Mathematics Trigonometric Functions
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### Topics Covered

star Trigonometric Functions
star Sign of trigonometric functions
star Domain and range of trigonometric functions

### Trigonometric Functions

We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions.

Consider a unit circle with centre at origin of the coordinate axes.

Let P (a, b) be any point on the circle with angle AOP = x radian, i.e., length of arc AP = x (Fig 3.6).

We define cos x = a and sin x = b Since ΔOMP is a right triangle, we have

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(OM^2 + MP^2 = OP^2)  or color(green)(a^2 + b^2 = 1)

Thus, for every point on the unit circle, we have

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color(green)(a^2 + b^2 = 1) or color(red)(cos^2 x + sin^2 x = 1)

=> Since one complete revolution subtends an angle of 2π radian at the centre of the circle, color(purple)(∠AOB =π/2), color(purple)(∠AOC = π) and color(purple)(∠AOD = 3π /2) .

color(blue)(=>"All angles which are integral multiples of"\ \ π /2 "are called quadrantal angles.")

The coordinates of the points A, B, C and D are, respectively, (1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have

 \ \ \ \ \ \ \ \ \ color(green)(cos 0° = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin 0° = 0,)

 \ \ \ \ \ \ \ \ \color(green)( cos (π/ 2) = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin (π /2) = 1)

 \ \ \ \ \ \ \ \ \color(green)(cosπ = − 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sinπ = 0)

 \ \ \ \ \ \ \ \ \color(green)(cos ((3π)/2 )= 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin((3π)/2) = –1)

 \ \ \ \ \ \ \ \ \color(green)( cos 2π = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin 2π = 0)

Now, if we take one complete revolution from the point P, we again come back to same point P. Thus, we also observe that if x increases (or decreases) by color(blue)("any integral multiple of 2π") color(red)("the values of sine and cosine functions do not change.") Thus,

 \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)(sin (2nπ + x) = sin x, n ∈ Z , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cos (2nπ + x) = cos x , n ∈ Z)

Further, color(green)(sin x = 0), if color(green)(x = 0, ± π, ± 2π , ± 3π, ........,) i.e., when x is an integral multiple of π

and color(green)(cos x = 0) , if color(green)(x = pm pi/2 , pm (3 pi)/2 , pm (5 pi)/2),.... i.e., cos x vanishes when x is an odd multiple of π/2 . Thus

 color(red)(sin x = 0) implies color(red)(x = nπ), where n is any integer

 color(red)(cos x = 0) implies  color(red)(x = (2n + 1)π/2) , where n is any integer.

We now define other trigonometric functions in terms of sine and cosine functions:

 color(red)(cosec x = 1/(sin x)) , \ \ color(blue)(x ≠ nπ) , where n is any integer.

 color(red)(sec x = 1/(cos x)) , \ \ color(blue)(x ≠ (2n +1 ) pi/2) , where n is any integer

 color(red)(tan x = (sin x)/(cos x) ), \ \ color(blue)(x ≠ (2n +1) pi/2 ), where n is any integer.

 color(red)(cot x = ( cos x)/(sin x)), \ \ color(blue)(x ≠ n π), where n is any integer

We have shown that for all real x

color(purple)( sin^2 x + cos^2 x = 1)

color(purple)(1 + tan^2 x = sec^2 x)

color(purple)(1 + cot^2 x = cosec^2 x)
Q 3126501471

Find the value of sin( (31pi)/3)

Solution:

We know that values of sin x repeats after an interval of 2π. Therefore

sin( 31pi/3 )= sin (10pi+pi/3) = sin = sqrt3/2
Q 3166501475

Find the value of cos (–1710°).

Solution:

We know that values of cos x repeats after an interval of 2π or 360°.
Therefore, cos (–1710°) = cos (–1710° + 5 × 360°)
cos (–1710° + 1800°) = cos 90° = 0.

### Sign of trigonometric functions

Let P (a, b) be a point on the unit circle with centre at the origin such that ∠AOP = x. If ∠AOQ = – x, then the coordinates of the point Q will be (a, –b) (Fig 3.7). Therefore

color(blue)(cos (– x) = cos x)

and color(blue)(sin (– x) = – sin x)

Since for every point P (a, b) on the unit circle, – 1 ≤ a ≤ 1 and – 1 ≤ b ≤ 1,

we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x.

=> In the color(green)(ul"first quadrant"\ \ ( 0 < x < pi/2)) ,  color(red)(a) and color(red)(b) are both positive,

=> In the color(green)(ul"second quadrant" \ \ ( pi/2 < x < pi )) ,  color(red)(a) is negative and color(red)(b) is positive,

=> In the color(green)(ul"third quadrant" \ \ (pi < x < (3pi)/2 )),  color(red)(a) and color(red)(b) are both negative and

=> In the color(green)(ul"fourth quadrant" \ \ ( (3 pi)/2 < x < 2 pi )),  color(red)(a) is positive and color(red)(b) is negative.

Therefore, sin x is positive for 0 < x < π, and negative for π < x < 2π.

Similarly, cos x is positive for 0 < x < pi/2 , negative for  pi/2< x < (3 pi)/2 and also positive for (3 pi)/2 < x < 2 pi .

Likewise, we can find the signs of other trigonometric

functions in different quadrants. In fact, we have the following table.
Q 3126401371

If cos x = – 3/5 x lies in the third quadrant, find the values of other five trigonometric functions.

Solution:

Since cos x = - 3/5 , we have sec x = -5/3

Now sin^2 x + cos^2 x = 1, i.e., sin^2 x = 1 – cos^2 x

or sin_2 x = 1 – 9/25 = 16/25

Hence sin x = ± 4/5

Since x lies in third quadrant, sin x is negative. Therefore

sin x = – 4/5

which also gives

cosec x = 5/4

Further, we have

tan x = (sinx)/cosx = 4/3 and cot x = (cosx)/(sinx) = 3/4
Q 3176401376

If cot x = – 5/12, x lies in second quadrant, find the values of other five trigonometric functions.

Solution:

Since cot x = – 5/12 we have tan x = - 12/5

Now sec^2 x = 1 + tan^2 x = 1 + 144/25 = 169/25

Hence sec x = ± 13/5

Since x lies in second quadrant, sec x will be negative. Therefore

sec x = – 13/5

which also gives

cos x = – 5/13

Further, we have

sin x = tan x cos x = (-12/5)xx (-5/13) = 12/13

and cosec x = 1/ sin x = 13/12

### Domain and range of trigonometric functions

From the definition of sine and cosine functions, we observe that they are defined for all real numbers.

=> Further, We observe that for each real number x,

color(purple)(– 1 ≤ sin x ≤ 1) and color(purple)(– 1 ≤ cos x ≤ 1)

=> Thus, color(purple)(ul"domain") of color(red)(y = sin x) and color(red)(y = cos x) is the color(blue)("set of all real numbers") and color(purple)(ul"range") is the interval color(blue){([–1, 1])}, i.e., – 1 ≤ y ≤ 1.

=> Since color(red)(cosec x =1/(sin x )) , the domain of y = cosec x is the set { x : x ∈ R and x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}.

Similarly, the domain of

=> color(red)(y = sec x) is the set {x : x ∈ R and x ≠ (2n + 1) π/2 , n ∈ Z} and color(purple)("range") is the set {y : y ∈ R, y ≤ – 1 or y ≥ 1}.

=> The domain of color(red)(y = tan x) is the set {x : x ∈ R and x ≠ (2n + 1) π/2 , n ∈ Z} and color(purple)("range") is the color(blue)("set of all real numbers").

=> The domain of color(red)(y = cot x) is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the color(purple)("range") is the color(blue)("set of all real numbers").