 ## 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.

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.
##### Videos for Trigonometric Ratios ##### Simple Trick to Write Trigonometry Table - Nothing to Remember

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

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