**What is Trigonometry?**

Trigonometry is an important branch of mathematics that is used to study the relationship between ratios of the sides of a right-angled triangle with its angles. The word “trigonometry” is derived from the Greek words, “trigon” and “metron”. The word “trigon” means a triangle and the word ”metron” means a measure. Hence, trigonometry means the science of measuring triangles. In broader cases it is that branch of mathematics which deals with the measurement of the sides and the angles of a triangle and the problems allied with the angles.

Though the field emerged in Greece during the third century B.C., some of the most important contributions (such as the sine function) came from India in the fifth century A.D. Because early trigonometric works of Ancient Greece have been lost, it is not known whether Indian scholars developed trigonometry independently or after Greek influence. According to Victor Katz in “A History of Mathematics (3rd Edition)” (Pearson, 2008), trigonometry developed primarily from the needs of Greek and Indian astronomers.

Before moving on to understand the concepts in trigonometry, it is important to recall what we mean by an angle.

An angle is considered as the figure obtained by rotating a given ray about its end point. The measure of an angle is the amount of rotation from the initial side to the terminal side.

In the above figure, QR is the initial side and PQ is the terminal side.

Now let us learn about trigonometric ratios.

**What are Trigonometric Ratios?**

Consider an acute angle ∠YAX = θ with initial side AX and terminal side AY. Let P be any point on the terminal side AY. Draw PM perpendicular from P on AX to get the right angled triangle AMP in which ∠PAM = θ.

Now, in the right angled triangle AMP, Base = AM = x and Perpendicular = PM = y and Hypotenuse = AP = r

We define the following six trigonometric ratios

**Sin θ = $\frac{Perpendiculur}{Hypotenuse}=\frac{y}{r}$, written as sin θ****Cos θ = $\frac{Base}{Hypotenuse}=\frac{x}{r}$, written as cos θ****Tan θ = $\frac{Perpendiculur}{Base}=\frac{y}{x}$, written as tan θ****Cosecant θ = $\frac{Hypotenuse}{Perpendicular}=\frac{r}{y}$, written as cosec θ****Secant θ = $\frac{Hypotenuse}{Base}=\frac{r}{x}$, written as sec θ****Cotangent θ = $\frac{Base}{Perpendicular}=\frac{x}{y}$, written as cot θ**

It is important to note here that by sin θ we mean “sine of angle θ” and not the product of sin and θ. Similar is the case for other trigonometric ratios.

Does the value of trigonometric ratios change with change of points on the angle? Let us find out.

The above trigonometric ratios, as we can see depend upon the value of angle θ and are independent of the position of the point P on the terminal side AY of the acute angle ∠XAY. This means that if we choose P somewhere else on AY, then the lengths PM, AM and AP will change. However, the trigonometric ratios will remain same. Hence we can say that –

**The trigonometric ratios are same for the same angle.**

**Relation between Trigonometric Ratios**

The trigonometric ratios sin θ, cos θ and tan θ of an angle θ are very closely connected by a relation. If any one of them is known, the other two can be easily calculated. Let us see how.

Let us again consider the following figure –

We have learnt above that

Sin θ = $\frac{PM}{AP}$

There are only three values that constitute all the angles Base, Perpendicular and the Hypotenuse. Hence, if we know the value of one of the angles, this means we know two of these three values using which we can find the remaining value. Suppose, if we know the value of sin θ, we can easily calculate the value of the base using the Pythagoras theorem. We have,

PM^{2} + AM^{2} = AP^{2}

Now, since we have the value of sin θ, we know the value of the perpendicular and the hypotenuse using which we will get the value of the base.

Therefore, it is clear from the definitions of the trigonometric ratios that for any acute angles θ, we have,

- cosec θ = $\frac{1}{sinsinθ}$
- sec θ = $\frac{1}{coscosθ}$
- cot θ = $\frac{1}{tantanθ}$
- cot θ = $\frac{coscosθ}{sinsinθ}$
- tan θ = $\frac{sinsinθ}{coscosθ}$
- tan θ. Cot θ = 1

Let us understand this with an example.

**Example** If sin A = $\frac{3}{5}$, find the value of cos A and tan A

**Solution** We have been given that sin A = $\frac{3}{5}$

We have also learnt that sin θ = $\frac{Perpendiculur}{Hypotenuse}$

This means that $\frac{Perpendiculur}{Hypotenuse}= \frac{3}{5}$

So, we draw a triangle ABC, right angled at B such that Perpendicular = BC = 3 units and Hypotenuse = Ac = 5 units.

By Pythagoras theorem we have,

AC^{2} = AB^{2} + BC^{2}

5^{2} = AB^{2} + 3^{2}

AB^{2} = 5^{2} – 3^{2}

AB^{2} = 25 – 9

AB^{2} = 16

⇒ AB = 4

Now, that we know the value of AB, we can find the value of cos A

cos A = ** **$\frac{Base}{Hypotenuse}=\frac{4}{5}$

Also, tan A = $\frac{Perpendiculur}{Base}=\frac{3}{4}$

**Trigonometric Ratios of Some Specific Angles **

Now, we shall learn about the trigonometric ratios of some standards angles which have been derived using some elementary knowledge of geometry.

**Trigonometric Ratios of 0**^{o}** and 90**^{o}

^{o}

^{o}

Let XAY = θ be an acute angle and let P be a point on its terminal side AY. Draw a perpendicular PM from P to AX.

In Δ AMP, we have,

sin θ = $\frac{PM}{AP}$

cos θ = $\frac{AM}{AP}$

tan θ = $\frac{PM}{AM}$

It is evident from Δ AMP that as θ becomes smaller and smaller line segment Pm also becomes smaller and smaller and finally when θ becomes 0^{o}, the point P will coincide with M. Consequently, we have,

PM = 0 and AP = AM

Therefore,

sin 0^{0} = $\frac{PM}{AP}=\frac{0}{AP}$ = 0

Similarly,

cos 0^{0} = $\frac{AM}{AP}=\frac{AP}{AP}$ = 1

and

tan 0^{0} = $\frac{PM}{AM}=\frac{0}{AM}$ = 0

Thus, we have,

**sin 0**^{0}** = 0, cos 0**^{0}** = 1 and tan 0**^{0}** = 0**

**Similarly, cosec 0**^{0}** = Not defined sec 0**^{0}** = 1 and cot 0**^{0}** = Not Defined**

From Δ AMP, it is evident that as θ increases, the lien segment AM becomes smaller and smaller and finally when θ becomes 90^{o}, the point M will coincide with A. Consequently, we have,

AM = 0, AP = PM

Therefore,

sin 90^{0} = $\frac{PM}{AP}=\frac{PM}{PM}$ = 1

Similarly,

cos 90^{0} = $\frac{AM}{AP}=\frac{0}{AP}$ = 0

and

tan 90^{0} = $\frac{PM}{AM}=\frac{PM}{0}$ = Not Defined

Thus, we have,

**sin 90**^{0}** = 1, cos 90**^{0}** = 0 and tan 0**^{0}** = Not Defined**

**Similarly, cosec 90**^{0}** = 1sec 90**^{0}** = Not Defined, cot 90**^{0}** = 0**

**Trigonometric Ratios of 30**^{o}** and 60**^{o}

^{o}

^{o}

Consider an equilateral triangle ABC with each side of length 2a. Since each angle of an equilateral triangle is 60^{o}, therefore, each angle of the triangle ABC will be equal to 60^{o}. Let AD be perpendicular from A on BC. Since the triangle is equilateral, therefore, AD is the bisector of angle A and D is the mid-point of BC.

Therefore,

BD = DC = a and ∠BAD = 30^{o}

Thus in triangle ABD, ∠D is a right angle, hypotenuse AB = 2a and BD = a

So, by Pythagoras theorem, we have

AB^{2} = AD^{2} + BD^{2}

⇒ (2a)^{2} = AD^{2} +a^{2}

⇒ AD^{2} = 4a^{2} – a^{2}

⇒ AD = $\sqrt{3}$a

**Trigonometric Ratios of 30**^{o}

^{o}

So, now we have,

Base = AD = $\sqrt{3}$a, Perpendicular = BD = a, Hypotenuse = AB = 2a and ∠DAB = 30^{o}

Therefore,

sin 30^{0} = $\frac{BD}{AB}=\frac{a}{2a} =\frac{1}{2}$

Similarly,

cos 30^{0} = $\frac{AD}{AB}=\frac{\sqrt{3a}}{2a}=\frac{\sqrt{3}}{2}$

and

tan 30^{0} = $\frac{BD}{AD}=\frac{a}{\sqrt{3a}}=\frac{1}{\sqrt{3}}$

Thus, we have,

**sin 30 ^{0} = **$\frac{1}{2}$

**, cos 30**$\frac{\sqrt{3}}{2}$

^{0}=**and tan 30**$\frac{1}{\sqrt{3}}$

^{0}=**Similarly, cosec 30 ^{0} = 2, sec 30^{0} = **$\frac{2}{\sqrt{3}}$

**and cot 30**$\sqrt{3}$

^{0}=**Trigonometric Ratios of 60**^{o}

^{o}

In right triangle ADB, we have,

Base = BD = a, Perpendicular = AD = $\sqrt{3}$a, Hypotenuse = AB = 2a and ∠DAB = 60^{o}

sin 60^{0} = $\frac{AD}{AB}=\frac{\sqrt{3a}}{2a}=\frac{\sqrt{3}}{2}$

cos 60^{0} = $\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$

tan 60^{0} = $\frac{AD}{AB}=\frac{\sqrt{3a}}{a}=\sqrt{3}$

Thus, we have,

**sin 60 ^{0} =**$\frac{\sqrt{3}}{2}$

**, cos 60**$\frac{1}{2}$

^{0}=**, and tan 60**$\sqrt{3}$

^{0}=

**Similarly, cosec 60 ^{0} = **$\frac{2}{\sqrt{3}}$

**, sec 60**$\frac{1}{\sqrt{3}}$

^{0}= 2 and cot 60^{0}=**Trigonometric Ratios of 45**^{o}

^{o}

Consider a right angled triangle ABC with right angle at B such that ∠A = 45^{o}. Then,

Therefore, we have,

∠A + ∠B +∠C = 180^{o}

⇒ 45^{o} + 90^{o} + ∠C = 180^{o}

⇒ ∠C = 45^{o}

Hence, we can say that ∠A = ∠C

This means that AB = BC

Let AB = BC = a. then by Pythagoras theorem, we have,

AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = a^{2} + a^{2}

⇒ AC^{2} = 2a^{2}

⇒ AC = $\sqrt{2}$a

Therefore, we have

sin 45^{0} = $\frac{BC}{AC}=\frac{a}{\sqrt{2a}}=\frac{1}{\sqrt{2}}$

Similarly,

cos 45^{0} = $\frac{AB}{AC}=\frac{a}{\sqrt{2a}}=\frac{1}{\sqrt{2}}$

and

tan 45^{0} = $\frac{BC}{AB}=\frac{a}{a}$= 1

Thus, we have,

**sin 45 ^{0} = **$\frac{1}{\sqrt{2}}$

**, cos45**$\frac{1}{\sqrt{2}}$

^{0}=**and tan 45**

^{0}= 1**Similarly, cosec 45 ^{0} = **$\sqrt{2}$

**, sec 45**$\sqrt{2}$

^{0}=**, and cot 45**

^{0}= 1The following table gives us the values of various trigonometric ratios of 0^{o}, 30^{o}, 45^{o}, 60^{o} and 90^{o} for ready reference.

Angles/T. Ratios | 0^{o} | 30^{o} | 45^{o} | 60^{o} | 90^{o} |

sin θ | 0 | $\frac{1}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{\sqrt{3}}{2}$ | 1 |

cos ͔͔θ | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{2}$ | 0 |

tan θ | 0 | $\frac{1}{\sqrt{3}}$ | 1 | $\sqrt{3}$ | Not Defined |

cosec θ | Not Defined | 2 | $\sqrt{2}$ | $\frac{2}{\sqrt{3}}$ | 1 |

sec θ | 1 | $\frac{2}{\sqrt{3}}$ | $\sqrt{2}$ | 2 | Not Defined |

cot θ | Not Defined | $\sqrt{3}$ | 1 | $\frac{1}{\sqrt{3}}$ | 0 |

Let us understand their use through an example.

**Example 1** Evaluate the value of sin60^{o}cos30^{o} + cos60^{o}sin30^{o}

**Solution**** **We have,

sin60^{o}cos30^{o} + cos60^{o}sin30^{o}

From the above table, we can obtain the values of the given trigonometric ratios. So, we get –

sin60^{o} = $\frac{\sqrt{3}}{2}$

cos30^{o} = $\frac{\sqrt{3}}{2}$

cos60^{o} = $\frac{1}{2}$

sin30^{o} = $\frac{1}{2}$

Replacing these values in the given express we have,

sin60^{o}cos30^{o} + cos60^{o}sin30^{o}

= $\frac{\sqrt{3}}{2}$ x $\frac{\sqrt{3}}{2}$ + $\frac{1}{2}$ x $\frac{1}{2}$

= $\frac{3}{4}$ + $\frac{1}{4}$ = 1

**Hence, sin60**^{o}**cos30**^{o}** + cos60**^{o}**sin30**^{o}** = 1**

**Trigonometric Ratios of Complementary Angles**

Now that we have learnt the values of the trigonometric ratios on various degrees, we shall obtain the trigonometric ratios of complementary angles in terms of the trigonometric ratios of the given angles.

Let us recall what we mean by complementary angles. **Two angles are said to be complementary if their sum is 90 ^{o}. **It follows from this definition that θ and ( 90

^{o}– θ ) are complementary angles for an acute angles θ.

So, we have

**sin ( 90**^{o}** – θ ) = cos θ**

**cos ( 90**^{o}** – θ ) = sin θ**

**tan ( 90**^{o}** – θ ) = cot θ**

**cot ( 90**^{o}** – θ ) = tan θ**

**sec ( 90**^{o}** – θ ) = cosec θ**

**cosec ( 90**^{o}** – θ ) = sec θ**

Let us understand these formulae with an example

**Example** Find the value of $\frac{coscos 37^{0}}{sinsin 53^{0}}$

**Solution**** **We have,

$\frac{coscos 37^{0}}{sinsin 53^{0}}$

= $\frac{coscos 90^{0} – 53^{0}}{sinsin 53^{0}}$

= $\frac{sinsin 53^{0}}{sinsin 53^{0}}$ = 1

**Hence, **$\frac{coscos 37^{0}}{sinsin 53^{0}}$ **= 1**

**Solved Examples**

**Example 1**** **If cosec θ = √10, then find the values of all T-ratios of θ.

**Solution** we have been given that cosec θ = √10. We are required to find the value of the remaining five trigonometric ratios.

Now, we have learnt that

cosecant θ = $\frac{Hypotenuse}{Perpendicular}$

Since, we have

cosec θ = √10, therefore,

$\frac{Hypotenuse}{Perpendicular} = \frac{\sqrt{10}}{1}$

So, we draw a right triangle ABC, right angled at B such that,

Perpendicular = BC = 1 unit and

Hypotenuse = AC = $\sqrt{10}$ units

Now, using the Pythagoras theorem we have,

AC^{2} = AB^{2} + BC^{2}

⇒ (√10 )^{2} = AB^{2} + 1

⇒ AB^{2} = 10 – 1 = 9

⇒ AB = 3

When we consider the trigonometric ratios of A, we have,

Base = AB = 3 units, Perpendicular = BC = 1 unit and Hypotenuse = AC = $\sqrt{10}$ units

So,

**sin θ = **$\frac{Perpendiculur}{Hypotenuse}=\frac{1}{\sqrt{10}}$

**cos θ = **$\frac{Base}{Hypotenuse}=\frac{3}{\sqrt{10}}$

**tan θ = **$\frac{Perpendiculur}{Base}=\frac{1}{3}$

**sect θ = **$\frac{Hypotenuse}{Base}=\frac{\sqrt{10}}{3}$

**cot θ = **$\frac{Base}{Hypotenuse}=\frac{3}{1}$** = 3**

**Example 2**** **cot 12° cot 38° cot 52° cot 60° cot 78°

**Solution** We are required to find the value of cot12°cot38°cot52°cot60°cot78°

In order to find the value of the given expression, we shall first group the terms in such a way that the angles involved are complementary i.e. their sum is 90^{o}, we have,

= ( cot 12° cot 78° ) (cot 38° cot 52° ) cot 60°

= (cot 12° tan 12° ) (cot 38° tan 38° ) cot 60°

Now, we know that

cot 78° = cot ( 90 – 12 )^{o} = tan 12°

Also,

cot 52° = cot ( 90 – 52 )^{o} = tan 38°

Therefore, the given expression can be written as

= (cot 12° tan 12° ) (cot 38° tan 38° ) cot 60°

= 1 x 1 x $\frac{1}{\sqrt{3}}$ = $\frac{1}{\sqrt{3}}$

**Hence, cot 12° cot 38° cot 52° cot 60° cot 78° = **$\frac{1}{\sqrt{3}}$

**Example 3** Evaluate the following expression

2 sin^{2} 30^{o} tan 60^{o} – 3 cos^{2} 60^{o} sec^{2} 30^{o}

**Solution** We are required to find the value of the expression

2 sin^{2} 30^{o} tan 60^{o} – 3 cos^{2} 60^{o} sec^{2} 30^{o}

In order to find this value, we will first need to recall the values of the given trigonometric ratios in the expression

We have learnt that

sin 30^{o} = $\frac{1}{2}$

tan 60^{o} = $\sqrt{3}$

cos 60^{o} = $\frac{1}{2}$

sec 30^{o} = $\frac{2}{\sqrt{3}}$

Substituting the above values in the given expression we have,

2 sin^{2} 30^{o} tan 60^{o} – 3 cos^{2} 60^{o} sec^{2} 30^{o}

= 2 x ($\frac{1}{2}$)^{2} x $\sqrt{3}$ – 3 x ($\frac{1}{2}$) ^{2} x ($\frac{2}{\sqrt{3}}$) ^{2}

= 2 x $\frac{1}{4}$ x $\sqrt{3}$ – 3 x $\frac{1}{4}$ x $\frac{4}{3}$

= $\frac{\sqrt{3}}{2}$ – 1 = $\frac{\sqrt{3}-2}{2}$

**Hence, the value of the given expression 2 sin ^{2} 30^{o} tan 60^{o} – 3 cos^{2} 60^{o} sec^{2} 30^{o} is** $\frac{\sqrt{3}-2}{2}$

**Remember**

- Trigonometry means the science of measuring triangles.
- The trigonometric ratios are the same for the same angle.
- Sin θ = $\frac{Perpendiculur}{Hypotenuse}=\frac{y}{r}$, written as sin θ
- Cos θ = $\frac{Base}{Hypotenuse}=\frac{x}{r}$, written as cos θ
- Tan θ = $\frac{Perpendiculur}{Base}=\frac{y}{x}$, written as tan θ
- Cosecant θ = $\frac{Hypotenuse}{Perpendicular}=\frac{r}{y}$, written as cosec θ
- Secant θ = $\frac{Hypotenuse}{Base}=\frac{r}{x}$, written as sec θ
- Cotangent θ = $\frac{Base}{Perpendicular}=\frac{x}{y}$, written as cot θ
- cosec θ = $\frac{1}{sinsinθ}$
- sec θ = $\frac{1}{coscosθ}$
- cot θ = $\frac{1}{tantanθ}$
- cot θ = $\frac{coscosθ}{sinsinθ}$
- tan θ = $\frac{sinsinθ}{coscosθ}$
- tan θ. Cot θ = 1
- sin ( 90
^{o}– θ ) = cos θ - cos ( 90
^{o}– θ ) = sin θ - tan ( 90
^{o}– θ ) = cot θ - cot ( 90
^{o}– θ ) = tan θ - sec ( 90
^{o}– θ ) = cosec θ - cosec ( 90
^{o}– θ ) = sec θ - sin 0
^{0}= 0, cos 0^{0}= 1 and tan 0^{0}= 0 - cosec 0
^{0}= Not defined sec 0^{0}= 1 and cot 0^{0}= Not Defined - sin 30
^{0}= $\frac{1}{2}$, cos 30^{0}= $\frac{\sqrt{3}}{2}$ and tan 30^{0}= $\frac{1}{\sqrt{3}}$ - cosec 30
^{0}= 2, sec 30^{0}= $\frac{2}{\sqrt{3}}$ and cot 30^{0}= $\sqrt{3}$ - sin 45
^{0}= $\frac{1}{\sqrt{2}}$, cos45^{0}= $\frac{1}{\sqrt{2}}$ and tan 45^{0}= 1 - cosec 45
^{0}= $\sqrt{2}$, sec 45^{0}= $\sqrt{2}$, and cot 45^{0}= 1 - sin 60
^{0}=$\frac{\sqrt{3}}{2}$ , cos 60^{0}= $\frac{1}{2}$, and tan 60^{0}=$\sqrt{3}$ - cosec 60
^{0}= $\frac{2}{\sqrt{3}}$, sec 60^{0}= 2 and cot 60^{0}= $\frac{1}{\sqrt{3}}$ - sin 90
^{0}= 1, cos 90^{0}= 0 and tan 0^{0}= Not Defined - cosec 90
^{0}= 1sec 90^{0}= Not Defined and cot 90^{0}= 0

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