## Trigonometric Ratios

Subject: Optional Maths

#### Overview

This note includes the information about the trigonometric ratios and their values.

#### Trigonometric Ratios:

As we know, in the earlier chapter that right angled triangle consists of three sides i.e ; Perpendicular , Base and Hypotenuse then as we talk about ratios 6 ratios each of them obtained from these 3 sides they are :

1. $\frac{p} {h}$
2. $\frac{b}{h}$
3. $\frac{p}{b}$
4. $\frac{b}{p}$
5. $\frac{h}{b}$
6. $\frac{h}{p}$

Above ratios are called trigonometric ratios and they are given certain names .Detail information is given below ;

Here,

Let Δ MNO be right angled triangle where ∠MNO = 90 ,

∠MNO = θ be the reference angle .

Then,

Side MN = Perpendicular ( p )

Side NO = Base ( b )

Side MO= Hypotenuse ( h )

Now , we can get six ratios here ;

1. $\frac{p}{h}$ = $\frac{MN}{MO}$
2. $\frac{b}{h}$ = $\frac{NO}{MO}$
3. $\frac{p}{b}$ = $\frac{MN}{NO}$
4. $\frac{b}{p}$ = $\frac{NO}{MN}$
5. $\frac{h}{b}$ = $\frac{MO}{NO}$
6. $\frac{h}{p}$ = $\frac{MO}{MN}$

Now , lets introduce the names for these ratios.

 S.No. Ratio Nomenclature Abbreviation 1. $\frac{p}{h}$ sine sin 2. $\frac{b}{h}$ cosine cos 3. $\frac{p}{b}$ tangent tan 4. $\frac{b}{p}$ cotangent cot 5. $\frac{h}{b}$ secant sec 6. $\frac{h}{p}$ cosecant cosec/csc

Again , In the earlier statement if we try to link these ratio we get,

Sinθ = $\frac{p}{h}$ =$\frac{MN}{MO}$

cosθ = $\frac{b}{p}$ =$\frac{NO}{MO}$

tanθ = $\frac{p}{b}$ =$\frac{MN}{NO}$

Cotθ =$\frac{b}{p}$ =$\frac{NO}{MN}$

Secθ = $\frac{h}{b}$ =$\frac{MO}{NO}$

Cosecθ=$\frac{h}{p}$ = $\frac{MO}{MN}$

Operation Of Trignometric Ratios

Here, in trignometry ratio also, addition , subtraction , multiplication and division take place in sameway as in algebra .It is important to have the prior knowledge on algebra before operating trignometric ratios.

In algebra we can add terms just adding their coefficient .

for eg;

1. a + a = 2a

sin θ+ sinθ = 2sinθ

2. a + b = a + b

sinθ + cosθ = sinθ + cosθ

3. a2+ a = a2 + a

sec2 θ+ secθ = sec2θ + secθ

Subtraction

We can subtract the trigonometric ratios as we subtract in algebra.

1. 2a - a = a

2tanθ - tanθ = tanθ

2. a - b = a - b

cosecθ - cotθ = cosecθ - cotθ

3. a2- a = a2- a

sec2θ - secθ = sec2θ - secθ

Multiplication and Division

For, multiplication trigonometric ratios will also follow the laws of indices. Let's review laws of indicesonce .

a) a2× a3 = a ×a ×a ×a ×a = 5

Generalizing this statement : ap×aq= ap+q

b) a6÷ a2 = $\frac{a×a ×a×a×a×a}{a×a}$

= a4

Generaliziting ,

ap÷ aq = apq

c) ( a2)3= a2×a2×a2

= a2+2+2

=a6

Similary,

sin θ× sinθ = sin2θ

2sin2θ ×3sin3θ = 6sin5θ [ ap × aq = ap+q ]

6cos4θ ÷ 2cos2θ =$\frac{6cos^4θ}{2cos^2θ}$ [ ap÷ aq = apq ]

( cot2θ)3 = cotθ6θ [ (ap)q÷ aq= ap×q]

3cosecθ× 4secθ = 12cosecθ.secθ

Some basic algebraical formula

 S.No. Formula Expanded form Factorized form 1. (a + b)2 a2+2ab+b2 (a+b) (a+b) 2. (a - b)2 a2-2ab+b2 (a-b) (a-b) 3. a2 -b2 (a+b) (a-b) 4. a2+b2 (a+b)2 - 2ab (a-b)2 +2ab 5. (a + b )3 a3+3a2b+3ab2+b3 a3+b3+3ab(a+b) (a+b)(a+b)(a+b) 6. ( a- b)3 a3-3a2b+3ab2-b3 a3-b3-3ab(a-b) (a-b)(a-b)(a-b) 7. a3+ b3 (a+b)3-3ab(a+b) (a+b) (a2-ab+b2) 8. a3- b3 (a-b)3+3ab(a-b) (a-b)(a2+ab+b2)

##### Things to remember
• Multiplication trigonometric ratios follow the laws of indices.
• We can subtract the trigonometric ratios as we subtract in algebra.
• 6 ratios each of them obtained from these 3 sides they are :

1. $\frac{p} {h}$
2. $\frac{b}{h}$
3. $\frac{p}{b}$
4. $\frac{b}{p}$
5. $\frac{h}{b}$
6. $\frac{h}{p}$
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Questions and Answers

The full form of the ratio cosec is $\frac{h}{p}$ .

Solution ;

Here, In ΔABC, ∠ABC = 90 ∠BAC = θ be the referance angle .

Then ,

Side BC = perpendicular ( p )

Side AC =  hypotenuse ( h )

Side AB = base  (b )

Now, using trignometry ratios formula we get;

sinθ = $\frac{p}{h}$                                        coscθ = $\frac{b}{h}$

tanθ =  $\frac{p}{b}$                                       cotθ =  $\frac{b}{p}$

secθ = $\frac{h}{b}$                                    cosecθ = $\frac{h}{p}$

Solution ;

Here , ΔPQR is a right-angled triangle , ∠PQR = 90

Taking, ∠PQR = β as the reference angle ,

Perpendicular ( p ) = PQ

Base ( b ) = QR

Hypotenuse ( h ) =PR

Now, using a trignometric ratios we get;

sin β = $\frac{p}{h}$ =  $\frac{PQ}{PR}$
cos β  = $\frac{b}{h}$ =$\frac{QR}{PR}$

tan β =$\frac{p}{b}$  = $\frac{PQ}{QR}$

cot β  = $\frac{b}{p}$ = $\frac{QR}{PQ}$

sec β =$\frac{h}{p}$  =  $\frac{PR}{QR}$

cosec β =$\frac{h}{b}$ =$\frac{PR}{PQ}$

Again ,

Taking ∠QPR = α be the referance angle,

Perpendicular ( p ) = QR

Base ( b ) = PQ

Hypotenuse ( h ) = PR

then, using tregnometric ratios we get,

sinα  = $\frac{p}{h}$ = $\frac{QR}{PR}$

cosα =  $\frac{b}{h}$ = $\frac{PQ}{PR}$

tanα  = $\frac{p}{b}$ = $\frac{QR}{PQ}$

cotα =  $\frac{b}{p}$ = $\frac{PQ}{QR}$

secα  = $\frac{h}{b}$ = $\frac{PR}{PQ}$

cosecα = $\frac{h}{b}$ = $\frac{PR}{QR}$

Solution;

Here, ΔABC and ΔADC are two right-angled triangles having reference angles θ and α within them respectively .

Now, Taking ∠BAC = θ as the reference angle for ΔABC,

Perpendicular ( p ) = BC

Hypotenuse ( h ) = AC

Base ( b ) = AB

Now , using tregnometric ratios  we get ,

sinθ = $\frac{p}{h}$ =  $\frac{BC}{AC}$

cosθ = $\frac{b}{h}$ = $\frac{AB}{AC}$

tanθ = $\frac{p}{b}$ =  $\frac{BC}{AB}$

cotθ = $\frac{b}{p}$ = $\frac{AB}{BC}$

secθ = $\frac{h}{b}$ =  $\frac{AC}{AB}$

cosecθ = $\frac{h}{b}$ = $\frac{AC}{BC}$

Again,

Taking ∠ADC = α as the reference angle , for ΔADC , we get ,

Perpendicular ( p ) = AC

Base  (b) = AD

Hypotenuse ( h ) = DC

Then,

sinα = $\frac{p}{h}$ = $\frac{AC}{DC}$

cosα = $\frac{b}{h}$ = $\frac{AD}{DC}$

tanα = $\frac{p}{b}$ = $\frac{AC}{AD}$

cotα = $\frac{b}{p}$ = $\frac{AD}{AC}$

secα = $\frac{h}{b}$ = $\frac{DC}{AD}$

cosecα = $\frac{h}{b}$ = $\frac{DC}{AC}$

a )

Solution;

sinθ + 2sinθ

=sinθ ( 1 + 2 )

= sinθ ( 3 )

=3 sinθ

b )

Solution;

5tanθ - 2tanθ

=tanθ ( 5 - 2 )

=tanθ ( 3 )

=3 tanθ

a )sin θ × sin 2θ

= sin θ × sin 2θ                     (   a× a n= am + n  )

=sin3 θ

b ) ( cosec2θ - sec2θ )   by ( cosecθ +  secθ )

= $\frac{cosec^2θ-sec^2θ}{cosecθ+secθ}$

=  $\frac{(cosecθ)-(secθ)^2}{cosecθ+secθ}$

=$\frac{(cosecθ+secθ)(cosecθ-secθ)}{cosecθ+secθ}$

=cosecθ-secθ

The methods of providing a trigonometric identity are as follows:

a. Take the identity on the left-hand side(L.H.S) and show it equal  to right-hand side(R.H.S).

b. Take the identity on the R.H.S and show it equal to L.H.S.

Solution

sinθ + 3 sinθ

= sinθ(1 + 3)

= sinθ(4) = 4 sinθ

Solution

4 tan θ - 2tanθ

= tan θ (4 - 2)

= tan θ(2)

= 2 tanθ

Solution

= cos θ ( cosθ + sinθ)  - sinθ ( cosθ + sinθ)

= cos 2 θ + cosθ. sinθ - sinθ . cosθ - sin2θ

= cos 2 θ + cosθ. sinθ - cosθ . sinθ - sin2θ

= cos2θ - sin2θ