((θ – α)/2) = integral multiple of π

∴ θ – α = 2nπ

i.e. θ = 2nπ + α

θ = 2nπ + (–1)2n α where n = 0, ±1, ±2  … (1) 

or, cos {(θ + α)/2} = 0

i.e. {(θ + α)/2} = any odd multiple of π/2

i.e. {(θ + α)/2} = (2n + 1)π/2 

i.e. θ = (2n + 1)π – α 

⇒ θ = (2n +1)π + (–1)2n+1 α … (2)

From (1) and (2), we conclude that

θ = nπ + (–1)n α, where n is integral multiple, is the general solution of the equation sin θ = k 

Trigonometric Equations with their general Solutions:

Trigonometrical equation

General Solution

sin θ = 0 

 Then θ = nπ

cos θ = 0

 θ = (nπ + π/2)

tan θ = 0

 θ = nπ

sin θ = 1

 θ = (2nπ + π/2) = (4n+1)π/2

cos θ = 1

 θ = 2nπ

sin θ = sin α

 θ = nπ + (-1)nα, where α ∈ [-π/2, π/2]

cos θ = cos α

 θ = 2nπ ± α, where α ∈ (0, π]

tan θ = tan α

 θ = nπ + α, where α ∈ (-π/2 , π/2]

sin2 θ = sin2 α

 θ = nπ ± α

cos2 θ = cos2 α

 θ = nπ ± α

tan2 θ = tan2 α

 θ = nπ ± α


Hope it helps♥

Answer 2

heya..

here is your answer..

The 6 Trig Ratios

For any right triangle, there are six trig ratios: Sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot).  

Here are the formulas for these six trig ratios:

Given a triangle, you should be able to identify all 6 ratios for all the angles (except the right angle).

Let's start by finding all 6 ratios for angle A.  

Notice that the csc, sec, and cot can be found simply by flipping the ratio they are associated with. Or,

you can use the formulas.

To find the 6 ratios for angle B, just start over again and rethink them looking at angle B instead of angle A. This means that the opposite and the adjacent sides switch while the hypotenuse stays the same. Here are the 6 ratios for angle B:

sin() = y/z and cos() = x/z.

Four other trig functions are defined by

tan() = sin()/cos() cot() = cos()/sin()

sec() = 1/cos() csc() = 1/sin().

it may help you..