Subject: Optional Maths
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 :
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 ;
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.
Addition
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) |
6 ratios each of them obtained from these 3 sides they are :
What is the abbreviated form of cotangent ?
The abbreviated form of cotangent is cot.
What is the full form of the ratio cosec ?
The full form of the ratio cosec is \(\frac{h}{p}\) .
In ΔABC given alongside find , all trigonometric ratios for reference angle θ .
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 = \(\frac{h}{p}\)
Find all six trigonometric ratios for the adjoining figure with respect to α and β .β AS
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 = \(\frac{h}{b}\) = \(\frac{PR}{QR}\)
Find the trigonometric ratios for the reference angles mentioned in Greek Alphabets .
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 = \(\frac{h}{b}\) = \(\frac{DC}{AC}\)
Add and subtract the following .
a ) sinθ + 2sinθ
b ) 5tanθ - 2tanθ
a )
Solution;
sinθ + 2sinθ
=sinθ ( 1 + 2 )
= sinθ ( 3 )
=3 sinθ
b )
Solution;
5tanθ - 2tanθ
=tanθ ( 5 - 2 )
=tanθ ( 3 )
=3 tanθ
Find the product of :
a )sin θ × sin 2θ
b ) ( cosec2θ - sec2θ ) by ( cosecθ + secθ )
a )sin θ × sin 2θ
= sin θ × sin 2θ ( am × 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
Write the methods of providing a trigonometric identity?
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.
Add
sinθ + 3 sinθ
Solution
sinθ + 3 sinθ
= sinθ(1 + 3)
= sinθ(4) = 4 sinθ
Subtract
4 tan θ - 2tanθ
Solution
4 tan θ - 2tanθ
= tan θ (4 - 2)
= tan θ(2)
= 2 tanθ
Find the product of:
sinθ × sin3θ
Solution
sinθ × sin3θ
=sin4θ
Find the product of:
( cosθ - sinθ) (cosθ + sinθ)
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
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