 # Introduction to Trigonometric Functions

The trigonometric ratios we have studied till now are the ratios of acute angles in any right-angled triangle. What about trigonometric ratios for any angled triangle. When we talk about a normal triangle with any degree of angle, we extend our term to trigonometric functions. The following chapter shall extend your knowledge of Trigonometric functions for a triangle with any degree of angle.

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The angle measured in a triangle for trigonometric functions is taken in terms of Radian measures. For understanding the concept of radian measure and trigonometric functions, we need to first analyse the figure given below: The figure above is of a circle with centre O, that lies at the origin of coordinate axes. Q(a,b) is any point on the circle with angle SOQ = x radian. The length of arc SQ here from ∠SOQ is x. Now we define cos x=a and sin x = b.

In the figure, we can see that ONQ is a right-angled triangle, therefore,

ON2 + NQ2 = OQ2
or a2 +b= 1
or, cos2x +sin2x = 1

This signifies every point on the unit circle is cos2x +sin2x. Now, we already know that one complete revolution subtends the angle of 2π radian at the centre of the circle, so      ∠ SOT = π/2, similarly, ∠ SOU = π and ∠ SOV =  3π/2.

The angles discussed above are called quadrantal angles because these are integral multiples of π/2.  The coordinate value of the points S,T,U and V are respectively, (1,0), (0,1), (-1,0) and (0,-1). The value of the quadrantal angles in the above circle shall be: When we take one complete revolution from point Q, we come back to the same starting point Q.  Also, the increase or decrease in the value of x by an integral multiple of 2π does not affect the values of sine and cosine functions.

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

From this we know that, sin x = 0,  only if x = 0, ±π, ±2π, ±3π, …….. This implies that sine x shall be 0 if is an integral multiple of π. Further, cos = 0, if x = ±π/2, ±3π/2, ±5π/2. This means that the value of cos x becomes 0 or the value of cos x vanishes when happens to be the odd multiple of π/2From the above discussion we can hence conclude that sin x = 0 when x = nπ, and cos = o when = (2n+1)π/2,  where n is an integer.

## Trigonometric Functions in terms of Sine and Cosine Functions

Let’s use sine and cosine functions to determine the other trigonometric functions.

• cosec  x = 1/sin x, where ≠ nπ
• sec x = 1/cos  where ≠ (2n +1)π/ 2
• tan = sin x /cos x , where x ≠ (2n +1)π/ 2
• cot = cos x/ sin x, where ≠ nπ

In all the above functions, n is an integer. For all the real values of x, we already know that,

• sin2 x + cos2 x = 1,

This lets us know that,

• 1 + tan2 x = sec2 x
1 + cot2 x = cosec2 x

We already know the values of trigonometric ratios for the angles of 0°, 30°, 45°,60° and 90°. We use the same values for trigonometric functions as well. The values of trigonometric functions thus are as shown in the table below: For knowing the values of cosec x, sec x and cot x we reciprocate the values of sin x, cos x and tan x, respectively. From the above discussion, we can now calculate the values of the various trigonometric functions by using the respective trigonometric ratios, as stated in the table above.

## Solved Examples For You

Question: If $$\displaystyle \frac{\pi}{2}<\alpha<\pi$$, then
$$\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}+\sqrt{\frac{1+\cos\alpha}{1-\cos\alpha}}=?$$

Solution:

$$\sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}} + \sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}$$ $$= \sqrt{\frac{(1-\cos \alpha)^{2}}{\sin ^{2}\alpha}} + \sqrt{\frac{(1+\cos \alpha)^{2}}{\sin ^{2}\alpha}}$$
$$= |\mathrm{cosec}\alpha – cot\alpha| + |\mathrm{cosec}\alpha + \cot\alpha|$$
$$= 2\mathrm{cosec}\alpha$$ because $$\sin \alpha > 0,\alpha \in(\frac{\pi}{2},\pi)$$

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