Trigonometric Functions: Angles & Ratios

Written ByVaibhav_Raj_AsthanaLast Modified 08-06-2022

Trigonometric function is one of the most important topics taught in class 9th, which will be useful throughout kanalbdg.com life. Trigonometric functions are formed when trigonometric ratios are studied in terms of radian measure for any angle (0, 30, 90, 180, 270..). These are also defined in terms of sine and cosine functions. Students need to have basic knowledge of triangles and their angles to understand trigonometric ratios clearly.

In kanalbdg.com this article, students will find all the details on trigonometric functions such as value in degree, radians, complete trigonometric table and other relevant information. Students need to follow this article to develop basic knowledge of trigonometry and trigonometric functions.

Embibe offers a range of study materials that include sample test papers, mock tests, PDF of NCERT books and previous year question papers. Students can practice from these study materials for free. It will expose them to more number of questions and will further strengthen their ability to perform in the boards. Trigonometric Functions: Class 11

Here, students will learn how trigonometric functions like sin, cos, tan, cosec, sec, cot are calculated at different values of θ.

Let us take a circle with the centre at the origin of the x-axis. Let P (a, b) be any point on the circle with angle AOP = x radian, i.e., AP = x.

Here cos x = a and sin x = b. Since ∆OMP is a right triangle, we have OM2 + MP2 = OP2 or a2 + b2 = 1.

So, for every point on circle, we have a2 + b2 = 1 or cos2 x + sin2 x = 1. Since a complete revolution creates an angle of 2π radian = 360°, we have ∠AOB = π/2, ∠AOC = π and ∠AOD = 3π/2. Angles that are multiples of π/2 are called quadrantal angles.

The coordinates of the points A, B, C and D are already provided in the figure. Hence, for quadrantal angles, we have the following:cos 0° = 1sin 0° = 0cos π/2 = 0sin π/2 = 1cos π = − 1sin π = 0cos 3π/2 = 0sin 3π/2 = –1cos 2π = 1sin 2π = 0

Important Note: Sin x = 0 for x = nπ, where n is any integer. Cos x = 0 for x = (2n + 1) π 2 , where n is any integer.Six Trigonometric Functions

Now that you are aware of how the value of sin and cos at different angles is calculated. Let us now introduce you to other functions like tan, cosec, sec, and cot.

For this, we will require a right-angled triangle.

Learn 10th CBSE Exam Concepts

Sin or sine function is expressed as the ratio of perpendicular to hypotenuse i.e sin θ = a/h.

Cosine Function or Cos Function

Cos or cosine is written as the ratio of Base to hypotenuse i.e cos θ = b/h.

Tangent Function or Tan Function

Tan or tangent function in trigonometry is expressed as the ratio of perpendicular to base i.e tan θ = a/b.

Cosecant Function or Cosec Function

Cosec or cosecant function is the inverse of sin function as has the inverse value of sine i.e cosec θ = h/a.

Secant Function or Sec Function

Sec or secant is the inverse of cos and has the inverse value of cosine function i.e sec θ = h/b.

Cotangent Function or Cot Function

Cot or Cotangent is the inverse of tan function and can also be expressed as cos/sin i.e cot θ = b/a.Trigonometric Functions Table

We can define other trigonometric functions in terms of sin and cos.cosec x = 1/sin x or \(\frac{1}{sin\,x}\), x ≠ nπ, where n is any integer.sec x = 1/cos x or \(\frac{1}{cos\,x}\), x ≠ (2n + 1) π 2 , where n is any integer.tan x = sin x/cos x or \(\frac{sin\,x}{cos\,x}\), x ≠ (2n +1) π 2 , where n is any integer.cot x = cos x/sin x or \(\frac{cos\,x}{sin\,x}\), x ≠ n π, where n is any integer.Angles0°30° or π/645° or π/460° or π/390° or π/2180° or π270° or 3π/2360° or 2πSin01/21/√2√3/210−10Cos1√3/21/√21/20−101Tan01/√31√3Not Defined0Not Defined0CotNot Defined√311/√30Not Defined0Not DefinedCosecNot Defined2√22/√31Not Defined−1Not DefinedSec12/√3√22Not Defined−1Not Defined1Sign of Trigonometric Functions

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

So, cos (– x) = cos x and sin (– x) = – sin x

Practice 10th CBSE Exam Questions

In the same way, we can define other functions as well. Below we have provided the table for the negative or positive value of functions in all four quadrants.Functions (↓) And Quadrants (→)IIIIIIIVSin x++––Cos x+––+Tan x+–+–Cosec x++––Sec x+––+Cot x+–+–Sine, Cosine, Tangent Graphs

There might be instances where a question involves drawing the graph for the trigonometric function. You can use the table below to draw the graphs.FunctionDefinitionDomain RangeSine Functiony=sin xx ∈ R− 1 ≤ sin x ≤ 1Cosine Functiony = cos xx ∈ R− 1 ≤ cos x ≤ 1Tangent Functiony = tan xx ∈ R , x≠(2k+1)π/2,− ∞ < tan x < ∞Cotangent Functiony = cot xx ∈ R , x ≠ k π− ∞ < cot x < ∞Secant Functiony = sec xx ∈ R , x ≠ ( 2 k + 1 ) π / 2sec x ∈  ( − ∞ , − 1 ] ∪ [ 1 , ∞ )Cosecant Functiony = csc xx ∈ R , x ≠ k πcsc x ∈  ( − ∞ , − 1 ] ∪ [ 1 , ∞ )Source: NCERT Text BookFunctions of Negative Angles

If θ is the angle then we can define all the trigonometric functions as given below:(i) sin (– θ) = – sin θ(ii) cos (–θ) = cos θ(iii) tan (– θ) = – tan θ(iv) cot (–θ) = – cot θ sec (–θ) = sec θ(v) cosec (– θ) = – cosec θTrigonometric Ratios of Complementary Angles

In this section, we have provided you with the complementary angle ratios for all 4 quadrants.

sin(π/2−\(\theta\)) = \(\cos \theta\)

cos(π/2−\(\theta\)) = \(\sin \theta\)

tan(π/2−\(\theta\)) = \(\cot \theta\)

cot(π/2−\(\theta\)) = \(\tan \theta\)

sec(π/2−\(\theta\)) = cosec\(\theta\)

cosec(π/2−\(\theta\)) = \(\sec \theta\)

sin(π−\(\theta\)) = \(\sin \theta\)

cos(π−\(\theta\)) = -\(\cos \theta\)

tan(π−\(\theta\)) = -\(\tan \theta\)

cot(π−\(\theta\)) = -\(\cot \theta\)

sec(π−\(\theta\)) = -sec\(\theta\)

cosec(π−\(\theta\)) = cosec\(\theta\)

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sin(π+\(\theta\)) = -\(\sin \theta\)

cos(π+\(\theta\)) = -\(\cos \theta\)

tan(π+\(\theta\)) = \(\tan \theta\)

cot(π+\(\theta\)) = \(\cot \theta\)

sec(π+\(\theta\)) = -sec\(\theta\)

cosec(π+\(\theta\)) = -cosec\(\theta\)

sin(2π−\(\theta\)) = -\(\sin \theta\)

cos(2π−\(\theta\)) = \(\cos \theta\)

tan(2π−\(\theta\)) = -\(\tan \theta\)

cot(2π−\(\theta\)) = -\(\cot \theta\)

sec(2π−\(\theta\)) = sec\(\theta\)

cosec(2π−\(\theta\)) = -cosec\(\theta\)Trigonometric Functions Formulas

Here are all the formulas related to trigonometric functions that will aid you in your preparation.(i) sin (A + B) = sin A cos B + cos A sin B(ii) sin (A – B) = sin A cos B – cos A sin B(iii) cos (A + B) = cos A cos B – sin A sin B(iv) cos (A – B) = cos A cos B + sin A sin B(v) Sin 2A = 2 Sin A Cos A(vi) Cos 2A = Cos2 A – Sin2 A = 1 – 2 Sin2 A = 2 Cos2 A – 1(vii) Cos 3A = 4Cos3 A – 3Cos A(viii) 2Sin A Cos B = Sin (A + B) + Sin (A – B)(ix) 2Cos A Sin B = Sin (A + B) – Sin (A – B)(x) 2Cos A Cos B = Cos (A + B) + Cos (A – B)(xi) 2Sin A Sin B = Cos (A – B) – Cos (A + B)(xii) Tan (A + B) = (Tan A + Tan B) ÷ (1 − Tan A tan B)(xiii) Tan (A – B) = (Tan A − Tan B) ÷ (1 + Tan A Tan B)(xiv) Tan 2A = (2 Tan A) ÷ (1 – Tan2 A)(xv) Tan 3A = (3 Tan A – Tan3 A) ÷ (1 – 3Tan2 A)(xvi) Cot (A + B) = (Cot A Cot B − 1) ÷ (Cot A + Cot B) (xvii) Cot (A − B) = (Cot A Cot B + 1) ÷ (Cot B − Cot A) Trigonometric Equations

Equations that involve trigonometric functions of a variable are called trigonometric equations. The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions.

The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. The best way to explain this topic is via examples.

Example 1: Find the principal solutions of the equation sin x = 3/2.

Solution: From the trigonometric table we know that, sin π/3 = √3/2 and sin 2π/3 = sin (π − π/3) = sin (π/3) = √3/2.

Therefore, principal solutions are x = π/3 and 2π/3.Sample Questions on Basic Trigonometric Functions

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