Notes on Trigonometric Ratios | Grade 7 > Optional Maths > Trigonometry | KULLABS.COM

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#### 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)

• 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. $$\frac{h}{p}$$
.

#### Click on the questions below to reveal the answers

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 = $$\frac{h}{p}$$

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

= 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

0%

two
six
three
five

sine
tan
cos

cos
sin
tan
cot

cos
tan
cot
sin

(frac{h}{b})
(frac{p}{h})
(frac{b}{h})
(frac{p}{b})

(frac{b}{h})
(frac{b}{p})
(frac{h}{p})
(frac{p}{h})

(frac{p}{h})
(frac{h}{p})
(frac{h}{b})
(frac{p}{b})

sec
sin
cot
tan

sec
sin
cosec
cos

cosine
sine
secant
cot

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