Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.
Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.
As given in the above figure in a right-angled triangle:
- Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
- Base: The side on which angle C lies is known as the base.
- Perpendicular: It is the side opposite to angle C in consideration.
Trigonometric Functions
Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,
sine: It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin θ
cosine: It is defined as the ratio of base and hypotenuse and it is represented as cos θ
tangent: It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
secant: It is the reciprocal of cos θ and is represented as sec θ.
cotangent: It is the reciprocal of tan θ and is represented as cot θ.
According to the above image, Trigonometric Ratios are
Sin θ = Perpendicular / Hypotenuse = AB/AC
Cosine θ = Base / Hypotenuse = BC/AC
Tangent θ = Perpendicular / Base = AB/BC
Cosecant θ = Hypotenuse / Perpendicular = AC/AB
Secant θ = Hypotenuse / Base = AC/BC
Cotangent θ = Base / Perpendicular = BC/AB
Reciprocal Identities
Sin θ = 1/ Cosec θ OR Cosec θ = 1/ Sin θ
Cos θ = 1/ Sec θ OR Sec θ = 1 / Cos θ
Cot θ = 1 / Tan θ OR Tan θ = 1 / Cot θ
Cot θ = Cos θ / Sin θ OR Tan θ = Sin θ / Cos θ
Tan θ.Cot θ = 1
Values of Trigonometric Ratios
| 0° | 30° | 45° | 60° | 90° | |
|---|---|---|---|---|---|
| Sin θ | 0 | 1/2 | 1√2 | √3/2 | 1 |
| Cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| Tan θ | 0 | 1/√3 | 1 | √3 | Not Defined |
| Cosec θ | Not Defined | 2 | √2 | 2/√3 | 1 |
| Sec θ | 1 | 2/√3 | √2 | 2 | Not Defined |
| Cot θ | Not Defined | √3 | 1 | 1/√3 | 0 |
Trigonometric Identities of Complementary and Supplementary Angles
- Complementary Angles: Pair of angles whose sum is equal to 90°
- Supplementary Angles: Pair of angles whose sum is equal to 180°
Identities of Complementary angles are
sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cot (90° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (90° – θ) = sec θ
Identities of supplementary angles
sin (180° – θ) = sin θ
cos (180° – θ) = – cos θ
tan (180° – θ) = – tan θ
cot (180° – θ) = – cot θ
sec (180° – θ) = – sec θ
cosec (180° – θ) = – cosec θ
Who is the founder of trigonometry?
Solution:
The first trigonometric table was apparently formed by Hipparchus, who is consequently now known as “the father of trigonometry”.
Hipparchus of Nicaea, Hipparkhos; c. 190 – c. 120 BC was a Greek astronomer, geographer, and mathematician.
He is considered the founder of trigonometry, but is most famous for his incidental discovery of precession of the equinoxes.
Hipparchus was born in Nicaea, Bithynia, and probably died on the island of Rhodes, Greece.
He is known to have been a working astronomer between 162 and 127 BC.
He is the first who created the first trigonometric table and solved several problems of spherical trigonometry by representing the corresponding values of arc and chord for a series of angles such as 0°, 30°, 45°, 60° and 90°, etc.
Sample Questions
Question 1: If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ – y cos θ = 0, then prove that x2 + y2 = 1, (where, sin θ ≠ 0 and cos θ ≠ 0).
Solution:
Here we have,
x sin3 θ + y cos3 θ = sin θ cos θ
Given:
x sin3 θ + y cos3 θ = sin θ cos θ
⇒ (x sin θ) sin2 θ + (y cos θ) cos2 θ = sin θ cos θ
⇒ (x sin θ) sin2 θ + (x sin θ) cos2 θ = sin θ cos θ (∵ y cos θ = x sin θ)
⇒ x sin θ(sin2 θ + cos2 θ) = sin θ cos θ (sin2 θ + cos2 θ = 1)
⇒ x sin θ = sin θ cos θ
⇒ x = cos θ ….(eq. 1)
now another trigono eq we have x sin θ – y cos θ = 0
we can write it as
x sin θ = y cos θ
from eq. 1 we have x = cos θ
so put in above eq. x sin θ = y cos θ
So, x sin θ = y cos θ
cos θ sin θ = y cos θ
y = sin θ eq. 2
Now by squaring and adding both the equation 1 &
x = cos θ & y = sin θ
x2 = cos2 θ & y2 = sin2 θ
So now
x2 + y2 = cos2 θ + sin 2θ { cos2 θ + sin 2θ = 1 }
x2 + y2 = 1
Hence proved
Question 2: If tan A = 4 and tan B = 2, find the value for tan(A – B).
Solution:
As per the formula
tan(A + B) = (tanA-tanB)/(1+tanAtanB)
= (4 – 2)/(1 + 4 × 2)
= 2/9
Therefore, the value of tan(A – B) is 2/19.
Question 3: If sin(y) = 10/29 then find the value of csc(-y)
Solution:
Here we have
sin(y) = P/B = 10/29
So
cosec(-y) = 1/sin(-y)
= 1/(-siny)
= – (1/siny)
= – cosec(y) {Cosecant θ = Hypotenuse / Perpendicular = AC/AB}
= -{1/(10/29)}
= -29/10
Question 4: Perform the indicated operation and simplify the result. {sec X}/{csc X} + {csc X}/{sec x}
Solution:
We have
{sec X}/{csc X} + {csc X}/{sec x}
here we can write csc x = 1/sin x and sec x = 1/cos x
= {(1/cos x )/ (1/ sin x) } + {(1 /sin x) / ( 1/cos x)}
= (sin x / cos x ) + ( cos x / sin x )
= tan x + cot x {Tan x = Sin x / Cos x and Cot x = Cos x / Sin x}
Therefore, {sec X}/{csc X} + {csc X}/{sec x} = tan x + cot x