In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of sides of a right angled triangle. 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

OM^{2} + MP^{2} = OP^{2} or a^{2} + b^{2} = 1

Thus, for every point on the unit circle,

we have

a^{2} + b^{2} = 1 or cos^{2} x + sin^{2} x = 1

Since one complete revolution subtends an angle of 2π radian at

the centre of the circle, ∠AOB = π / 2,

∠AOC = π and ∠AOD = 3π / 2 . 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

cos 0° = 1sin 0° = 0,

cos π/2 = 0sin π/2 = 1

cos π = −1sin π = 0

cos 3π/2 = 0sin 3π/2 = -1

cos 2π = 1sin 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 any integral multiple of 2π, the values of sine and cosine functions do not change. Thus,

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

Further, sin x = 0, if x = 0, ± π, ± 2π, ± 3π, ..., i.e., when x is an integral multiple of π and cos x = 0, if x = ± π/2, ± 3π/2, ± 5π/2, ... i.e., cos x vanishes when x is an odd multiple of π/2 . Thus

sin x = 0 implies x = nπ, where n is any integer

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

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

cosec x = 1/sin x , x ≠ nπ where n is any integer.

sec x =1 / cos x , x ≠ (2n + 1)π/2 where n is any integer.

tan x = sin x/cos x, x ≠ (2n + 1)π/2 where n is any integer.

cot x = cos x/sin x, x ≠ nπ where n is any integer.

We have shown that for all real x, sin^{2} x + cos^{2} x = 1

It follows that

1 + tan^{2} x = sec^{2} x (why?)

1 + cot^{2} x = cosec^{2} x (why?)

In earlier classes, we have discussed the values of trigonometric ratios for 0°, 30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same as that of trigonometric ratios studied in earlier classes. Thus, we have the following table:
The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cos x and tan x, respectively.

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

cos (– x) = cos x

and 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. We have learnt in previous classes that in the first quadrant (0 < x < π/2) a and b are both positive, in the second quadrant (π/2 < x < π) a is negative and b is positive, in the third quadrant (π < x < 3π/2 ) a and b are both negative and in the fourth quadrant (3π/2 < x < 2π) a is positive and 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 < π/2, negative for π/2 < x < 3π/2 and also positive for 3π/2 < x < 2π. Likewise, we can find the signs of other trigonometric functions in different quadrants. In fact, we have the following table.

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,

– 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1

Thus, domain of y = sin x and y = cos x is the set of all real numbers and range is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1.

Since 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 y = sec x is the set {x : x ∈ R and x ≠ (2n + 1)
π/2, n ∈ Z} and range is the set {y : y ∈ R, y ≤ – 1 or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and x ≠ (2n + 1)π/2 , n ∈ Z} and range is the set of all real numbers. The domain of y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real numbers.

We further observe that in the first quadrant, as x increases from 0 to π/2 , sin x increases from 0 to 1, as x increases from π/2 to π, sin x decreases from 1 to 0. In the third quadrant, as x increases from π to 3π/2, sin x decreases from 0 to –1 and finally, in the fourth quadrant, sin x increases from –1 to 0 as x increases from 3π/2 to 2π.

Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table:

Remark In the above table, the statement tan x increases from 0 to ∞(infinity) for 0 < x < π/2. Simply means that tan x increases as x increases for 0 < x < π/2 and assumes arbitraily large positive values as x approaches to π/2 . Similarly, to say that cosec x decreases from –1 to –∞(minus infinity) in the fourth quadrant means that cosec x decreases for x ∈ (3π/2, 2π) and assumes arbitrarily large negative values as x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behavior of functions and variables.

We have already seen that values of sin x and cos x repeats after an interval of 2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after an interval of π. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given below:

Example 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 = sin x/cos x = 4/3 and cot x = cos x/sin x = 3/4 .

Example Find the value of sin 31π/3 .

Solution We know that values of sin x repeats after an interval of 2π. Therefore Sin 31π/3 = sin (10π + π/3) = sin π/3 = √3 / 2.

Example 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.

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