Introduction to Trigonometry
Measurement of triangles is known as Trigonometry. Here we shall be dealing with the relation of sides and angles of a right-angled triangle.
In the above diagram Base = AB, Perpendicular = BC, Hypotenuse = AC and angle of reference = ∠A
The ratio between the lengths of two sides of a right-angled triangle is known as trigonometrical ratios. There are six trigonometrical ratios present for a right-angled triangle, they are given below.
Sin A = Perpendicular⁄Hypotenuse = BC⁄AC
Cos A = Base⁄Hypotenuse = AB⁄AC
Tan A = Perpendicular⁄Base = BC⁄AB
Cot A = Base⁄Perpendicular = AB⁄BC
Sec A = Hypotenuse⁄Base = AC⁄AB
Cosec A = Hypotenuse⁄Perpendicular = AC⁄BC
Each trigonometrical ratio is a real number and has no unit.
Example 1. From the below given figure find the value of Sin A, Cos A and Tan A.
Solution. As we can see, above figure is a wright angled triangle.
=> AC2 = AB2 + BC2
=> 132 = 52 + BC2
=> BC2 = 169 − 25
=> BC2 = 144
=> BC = √ 144
=> BC = 12
Sin A = BC⁄AC = 12⁄13
Cos A = AB⁄AC = 5⁄13
Tan A = BC⁄AB = 12⁄5
Example 2. In a right-angled triangle, if Tan A = 4/3 then find the value of Sin a, Cos A, Cot A, Sec A and Cosec A.
Tan A = 4⁄3
=> Perpendicular/Base = 4⁄3
=> BC⁄AB = 4⁄3
Let us assume AB = 3x and BC = 4x
AC2 = AB2 + BC2
=> AC2 = (3x)2 + (4x)2
=> AC2 = 9x2 + 16x2
=> AC = √ 25x2
=> AC = 5x
Sin A = 4x⁄5x = 4⁄5
Cos A = 3x⁄5x = 3⁄5
Cot A = 3x⁄4x = 3⁄4
Cosec A = 5x⁄4x = 5⁄4
Sec A = 5x⁄3x = 5⁄3
Example 3. If Cos A = 4x⁄5x, find the value of Tan A − Sec A.
Let's assume AB = 5x and AC = 13x.
AB2 + BC2 = AC2
=> BC2 = AC2 &miuns; AB2
=> BC2 = (13x)2 − (5x)2
=> BC2 = 169x2 − 25x2
=> BC = √ 144x2
=> BC = 12x
Tan A − Sec A
=> BC⁄AB − AC⁄AB
=> 12⁄5 − 13⁄5
Example 4. In the below given figure, ABC is a right-angled triangle. If BC = 5 cm and AC − BC = 1 cm, then find the value of Sin A and Sec A.
Solution. Here AB = 5 cm and AC − BC = 1 cm.
AC = 1 + BC
As we know, in right angle triangle,
AC2 = AB2 + BC2
=> (1 + BC)2 = 25 + BC2
=> 1 + BC2 + 2BC = 25 + BC2
=> 2BC = 24
=> BC = 12 cm
AC − BC = 1
=> AC = 13 cm
Sin A = BC⁄AC = 12⁄13
Sec A = AC⁄AB = 13⁄5
Example 5. In the below given triangle ABC, AD is perpendicular to BC. If BC = 112 cm, Tan B = 3/4 and Tan C = 5/12 . Find the length of BP and PC.
Solution. As given Tan B = 3⁄4
=> AP⁄BP = 3⁄4
We can assume AP = 3x and BP = 4x
Similarly, Tan C = 5⁄12
=> AP⁄PC = 5⁄12
Here, we can assume AP = 5x and PC = 12x
BP + PC = 112
=> 4x + 12x = 112
=> 16x = 112
=> x = 112⁄16
=> x = 7
BP = 4x = 4 × 7 = 28 cm
PC = 12x = 12 × 7 = 84 cm
Sin A = Perpendicular⁄Hypotenuse and Cosec A = Hypotenuse⁄Perpendicular
As we can see Sin A and Cosec A are reciprocal to each other.
Hence, Sin A = 1⁄Cosec A and Cosec A = 1⁄Sin A
Cos A = Base⁄Hypotenuse and Sec A = Hypotenuse⁄Base
Hence, Cos A = 1⁄sec A and Sec A = 1⁄Cos A
Tan A = Perpendicular⁄Base and Cot A = Base⁄Perpendicular
Here, we can see Tan A and Cot A are reciprocal to each other.
Hence Tan A = 1⁄Cot A and Cot A = 1⁄Tan A
Trigonometrical ratios of angles of 30o and 60o
In the below given diagram, lets assume ABC is an equilateral triangle having each side as 2a and AP is perpendicular to BC.
Here BP = PC = a In right angle triangle AB2 = BP2 + AP2
=> AP2 = AB2 − BP2
=> AP2 = (2a)2 − a2
=> AP2 = 4a2 − a2
=> AP2 = 3a2
=> AP = √3a
Sin 60o = √3a⁄2a = √3⁄2
Sin 30o = a⁄2a = 1⁄2
Cos 60o = a⁄2a = 1⁄2
Cos 30o = √3a⁄2a = √3⁄2
Tan 60o = √3a⁄a = √ 3
Tan 30o = a⁄√3a = 1⁄√3
Trigonometrical ratios of angles of 45o
Below given diagram is a right-angled isosceles triangle in which ∠B = 90o and AB = BC = a
Here ∠A = 45o
Sin 45o = a⁄√2a = 1⁄√2
Cos 45o = a⁄√2a = 1⁄√2
Tan 45o = a⁄a = 1
|Tan||0||1⁄√3||1||√ 3||Not Defined|
|Cot||Not Defined||√ 3||1||1⁄√3||0|
|Sec||1||2⁄√3||√ 2||2||Not Defined|
|Cosec||Not Defined||2||√ 2||2⁄√3||1|
From the above table, we can observe that as the angle increases from 0o to 90o
- Value of Sin increases from 0 to 1
- Value of Cos decreases from 1 to 0
- Value of Tan increases from 0 to infinite
Few formulas to remember for future problem solving.
- Sin2A + Cos2A = 1
- Sec2A − Tan2A = 1
- Cosec2A − Cot2A = 1
Example 1. Find the value of Sin230o + 4 Cos290o + 5 Tan345o.
Solution. Sin230o + 4 Cos290o + 5 Tan345o
= (1⁄2)2 + 4 × (0)2 + 5 × (1)3
= 1⁄4 + 0 + 5
Example 2. If A = 45o, then verify Sec2A − Tan2A = 1.
Solution. Sec245o − Tan245o
= (√ 2 )2 − (1)2
= 2 − 1
Example 3. If a = 15o, then find the value of 4 Sin 2a × Cos 3a × Cot 6a.
Solution. 4 Sin 2a × Cos 3a × Cot 6a
= 4 Sin 30o × Cos 45o × Cot 90o
= 4 × 1⁄2 × 1⁄√2 × 0
Example 4. Prove that Sin 60o = 2 Tan 30o⁄(1+Tan230o).
Solution. First consider LHS (Left Hand Side)
Sin 60o = √3⁄2
Now, lets consider RHS (Right Hand Side)
2 Tan 30o⁄(1+Tan230o)
= 2 (1⁄√3⁄1+1⁄3)
= 2⁄√3 × 3⁄4
= √ 3 ⁄2
Hence LHS = RHS.
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