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Reflection in 0

There is no change in function form. Function takes sign in accordance with sign rule.

sin 0 - x = - sin x ; cos 0 - x = cos x ; tan 0 - x = - tan x ; cosec 0 - x = - cosec x ; sec 0 - x = sec x ; cot 0 - x = - cot x

Reflection in π/2

Reflection in π/2 is also known as co-function identities. Functions are called co-functions when their compliments have same value. As such, sine and cosine are co-functions. In this case, there is change in function form as combination of angle involves π/2. Function takes sign in accordance with sign rule.

sin π 2 - x = cos x ; cos π 2 - x = sin x ; tan π 2 - x = cot x ; cosec π 2 - x = sec x ; sec π 2 - x = cosec x ; cot π 2 - x = tan x

Reflection in π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

sin π - x = sin x ; cos π - x = - cos x ; tan π - x = - tan x ; cosec π - x = cosec x ; sec π - x = - sec x ; cot π - x = - cot x

Shift by π/2

Shift refers to horizontal shift of graph. We shall explore this aspect of trigonometric function in detail in a separate module. From transformation point of view, there is change in function form as combination of angle involves π/2. Function takes sign in accordance with sign rule.

sin π 2 + x = cos x ; cos π 2 + x = - sin x ; tan π 2 + x = - cot x ; cosec π 2 + x = sec x ; sec π 2 + x = - cosec x ; cot π 2 + x = - tan x

Shift by π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

sin π + x = - sin x ; cos π + x = - cos x ; tan π + x = tan x ; cosec π + x = - cosec x ; sec π + x = - sec x ; cot π + x = cot x

Shift by 2π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

sin 2 π + x = sin x ; cos 2 π + x = cos x ; tan 2 π + x = tan x ; cosec 2 π + x = cosec x ; sec 2 π + x = sec x ; cot 2 π + x = cot x

Finding angles

Trigonometric functions are many-one relation. We are required to find angles corresponding to a given trigonometric value. For example, what are angles corresponding to sine value of -√3/2. In other words, we need to find angles whose sine evaluates to this value. Note that these values corresponds to intersection of parallel line y=-√3/2 with the graph of sine curve.

Graph of sine function

Intersection of sine function with parallel value line.

For the time being, let us concentrate the interval [0,2π], which corresponds to one cycle of four quadrants. We follow an algorithm as given here to find angles in this interval :

1 : Consider only numerical magnitude of the given value. Find acute angle whose trigonometric function value corresponds to the numerical magnitude of the given value.

2 : Use sign rule and identify quadrants in which trigonometric function has the sign that of given value.

3 : Use value diagram and determine the angles as required.

Trigonometric value diagram

Angles whose trigonometric function values are same in different quadrants(to be used in conjunction with sign diagram).

To see the working of the algorithm, let us consider sinx = -√3/2. Considering only the magnitude of numerical value, we have :

sin θ = 3 2 = sin π 3

Thus, required acute angle is π/3. Now, sine function is negative in third and fourth quadrants. Looking at the value diagram, the angle in third quadrant is :

x = π + θ = π + π 3 = 4 π 3

Similarly, angle in fourth quadrant is :

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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