Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

7.3 Solution of a system of linear Equation of in two variables by

Inverse Matrix Method

Let us consider two linear equations:

a1x + b1y = c1 ................. (i)

a2x + b2y = c2 ................. (ii)

Writing the above equations in matrix form,

a1x + b1y = c1
c2
( ( ( (a2x + b2y
( ( ( ( ( (or,a1 b1 x
a2 b2 y = c1
c2
or, AX = C ........... (iii)

a1( (where, A =b1 c1( ( ( (C = x
a2 b2 , c2 ,X= y

If A is a non - singular matrix, pre-multiplying (iii) by A-1, we get

A-1(AX) = A-1C

or, (A-1A) X = A-1C

or, I X = A-1C
or, X = A-1C,
? X = A-1 C
which gives the solution of above system of linear equations.
Note :
(a) A unique solution of given system of linear equations is possible if |A| ≠ 0.

If |A| = 0, A-1 does not exist and the solution of above system of equations is not
possible.
(b) Write the equation in the form of ax + by = c.

Steps in solving simultaneous equations of two variables by matrix method:
(a) Write the given equations in the form of ax + by = c , where a, b, and c are constants.

If coefficient of any variable is absent write it zero.
(b) Denote the coefficients of x and y by matrix A and constants in RHS by C and write in

the form AX = C.
(c) If |A| ≠ 0, find the inverse matrix of A., i.e. A-1
(d) Multiply by A-1 on both sides, we get, X = A–1C
Equate the corresponding elements in LHS and RHS which give the values of x and y.

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Example : Solve by matrix method.
Solution:
4x - 3y = 11

3x + 7y = - 1

Writing the given equations in the matrix form.

4 -3 x 11
37 y = -1

or, AX = C

or, X = A-1C ..........(i)

Where, A = 4 -3 ,X= x 11
3 7 y = -1

|A| = 4 -3 = 28 + 9 = 37 ≠ 0.
3 7
Since |A| ≠ 0, A-1 exists and given system of linear equations have unique

solution.

Adj.A = 7 3 A-1 = 1 Adj.A = 1 7 3
-3 4 |A| 37 -3 4

From (i), we get,

X = 1 7 3 11
37 -3 4 -1

= 1 77 – 3 = 1 74 = 2
37 –33 – 4 37 -37 -1

? x = 2 ? x = 2, y = –1
y -1

Worked out Examples

Example 1. Write the given system of linear equation in matrix form:
Solution:
(a) 4x + 3y = 4 and 2x + 7y = 5

(b) 3y - x + 7 = 0 and 2x - 4 = y

(a) Here, given equation are

4x + 3y = 4

and 2x + 7y = 5

writing the above equations in the matrix form,

43 x = 4
27 y 5

(b) Here, given equation are

3y - x + 7 = 0

2x - 4 = y

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These equation must be written in the form of ax + by = c

-x + 3y = -7

or, x - 3y = 7 .......... (i)

and 2x - y = 4 ........... (ii)

Writing the equations (i) and (ii) in the matrix form

1 -3 x = 7
2 -1 y 4

Example 2. Factorise 2x +3y
Solution: 4x - 7y
Example 3.
Here, 2x +3y = 23 x
Solution: 4x - 7y 4 -7 y

Solve by matrix method :

4x + 8y = 2

2x + 4y = 2

Given equations can be written as

4x + 8y= 2 or 2x + 4y = 1 ......(i)

and 2x + 4y = 2 or, x + 2y = 1 ..... (ii)

writing the equations (i) and (ii) in the matrix form,

24 x = 1
12 y 1

or, AX = C ........... (i)

Where A = 2 4 ,C= 1 ,X= x
1 2 1 y

|A| = 2 4 =4-4=0
1 2

Since |A| = 0, A-1 does not exist and given system of equations do not have

unique solution.

Example 4. Solve by matrix method :
Solution:
3 + 4y = 9 2 +y= 8
x x 3

Writing the equations in the matrix form.

34 19
21 x
y = 8
3

or, AX = C ..............(i)

3 4 1 9
2 1 x 8
where, A = , X= y , C= 3

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|A| = 3 4 = 3 - 8 = -5 ≠ 0
2 1

Since |A| ≠ 0, A-1 exists and the given system of linear equations have unique

solution.

Adj. A = 1 -4
-2 3

A-1 = 1 Adj.A = 1 1 -4
|A| -5 -2 3

Now from (i), X = A-1 C = 1 1 -4 9
-5 -2 3 8
3

1 9 - 32 1 27-32
-5 3 -5 3
= =
-18 +234 -18 + 8

= 1 -5 1
-5 3 =3

-10 2

11
? x =3

y2

i.e. 1 = 1 and y = 2 ? x = 3 and y = 2
x3
Example 5.
Solution Solve by matrix method.

2 + 5 =1 and 3 + 2 = 19
x y x y 20

Writing the given equations in matrix form

25 1 1
32 x = 19
1
y 20

or, AX = C .............. (i)

where, A = 2 5 , X= 1 , C= 1
|A| = 3 2 x 19
1 20
y

2 5 = 4- 15 = - 11 ≠ 0
3 2

Since |A| ≠ 0, A-1 exists and the given system of linear equations have unique
solution.

Adj. A = 2 -5 , A-1 = 1 Adj. A. = 1 2 -5
-3 2 |A| -11 -3 2

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Now, from (i)

X = A-1C = 1 2 -5 1
-11 -3 2 19
20

1 2 - 19 1 -141 1
11 4 11 -1110 4
=- =- = 1
-3 + 19 10
10
11
x 4
? 1 = 1

1 y 10
x 1
i.e. = 4 or, x=4
or, y = 10
and 1 = 1
y 10

? x = 4 and y = 10

Exercise 7.3 (b) B = 2x-4y

Short Questions
1. Factorize the given matrices:

(a) A = 4x+3y

x-y 5x + y

2. If A = a b , then answer the following questions:
c d

(a) Find determinant of matrix A..

(b) Under what condition A does not have its inverse.

3. If A, C, and X are three square matrices such that AX = C, write X in terms A-1 and C.

4. (a) If 1 0 X= 4 , find the value of X.
0 1 5

(b) If 32 X= 13 , find the value of X.
23 12

5. Check whether the system of linear equations have unique solution or not.

(a) x + y = 4 (b) 4x + 2y = 8 (c) 3x + 11y = 7

x-y=3 x-y=1 6x + 22y = 5

(d) x - 2y = 4 (e) 3x + 5y = 2 (f) 3 + 2 =1
x y
5 3
3x - 5y - 7 = 0 5y - 3x = 3 x - y =9

6. Solve for x and y in each of the following.

(a) 1 -8 x = -7 (b) 1 -2 x = -7
23 y 5 37 y 5

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(c) 57 x = 1 (d) 4 -3 x = 11
14 y -5 37 y -1

Long Questions

7. Solve the following system of linear equations by matrix method:

(a) x+ y = 5 (b) 9x - 8y = 12 (c) 3x + 5y = 21

x-y=3 2x + 3y =17 2x + 3y = 13

(d) x - 2y = -7 (e) 3x - 2y = 5 (f) 3x + y = 3

3x + 7y = 5 x+y=5 3x + 2y = 2

(g) 2x - 3y = 1 (h) 4x - 3y = 11 (i) x = 2y - 1

4y + 3x = 10 3x = 5 - y y = 2x

(j) 2x = 1 - y (k) 3x + 2y = 9 (l) 3x + 2y - 3 = 0

3x + 2y = 1 2x + 3y = 11 6x + 5y - 9 = 0

8. Solve the following equation by matrix method:

(a) x + y =4 (b) x + y = 7 (c) 5x + y = 5
3 3 2 2 2
x y 4
5x - 4y = - 3 2 + 3 = 3 3 x - y = -2

(d) x= 2 y (e) x - 2y = -7 (f) 3x + 2y = 1
3 2
3x 7y xy

4x - 3y = 1 5 + 5 =1 3 - 3 =1

9. Solve the following equations by matrix method:

(a) 4 + 3 =1 (b) 3 + 2 = 13 (c) 4 + 2y = 8
x y x y x
3 2 1 5 3 2
x - y = 24 x - y = 9 x + 3y = 10

(d) 3x + 4 = 10 (e) 3 + 1 = 5 (f) x + y =1
y x y 6 4
y
-2x + 3 = - 1 2 - 1 = 5 x - 4 =1
y x y 2
2 1 1
(g) x + y =2 (h) x + 2y =8 (i) 2x + 3y = 10xy
3 4 =5
y 1 1 5 4
x+ 2x - y = -1 x + y = 13

(j) 3y + 4x = 2xy (k) 3x + 5y = 7x + 3y = 4
4 5
7y + 5x = 29xy

(l) 2x + 4 =y= 40 - 3x (m) 2x + 5y = 1 = 3x - 2y
5 4 5
10. Equations of pair of lines are 2x - y = 5 and x - 2y = 1. Do the following:

(a) Write the equations in matrix form

(b) Is there unique solution of above given equations ?

(c) Solve the equations.

(d) check the solutions to show that the values x and y so obtained are true.

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1. (a) 43 x (b) 2 -4 x
1 -1 y 51 y

2. (a) ad - bc (b) |A| = 0 3. X = A-1C

4. (a) 4 (b) 3
5 2

5. (c) It does not have unique solution, others have solutions.

6. (a) (1, 1) (b) (-3, 2) (c) (3, -2) (d) (2, -1)
(d) (-3, 2)
7. (a) (4, 1) (b) (4, 3) (c) (2, 3) (h) (2, -1)
(l) (-1, 3)
(e) (3, 2) (f) 4 , -1 (g) (2, 1) (d) (-2, -3)
3
1 2
(i) 3 , 3 (j) (1, -1) (k) (1, 3)

8. (a) (5, 7) (b) (-10, 24) (c) 18 , 70
23 23

(e) (-3, 2) (f) (2, -1)

9. (a) (8, 6) (b) 1 , 1 (c) (2, 3) (d) (2, 1)
3 2 (g) (3, 2)
1 (k) (2, 2) 1 1
(e) 2 , -1 (f) (3, 2) (h) 6 , 4

(i) 1 , - 2 (j) 13 , - 13 (l) (8, 4)
7 11 106 73

(m) 15 , 13 10.(b) (3, 1)
19 19

7.4 Cramer’s Rule

We discuss the method of solving simultaneous equations consisting of two variables with
the help of determinants. It is also used to solve simultaneous equations with three variables.
But we limit it in solving simultaneous equations of two variables only.

Cramer's rule is the process of finding solution of simultaneous equations by determinant
method. It is also called determinant method of solving simultaneous equations of two or
three variables.

Solution of simultaneous equations with two variables by Cramer's Rule:
Let, a1x + b1y = d1 ........ (i)
and a2x + b2y = d2 ........ (ii)
be two linear equations, where x and y are variables and the remainings are constants.

Now, multiplying equation (i) by b2 and equation (ii) by –b1 and adding them, we get,
a1b2x + b1b2y = b2d1
–a2b1x – b1b2y = - b1d2

Adding, (a1b2 – a2b1)x = b2d1 – b1d2

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or, x = d1b2 – d2b1
a1b2 – a2b1
This equation can be put in the form of determinant

x= dd12 bb21 = D1
a1 b1 D

a2 b2

Where, D1 = add1 21 bbb121 , than D z 0, where D2 = aa21 dd21
Similarly, D= DDa22 , b2
y= provided

Note :

(i) D1 is also written as '1 or Dx

(ii) D2 is also written as '2 or Dy

(iii) If D = 0, there is no solution of given system of equations.

Steps to be used in Cramer's rule

The following steps are to be used in finding the solution of simultaneous linear equations

of two (or three) variables by using Cramer's rule:

(a) Arrange the given system of equations such that the terms containing the variables x
and y on left hand side and constant on the right hand side.

(b) Arrange the system of equations such that the term containing the variable x in the
first, then y.

(c) If any variable is missing in the equation, write the variable with coefficient zero.

(d) Collect the coefficients of the variables x and y, and the constant term and put them in
order.

(e) Write the determinant D with the elements of the coefficients of x and y.

(f) Write tfhirestdvetaerrimabilnea)nbtyDth1 ferocmonDstabnyt replacing the first column of D (i.e., the coefficient
of the column.

Similarly write determinant D2 replacing the second column of D by the constant column.

(g) The, use the formula:

x= D1 , y = D2
D D

Provided that D z 0.

Note :
If D = 0, there is no solution of given system of equations.

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Worked out Examples

Example 1: Solve the following simultaneous equations by using Cramer's rule:
Solution: x+y=5
x-y=3
Here, x + y = 5

x-y=3
Writing the coefficients of x, y and constants.

coefficient of x coefficient of y constant
5
11 3

1 -1

Now, D = 1 1 = -1 – 1 = -2
1 -1
5 1
D1 = 3 -1 = -5 – 3 = -8

D2 = 1 5 = 3 – 5 = -2
1 3

Using Cramer's rule,

x = D1 = -8 = 4

D -2
DD2 -2
y= = -2 =1

? x = 4 and y = 1

Example 2: Solve the given equations by using Cramer's rule:
Solution:
2x – 3y = 3 4x – y = 11

Here,

coefficient of x coefficient of y constant
3
2 -3 11

4 -1

Now, D = 2 –3 = –2 + 12 = 10
4 –1

D1 = 3 –3 = – 3 + 33 = 30
11 –1

D2 = 2 3 = 22 – 12 = 10
4 11

By using Cramer's rule,

x= DD1 = 30 =3
10

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y= D2 = 10 =1
D 10

? x = 3 and y = 1.

Example 3: Solve the given simultaneous equations by using determinant method.
Solution: 2(x + y) = 3x
7x = 2(3y + 7)
From the first equation,

2x + 2y = 3x
or, 2x – 3x + 2y = 0
? –x + 2y = 0 ........... (i)
From the second equation,

7x = 6y + 14
or, 7x – 6y = 14 ............ (ii)
Now,

coefficient of x coefficient of y constant
-1 2 0
7 -6 14

–1 2

Now, D = 7 –6 = 6 – 14 = –8
0 2
D1 = 14 –6 = 0 – 28 = –28

D2 = –1 0 = –14 – 0 = –14
7 14

By using Cramer's rule, we get

x= DD1 = –28 = 7
–8 2
D2 –14 7
y= D = –8 = 4

? x = 7 and y = 74.
2

Example 4: Solve the given simultaneous equations by using cramer's rule.
Solution: 3x + 4y – 4 = 0
2x + 5y + 7 = 0
From the first equation,

3x + 4y = 4
and from the second equation

2x + 5y = –7

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Now,

coefficient of x coefficient of y constant

344

2 5 -7

Now, D = 3 4 = 15 – 8 = 7
2 5
4 4
D1 = –7 5 = 20 + 28 = 48

D2 = 3 4 = –21 – 8 = –29
2 –7

By using Cramer's rule,

x= D1 = 48
D 7
D2
y= D = –29
7
48 -29
? x= 7 and y = 7

Example 5: A marketing boy sales 10 kg of rice, 12 kg of tea at cost Rs 1460 on Sunday.
Solution: On Monday he sales 5 kg of rice and 7 kg of sugar at cost Rs 810. Find the
costs of each kg of rice and sugar. (Use Cramer's rule)

Let Rs x and Rs y be the prices of per kg of rice and sugar respectively.

Then, we have,

10x + 12y = 1460
or, 5x + 6y = 730 ........ (i)
and 5x + 7y = 810 ........ (ii)
Now, to solve above equations using Cramer's rule,

coefficient of x coefficient of y constant

5 6 730

5 7 810

Now, D = 5 6 = 35 – 30 = 5
5 7
730 6
D1 = 810 7 = 5110 – 4860 = 250

D2 = 5 730 = 4050 – 3650 = 400
5 810

By using Cramer's rule,

x= D1 = 250 = 50
D 5
D2 400
y= D = 5 = 80

? The costs of per kg of rice and sugar are Rs 50 and Rs 80 respectively.

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Exercise 7.4

Very Short Questions :

1. (a) In equations a1x + b1y = c1 and a2x + b2y = C2, what are the determinants
representing D, Dx and Dy?

(b) If D = 20, Dx = 20, Dy = 40, write the values of x and y.

(c) If D = 4, Dx = 1 , Dy = 1 , find the values of x and y.
2 4

Long Questions :

2. Solve the following system of equations by using Cramer's rule:

(a) 3x – 2y = 1 and –x + 4y = 3

(b) 3x – 5y = 2 and 2x + y = 4

(c) 2x – 3y = 2 and 4x – y = 1

(d) 5x – 4y = –3 and 7x + 2y = 6

(e) 2x + 3y = 17 and 9x – 8y = 12

(f) 2x – y = 1 and 3x + y = 9

(g) 2x – 3y = 3 and 4x – y = 11

(h) 3x + 5y = 21 and 2x + 3y = 13

(i) 5x + 8y = 340 and 7x + 6y = 320

(j) 5x – 3y = 8 and 9x + 7y = 126

3. Solve the following system of equations by using Cramer's rule:

(a) 3 + 2 = 1 and 4 + 3 = 17
x y x y 6
3 5 4 3 29
(b) x + y = 1 and x + y = 30

(c) x – y = 2 and x + y = 59
3 4 8 3 24
2 5 3 6
(d) x + y = 30 and x + y = 27

(e) x – y = –6 and 3x – 1 = y
6 4

4. Solve the following equations by Cramer's rule:

(a) 2x – 5 = 28 and 4x + 3 = –9
y y
10 4
(b) x – 2y = –1 and x + 3y = 11

(c) 8 – 9 = 1 and 10 + 6 = 7
x y x y

(d) 2(3x – y) = 5(x – 2) and 3(x + 4y) = 2(y – 3)

(e) 7(x – y) = x + y and 5(x + y) = 35 (x – y)

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5. A helicopter has 4 seats for passengers. Those willing to pay first class fares can take 60
kg of baggage each, but tourist class passengers are restricted to 20 kg each. It can carry
only 120 kg of baggage altogether. To find the number of passengers of each kind, use
Cramer's rule.

6. If the total cost of 6 kg potatoes and 4 kg of tomatoes is Rs. 200 and the total cost of 1 kg
of potatoes and 1 kg tomatoes is 42. Find the price of per kg of each vegetable by using
determinant.

Project Work

7. Write two simultaneous equations related to cost of two daily used commodities and
solve them by using Cramer's rule.

1. (a) D = aa12 bb12 , Dx = cc12 bb21 , Dy = aa12 cc12 (b) (1, 2) (c) 1 , 1
8 16

2. (a) (1, 1) (b) 22 , 8 (c) 1 , - 3 (d) 9 , 51
13 13 10 5 19 38

(e) (4, 3) (f) (2, 3) (g) (3, 1) (h) (2, 3)

(i) (20, 30) (j) (7, 9) 3.(a) - 3 , 2 (b) (6, 10)
(c) (9, 4) (d) - 1 , 1 8 9

(e) (12, 8)

4. (a) 3 , - 1 15 12 (c) (2, 3) (d) -7, 3
2 5 (b) (2, 3) 6. (16, 26) 2
5. (1, 3)
(e) No solution

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Trigonometry

9.0 Review of Trigonometricy Ratios of Compound Angle

Let A and B be any two angles, the sum (A+ B) and the difference (A - B) are known as
compound angles. The following are the list of formula of compound angles.

(i) sin (A + B) = sinA.cosB + cosA. sinB

(ii) cos (A + B) = cosA. cosB - sinA. sinB

(iii) tan (A + B) = tanA + tanB
1 - tanA tanB
cotA.cotB-1
(iv) cot (A + B) = cotB + cotA

(v) sin (A -B) = sinA. cosB - cosA. sinB

(vi) cos (A - B) = cosA.cosB + cosA. sinB

(vii) tan (A -B) = tanA - tanB
1 + tanA.tanB
cotA.cotB + 1
(viii)cot (A-B) = cotB - cotA

(ix) sin (A + B). sin (A - B) = sin2A - sin2B = cos2B - cos2A

(x) cos(A + B).cos (A - B) = cos2A - sin2B = cos2B - sin2A

(xi) tan(A + B). tan(A - B) = tan2A – tan2B
1 - tan2A.tan2B
cot2A.cot2B-1
(xii) cot(A + B). cot (A - B) = cot2B - cot2A

9.1 Trigonometric Ratios of Multiple Angles

Let A be any angle. Then, 2A, 3A, 4A, ... etc. are called multiple angles of A. In this
sub-units, we discuss the trigonometric ratios of angles 2A and 3A.

I. Trigonometric Ratios of 2A
We know that sin (A + B) = sinA.cosB - cosA.sinB.

(a) sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA
+ tan2A + cot2A

Here, sin2A = sin(A + A) = sinA.cosA + cosA.sinA

? sin2A = 2sinA.cosA............ (i)

sin2A 2sinA.cosA = 2sinA.cosA
1 sin2A + cos2A

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2sinA.cosA
cos2A
= sinA +cosA = 2tanA
tan2A + 1
cos2A
2tanA
? sin2A = 1 + tan2A ...............(ii)

2tanA 2 1 1 2cot2A 2cotA
1 + tan2A cotA cotA cot2A + 1 + cot2A
Also, sin2A = = = . =
1 1
1 + cot2A
2cotA
? sin2A = 1 + cot2A ......... (iii)

Combining above (i), (ii) and (iii), we get,

sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA
+ tan2A + cot2A
1-tan2A cot2A -1
(b) cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1 + tan2A = 1 + cot2A

Here, cos(A + B) = cosA.cosB - sinA.sinB

Now, cos2A = cos(A + A) = cosA.cosA - sinA.sinA

? cos2A = cos2A - sin2A ............ (i)

cos2A = cos2A - sin2A = 1 - sin2A - sin2A

? cos2A = 1 - 2sin2A ............... (ii)

or, 2sin2A = 1 - cos2A.

? cos2A = 2cos2A - 1 .............(iii)

2cos2A = 1 + cos2A

Again, cos2A = cos2A - sin2A

cos2A-sin2A
cos2A
= cos2A - sin2A = sin2A + cos2A = 1- tan2A
sin2A + cos2A 1 + tan2A
cos2A
1-tan2A
? cos2A = 1+tan2A ........... (iv)

1 - 1 cot2A - 1
cot2A cot2A + 1
and cos2A = = .......... (v)
1
1 + cot2A

Combining (i), (ii), (iii), (iv), and (v), we get

cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1- tan2A = cot2A -1
1 + tan2A cot2A + 1
2tanA
(c) tan2A = 1 - tan2A

Here, tan(A + A) = tanA + tanA = 2tanA
1 - tanA.tanA 1 - tan2A
2 tanA
? tan2A = 1 – tan2A

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(d) cot2A = cot2A - 1
2cotA
cotA.cotB - 1
Here, cot(A + B) = cotB + cotA

Now, cot2A = cot(A + A) = cotA.cotA -1 = cot2A - 1
cotA + cotA 2cotA
cot2A - 1
? cot2A = 2cotA

II. Trigonometric Ratios of 3A.

(a) sin3A = 3sinA - 4sin3A
Now, sin3A = sin(A + 2A)

= sinA.cos2A + cosA.sin2A

= sinA (1 - 2sin2A) + cosA. 2sinA.cosA

= sinA - 2sin3A + 2sinA.cos2A

= sinA - 2 sin3A + 2sinA(1 - sin2A)

= sinA - 2sin3A + 2sinA - 2sin3A = 3sinA - 4sin3A

? sin3A = 3sinA - 4sin3A

Also, 4sin3A = 3sinA - sin3A

(b) cos3A = 4cos3A - 3cosA

Here, cos3A = cos(A + 2A)

= cosA.cos2A - sinA.sin2A

= cosA. (2cos2A - 1) - sinA.2sinA.cosA

= 2cos3A - cosA -2sin2A.cosA

= 2cos3A - cosA - 2(1 - cos2A) cosA

= 2cos3A - cosA - 2cosA + 2cos3A = 4cos3A - 3cosA.

Also, 4cos3A = 3cosA + cos3A

(c) tan3A = 3tanA - tan3A
1 - 3tan2A

Here, tan3A = tan (A +2A)

= tanA + tan2A = tanA + 2tanA
1-tanA.tan2A 1-tan2A
1- tanA.
tanA-tan3A +2tanA 2tanA
1-tan2A

= 1 - tan2A = 3tanA - tan3A
1 - tan2A -2tan2A 1 - 3tan2A
1-tan2A

? tan3A = 3tanA - tan3A
1- 3tan2A

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(d) cot3A = cot3A - 3cotA
3cot2A - 1

Here, cot3A = cot(A + 2A)

= cotA.cot2A - 1 = cotA. cot2A -1 - 1
cot2A + cotA 2cotA

cot2A -1 + cotA
2cotA

= cot3A - cotA - 2cotA . 2cotA
2cotA cot2A - 1+2cot2A
cot3A- 3cotA
= 3cot2A - 1

? cot3A = cot3A - 3cotA
3cot2A - 1

III. Geometrical Interpretation of Trigonometric Ratios of 2A.

Let a revolving line OP make an angle ‘XOP = 2A with an initial line OX.

Now, taking O as the centre and OP as radius of a circle, circle PMN is drawn. MN is the
diameter of the circle. OP, MP, PN are joined. PR is drawn perpendicular to MN. MPN
is semi-circle with a diameter MN, ‘MPN = 90° (angle at semi-circle is right angle).

‘MNP = 1 ‘NOP = 1 × 2A = A
2 2
(Circumference angle is half central angle standing on the same arc PN)

‘OPM = ‘OMP = A (? ∆ MOP is an isosceles base angles are equal).

‘MPR = 90° - A, ‘MPN = 90°, ‘RPN = 90° - ‘MPR = 90° - 90° + A = A

? ‘RPN = A Y

In right angled ∆ PRM, P

sinA = PR , cosA = RM , tanA = PR X' A 2A A X
PM PM MR M RN
In right angled ∆MPN, O

sinA = PN , cosA = PM
MN MN
Now, in right angled ∆ ORP, Y'

(i) sin2A = PR = PR = 2PR = 2PR . PM = 2sinA.cosA
OP MN PM MN
1 MN
2
OR 2OR OR + OR
(ii) cos2A = OP = 2OP = 2OP

= (ON - RN) + (RM - OM)
2OP
OM - RN + RM - OM
= 2OP

= RM - RN
MN
RM RN RM PM RN PN
= MN - MN = PM . MN – PN . MN

= cosA.cosA - sinA.sinA = cos2A - sin2A

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(iii) tan2A = PR
OR
PR PR
= 2PR = 2PR = 2. RM = 2. RM
2OR RM-RN
1- RN 1- RN . PR
RM PR MR
2tanA 2tanA
= 1 - tanA.tanA = 1 - tan2A

Worked Out Examples

Example 1. If sinA = 3 , find sin2A, cos2A and tan2A.
Solution: 5
3
Here, sinA = 5

cosA = 1 - sin2A = 1 - 9 = 4
25 5
3
5 3
tanA = sinA = 4 = 4
cosA
5
Now, sin2A = 2sinA. cosA = 2. 53 . 4 = 24
5 25

cos2A = cos2A - sin2A

= 16 - 9 = 16 - 9 = 7
25 25 25 25

24
25
tan2A = sin2A = 7 = 24
cos2A 7

25

Example 2. Prove that, cotT = ± 1+cos2T
Solution: 1-cos2T
We have 2cos2T = 1 + cos2T............ (i)
Example 3.
Solution and 2sin2T = 1 - cos2T ........... (ii)

Dividing (i) by (ii), we get,

2cos2T = 1 + cos2T
2sin2T 1-cos2T
1 + cos2T
or, cot2T = 1-cos2T

? cotT = ± 1+cos2T = RHS Proved.
1-cos2T
4
If sinA = 5 , then, find the values of sin3A, cos3A and tan3A.

Here, sinA = 4
5
16
Now, cosA = 1 - sin2A = 1 - 25 = 3
5

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4
5
tanA = sinA = 3 = 4
cosA 3

5
sin3A = 3sinA - 4sin3A

= 3. 4 - 4. 64
5 125

= 12 - 256 = 300 - 256 = 44
5 125 125 125

cos3A= 4cos3A - 3cosA

=4 27 - 3. 3 = 108 - 9 = 108-225 =- 117
125 5 125 5 125 125
3tanA - tan3A
tan3A = 1 - 3tan2A

3. 4 - 64 4– 64 108 - 64 44
3 27 1- 27 7(3 - 16) 117
= = 16 = = –
16 3
1- 3. 9

Example 4. (If = 1 m + 1 (, then, show that :
Solution: sinA 2 m ((

((a) 1 m2 (+1 ((b) = - 1 m3 + 1
cos2A = - 2 (m2 sin3A 2 m3
((
(Here, sinA = 1 m + ((1
2 m
(
(a) cos2A = 1 - 2sin2A
(
(= 1 - 2 . ( 1 m + 1 2
4 m
(= ( 1 1 1
((1 - 2. 4 m2 + 2.m. m + m2

=( 1- 1 m2 - 1 - 1 . 1
2 2 m2
(=
- 1 m2 + 1 = LHS. Proved.
2 m2

(b) sin3A = 3sinA - 4sin3A

(= 3 1 m+ 1 (- 4 .1 m + 1 3
2 m 8 m
[ ( ]3 -
(= 1 m+ 1 m + 1 2
2 m m
[ ]3
(= 1 m+ 1 - m2 - 2.m. 1 - 1
2 m m m2
[ ]-
(= 1 m+ 1 (m2 - 1 + 1
2 m m2
( (=
– 1 m + 1 m2 - m. 1 + 1
2 m m m2
(= –
1 m3 + 1 = RHS Proved.
2 m3

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Example 5. Prove the following:
Solution:
(a) sin2A - sinA = tanA (b) 1- cosA = tan2A
1-cosA + cos2A 1+ cos2A
(d) 1 + sec2A = cotA
(c) tanA + cotA = 2cosec2A tan2A

(a) LHS = sin2A - sinA
1-cosA + cos2A
2sinA.cosA - sinA sinA(2cosA - 1)
= 1 - cosA + 2cos2A - 1 = cosA(2cosA -1)

= sinA = tanA = RHS. proved
cosA
1 - cos2A 2sin2A sin2A
(b) LHS = 1 + cos2A = 2cos2A = cos2A

= tan2A = RHS Proved.

(c) LHS = tanA + cotA = sinA + cosA
cosA sinA
sin2A + cos2A 1 2
= sinA.cosA = 2. 2sinA.cosA = sin2A

= 2cosec2A = RHS Proved.

(d) LHS = 1 +sec2A = 1+ = cos2A + 1 × cos2A
tan2A cos2A cos2A sin2A

sin2A
cos2A
2cos2A cosA
= 2sinA.cosA = sinA = cotA = RHS Proved.

Example 6. Proved that:
Solution:
(a) sin6T + cos6T = 1 (1 + 3cos22T) (b) cos4T = 1 - 8sin2T + 8sin4T
4
(a) LHS = sin6T + cos6T

= (sin2T)3 + (cos2T)3

= (sin2T + cos2T) (sin4T - sin2T.cos2T + cos4T)

= 1.(sin4T + cos4T - sin2T.cos2T)

= (sin2T + cos2T)2 - 2sin2T.cos2T - sin2T.cos2T

= 12 - 3sin2T.cos2T

=1- 3 (2sinT . cosT)2
4
3
=1- 4 sin22T

= 1 (4 - 3sin22T)
4
1
= 4 (1 + 3 - 3sin22T)

= 1 {1 + 3 (1 - sin22T)} = 1 (1 + 3cos22T) = RHS. Proved.
4 4
(b) LHS = cos4T

= cos2(2T)

= 1 -2sin22T

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= 1 - 2[2sinT .cosT]2 = 1 – 8sin2T (1 - sin2T)]
= 1 - 8sin2T + 8sin4T = RHS. Proved.

Example 7. PLHroSved=thsaitn110s°in1-10ct°aons-6100c°°os130° =4
Solution:
= 1 - sin60°
Example 8. sin10° cos60°.cos10°
Solution: cos60°.cos10° - sin60°.sin10°
= sin10°.cos60°.cos10°
Example 9.
Solution: = cos (60° + 10°) = 2 . 2cos70° = 4cos70°
2 sin10°.cos10° sin20°
206 sin10° . 1 cos10°
2
4cos70° 4cos70°
= sin(90° -70°) = cos70° = 4 = RHS. Proved.

Proved that (a) cosec2T + cot4T = cotT - cosec4T

(b) 4cosec2T . cot2T =cosec2T - sec2T

(a) LHS = cosec2T + cot4T

= 2ssiicnn11o22sTT2T+++cs2coisnosi4sn4c2TT2o(T2s.4TcT)os2T
= 2sin2T . cos2T
=

= 2cos2T + 2cos22T - 1
2sin2T.cos2T
2cos2T(1 + cos2T) - 1
= 2sin2T.cos2T

= 2cos2T.2cos2T - 1 = 2cos2T – 1
= 2sin2T.cos2T sin2(2T) 2sinT.cosT sin4T
cosT 1
sinT - sin4T = cotT - cosec4T = RHS. Proved.

(b) LHS = 4cosec2T.cot2T

= 4. 1 . cos2T
sin2T sin2T
4cos2T
= 2sinT.cosT.2sinT.cosT

= cos2T - sin2T = cos2T - sin2T
sin2T.cos2T sin2T.cos2T sin2T.cos2T
1
= 1 - cos2T = cosec2T - sec2T = RHS. Proved.
sin2T

Proved that (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1) = 2cos8T + 1

LHS = (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1)

= {(2cosT)2 - 12} (2cos2T - 1) (2cos4T -1)

= (4cos2T –1) (2cos2T - 1) (2cos4T - 1)

= {2(2cos2T - 1) + 1} (2cos2T - 1) (2cos4T - 1)

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vedanta Excel In Opt. Mathematics - Book 10

= (2cos2T + 1) (2cos2T - 1) (2cos4T - 1)
= (4cos22T - 1) (2cos4T - 1)
= {2 (2cos22T - 1) + 1} (2cos4T - 1)
= (2cos4T + 1) (2cos4T – 1)
= 4cos24T - 1
= 2(2cos24T - 1) + 1
= 2cos8T + 1 = RHS. Proved.

Example 10 Prove that cos2D + sin2D . cos2E = cos2E + sin2E . cos2D
Solution: LHS = cos2D + sin2D . cos2E
= cos2D + sin2D(1 - 2sin2E)

= cos2D + sin2D - 2sin2E . sin2D
= 1 - 2sin2E . sin2D = cos2E + sin2E - 2sin2E . sin2D
= cos2E + sin2E(1 - 2sin2D)
= cos2E + sin2E . cos2D = RHS Proved.

Example 11. Prove that 2+ 2+ 2+ 2 + 2cos16T = 2cosT

Solution: LHS = 2+ 2+ 2+ 2 + 2cos16T

= 2+ 2+ 2+ 2(1+cos16T)

= 2+ 2+ 2+ 2.2cos28T

= 2+ 2+ 2 + 2cos8T

= 2+ 2+ 2(1+cos8T)

= 2+ 2+ 2 . 2cos24T

= 2+ 2+2cos4T

= 2+ 2(1+cos4T) = 2+ 2.2cos22T)
= 2(1+cos2T) = 2.2cos2T = 2cosT = RHS. Proved.

Example 12. Prove that 1 - 1 cotA = cot2A
tan3A - tanA cot3A -
1 1
Solution: LHS = tan3A - tanA - cot3A - cotA

= 1 - 1
tan3A -
tanA 1 - 1
tan3A tanA

= 1 tanA - tanA.tan3A
tan3A - tanA - tan3A

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vedanta Excel In Opt. Mathematics - Book 10

= 1 + tanA.tan3A = 1= 1
tan3A - tanA tan3A - tanA tan(3A - A)

= 1 = cot2A 1 +tanA.tan3A
tan2A = RHS. Proved.

Example 13. Proved that 4(cos320° + cos340°) = 3(cos20° + cos40°)

Solution: LHS = 4 (cos320 + cos340°)

= 4cos320 + 4cos340°

= 3cos20° + cos(3.20°) + 3cos40° + cos(3.40°)

= 3cos20° + cos60° +3cos40° + cos120°

= 3(cos20° + cos40°) + 1 - 1
2 2
= 3(cos20° + cos40°) = RHS. Proved.

Example 14. Prove that sin3T + sin3T = cotT
Solution: cos3T - cos3T
sin3T + sin3T 3sinT - 4sin3T + sin3T
LHS = cos3T - cos3T = cos3T - 4cos2T + 3cosT

= 3sinT - 3sin3T = 3sinT(1-sin2T) = sinT . cos2T
3cosT - 3cos3T 3cosT(1-cos2T) cosT . sin2T
cosT
= sinT = cotT = RHS. Proved

Example 15. Prove that : tanT + 2tan2T + 4tan4T + 8cot8T = cotT

Solution: LHS = tanT + 2tan2T + 4tan4T + 8cot8T

= tanT + 2tan2T + 4tan4T + 8
tan(2.4T)
8
= tanT + 2tan2T + 4tan4T + 2tan4T

1 - tan24T

(= tanT + 2tan2T + 8(1-tan24T)
2tan4T
4tan4T + (

(= tanT + 2tan2T + 4 tan4T + 1-tan24T(
tan4T(
(
(= tanT + 2tan2T + 4 tan24T + 1-tan24T
tan4T
4
= tanT + 2tan2T + tan4T

= tanT + 2tanT 4
2tan2T

(= tanT + 2 1 - tan22T
1-tan22T
tan2T + tan2T

= tanT + 2 tan22T +1- tan22T
tan2T

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= tanT + 2
2tanT

1 - tan2T

= tanT + 2(1 - tan2T)
2tanT
tan2T + 1 - tan2T 1
= tanT = tanT = cotT = RHS. Proved.

Example 16. Prove that

(a) 4sin3A.cos3A + 4cos3A.sin3A = 3sin4A

(b) cos3A.cos3A + sin3A. sin3A = cos32A

Solution: (a) We have, 4sin3A = 3sinA - sin3A

4cos3A = 3cosA + cos3A

Now, LHS = 4sin3A.cos3A + 4cos3A.sin3A

= (3sinA – sin3A). cos3A + (3cosA + cos3A).sin3A

=3sinA.cos3A - sin3A.cos3A + 3sin3A.cos3A + cos3A.sin3A

= 3(sinA.cos3A + sin3A.cosA)

= 3.sin(A + 3A) = 3sin4A = RHS Proved.

(b) LHS = cos3A cos3A + sin3A.sin3A

= 1 (3cosA + cos3A).cos3A + 1 (3sinA - sin3A).sin3A)
4 4
1
= 4 [3cosA.cos3A + cos23A + 3sinA.sin3A - sin23A]

= 1 [3(cosA.cos3A + sinA.sin3A) + (cos23A - sin23A)]
4
1
= 4 [3cos(A - 3A) + cos6A]

= 1 (3cos2A + cos6A)
4
1 1
= 4 [3cos2A + cos(3.2A)] = 4 [3cos2A + 4cos32A – 3cos2A]

= 4cos32A
4
= cos32A =RHS. Proved.

Example 17. Prove that cot(45° + T) + tan(45° - T) = 2cos2T
1 + sin2T

Solution: LHS = cot(45° + T) + tan(45 - T)

= cot45°.cotT - 1 + tan45° - tanT
cot45° + cotT 1 +tan45°.tanT
cotT - 1 1-tanT
= 1 + cotT + 1 + tanT

1 -1 1 - tanT
tanT 1 + tanT
= +
1 + 1
tanT

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= 1-tanT + 1-tanT
1 + tanT 1 + tanT
sinT
2(1-tanT) 1- cosT
1+tanT
= = 2. 1 + sinT
cosT
(cosT - sinT) cosT - sinT cosT + sinT
=2 cosT + sinT = 2. (cosT + sinT) × cosT + sinT

= 2. cos2T = 2cos2T = RHS. Proved.
sin2T + cos2T + 2sinT.cosT 1 + sin2T

Example 18. Prove that 2sinT + 2sin3T + 2sin9T = tan27T - tanT
cos3T cos9T cos27T
2sinT 2sin3T 2sin9T
Solution: LHS = cos3T + cos9T + cos27T

= 2sinT.cosT + 2sin3T.cos3T + 2sin9T.cos9T
cos3T.cosT cos9T.cos3T cos27T.cos9T

= sin2T + sin6T + sin18T
cos3T.cosT cos9T cos3T cos27T.cos9T

= sin(3T - T) + sin(9T - 3T) + sin(27T - 9T)
cos3T.cosT cos9T.cos3T cos27T.cos9T
sin3T.ccoossT3T-.ccoossT3T.sinT+ sin9T.cos3T - cos9T.sin3T
= cos9T.cos3T

+sin27T.ccooss99TT.-cocos2s277TT.sin9T

= tan3T - tanT + tan9T - tan3T + tan27T - tan9T

= tan27T - tanT = RHS. Proved.

Example 19. Without using table or calculator. Show that sin18° = 5-1 .
Solution: Let T = 18° 4

Then 5T = 90°
or, 2T + 3T = 90°
or, 3T = 90° - 2T
Taking cosine on both sides, we get

cos3T = cos(90° - 2T)
or, cos3T = sin2T
or, 4cos3T - 3cosT = 2sinT cosT
since sinT = sin18°≠ 0, dividing both sides by cosT.

4cos2T - 3 = 2sinT
or, 4 - 4sin2T - 2sinT - 3 = 0
or, -4sin2T - 2sinT + 1 = 0
or, 4sin2T + 2sinT - 1 = 0

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This equation is quadratic is sinT in the form of ax2 + bx + c = 0, where
a = 4, b = 2, c = -1

x= b± b2 - 4ac
2a

or, sinT = - 2 ± 22 - 4.4(-1) = 2± 4 + 16
2.4 8

= 2± 20 = 2±2 5 = -1 ± 5
8 8 4

since value of sin18° is positive. We take only positive sign.

Hence, sinT = sin18° = 5 -1 . Proved.
4

( ( (Prove that cosSc 2Sc 3Sc 1
Example 20. 7 (, cos 7 , cos 7 = 8
Solution:
( ( (LHS = cos Sc ( (( 2Sc 3Sc
7 ( (( , cos 7 , cos 7

2sin Sc . cos Sc . cos 2Sc . cos 3Sc
= 7 7 7 7
Sc
2sin 7

2.sin 2Sc . cos 2Sc . cos 2Sc
7 7 7
=
2.2sin Sc
7
4Sc 3Sc
sin 7 . cos 7

= 4sin Sc
7
3Sc 3Sc
2sin 7 . cos 7 [ ( ) ]
sin 4Sc Sc = sin 3Sc
= 4. 2sin Sc 7 = sin S - 7 7
7
(=
sin 6Sc sin Sc - Sc sin Sc 1
7 7 7 8
= = = = RHS. Proved.
8sin Sc 8sin Sc 8sin Sc
7 7 7

Exercise 9.1 211

Very Short Questions
1. (a) Define multiple angles with an example.

(b) Express sin2T in terms of sinT, cosT and tanT.
(c) Express cos2T in terms of sinT, cosT and tanT.
(d) Express cos2T in terms of cotT.

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vedanta Excel In Opt. Mathematics - Book 10

2. (a) Express sin3T in terms of sinT (b) Express cos3T in terms of cosT

(c) Express tan3T in terms of tanT. (d) Express tanT in terms of sin2T and cos2T

Short Questions

3. (a) If sinT = 1 , find the value of cos2T.
2
1
(b) If tanT 3 , find the value of tan3T

(c) If tanT = 1 , find the value of tan3T.
2
4
4. (a) If sinT = 5 , find the value of sin2T, cos2T, tan2T.

(b) If sinT = 1 , find the values of sin2T, cos2T, tan2T.
2
3
(c) If tanT = 4 , find the values of sin2T, cos2T and tan2T

(d) If sinT = 3 , find the value of sin3T and cos3T.
2
1 7
5. (a) If sinT = 4 , show that cos2T = 8

(b) If cosT = 6 , show that cos2T = 11
52 25

(c) If tanT = 5 , show that tan2T = 120
12 119

6. By using the formula of cos2T, establish the following :

(a) sinT = ± 1-cos2T (b) cosT = ± 1+cos2T
2 2

(c) tanT = ± 1-cos2$
1+cos2A
(7. 1 1
(a) If sinT = 2 p + P ((, show that :
(

((
(
( ((i) 1 1 (ii) 1 1
cos2T = – 2 p2 + P2 sin3T = – 2 p3 + P3

((b) 1 1 , show that :
If cosT = 2 p + P

((i) 1 1 ((ii) 1 1
cos2T = 2 p2 + P2 cos3T = 2 p3 + P3

8. Prove that :

(a) 1 - cos2T = tanT (b) sin2T = cotT
sin2T 1-cos2T
sin2T 1 - cos2T
(c) tanT = 1 + cos2T (d) 1 + cos2T = tan2T

(e) sin2T = 2cotT 1 (f) 1 - sin2T = 1 - tanT
cot2T + cos2T 1 +tanT
cos2T 1 - tanT cosT cosT
(g) 1 + sin2T = 1 + tanT (h) cosT - sinT - cosT + sinT = tan2T

(i) 1 + sin2A = sinA + cosA (j) cotT - tanT = cos2T
cos2A cosA - sinA cotT + tanT
4tanT cosT - sinT
(k) tan2T + sin2T = 1 - tan4T (l) cosT + sinT = sec2T - tan2T

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sin5T cos5T sin3T + cos3T 1
sinT cosT cosT + sinT 2
(m) - = 4 cos2T ((n) = 1 - sin2T

1 - sin2T Sc
cos2T 4
(o) cos4T - sin4T = cos2T ( (p) = tan - T
(
((q) Sc
cos2T = tan 4 - T
1 +sin2T
9. Prove that

(a) sin2T - cosT = cotT (b) 1 + sin2T - cos2T = tanT
1 - sinT - cos2T 1 + sin2T + cos2T

(c) sinT + sin2T = tanT (d) 1-cos2T + sin2T = tanT
1 + cosT + cos2T 1 + sin2T + cos2T

(e) (sinT + cosT)2 - (sinT - cosT)2 = 2 sin2T

(f) sinT + cosT + cosT - sinT = 2 sec2T
cosT - sinT cosT + sinT

(g) (1 + sin2T + cos2T)2 = 4cos2T(1 + sin2T)

(h) 1 - 1 cotT = cotT
tan2T - tanT cot2T -

10. Prove that following :

((a) cotT -1 2 (((b) 1 + tan2 Sc - T(
cotT+1 4 (
1-sin2T = ( = cosec2T
1+sin2T ( Sc -
((c) sin2 1 - tan2 4 T
Sc
4 - T = 1 (1 - sin2T)
2
b
11. If tanT = a , prove that a cos2T + b sin2T = a.

Long Questions

12. Prove that

(a) cos4T + sin4T = 1 (3 + cos4T) (b) cos6T - sin6T = cos2T (1 - 1 sin22T)
4 4
1 1
(c) sin4T = 8 (3 - 4 cos2T + cos4T) (d) cos8T + sin8T = 1 - sin22T + 8 sin42T.

(e) cos5T =16cos5T – 20cos3T + 5cosT (f) sin5T = 16sin5T - 20sin3T + 5sinT

13. Prove that

(a) 3 + 1 =4 (b) cosec10° - 3 sec10° = 4
sin40° cos40°
14. Prove that

(a) (2cosT + 1) (2cosT - 1) = 2cos2T + 1)

(b) sec8T - 1 = tan8T
sec4T - 1 tan2T
(c) tanT + 2tan2T + 4tan4T + 8cot8T = cotT

(d) sin2D - cos2D.cos2E = sin2E - cos2E.cos2D

(e) 2+ 2+ 2 + 2cos8T = 2cosT

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(f) sin2D - sin2E = tan (D + E)
sinD.cosD - sinE.cosE
15. Prove that

(a) 4(cos310° + sin320°) = 3(cos10° + sin20°)

(b) sin310° + cos320° = 3 (cos20° + sin10°)
4
(c) 4(cos320° + sin350°) = 3(cos20° + sin50°)

( ((d) Sc Sc
tanA + tan ( ( ((3 + $ – tan 3 - A = 3tan3A
(( ( (
16. Prove that ((( (

(a) cot (A + 45°) - tan (A - 45°) =(( 2cos2A
1 + sin2A
(b) tan(A + 45°) - tan (A - 45°) = 2sec2A.

(c) tan(A + 45°) + tan (A - 45°) = 2tan2A.

17. (a) If 2tanD = 3tanE, then prove that,

(i) tan (D - E) = sin2E (ii) tan(D + E) = 5sin2E 1
5 - cos2E 5cos2E -
1 1
(b) If tanT = 7 and tanE = 3 , prove that : cos2T = sin4E

18. Prove that

(a) cosA - 1 + sin2A = tanA (b) 1 - 1 = cot4T
sinA - 1 + sin2A tan3T + tanT cot3T + cotT

(c) cotA - tanA = 1.
cotA - cot3A tan3A - tanA
19. Prove that
( ( ( ((a)
8 1 + sin Sc 1+sin 3Sc 1- sin 5Sc 1-sin 7Sc =1
8 8 8 8
( ( ( ((b)
1-cos Sc 1- cos 3Sc 1- cos 5Sc 1- cos 7Sc = 1
8 8 8 8 8
( ( ( ((c)
sin4 Sc + sin4 3Sc + sin4 5Sc + sin4 7Sc = 3
8 8 8 8 2

20. Prove that cos233° – cos257° = 2
sin210.5° – sin234.5°
3
21. (a) Prove that cos3A + cos3(120° + A) + cos3(240° + A) = 4 cos3A

(b) Prove that sin3A + sin3(60° + A) + sin3(240° + A) = – 3 sin3A
4
Project Work

22. Prepare a report by calculating the values of sin18°, cos18°, sin36°, cos36°, cos54°,
tan18°, and tan54° by using relations of multiple angle ratios in trigonometry. For this
project work, the students of class can be divided in groups as required.

3. (a) 1 (b) f (c) 11 4.(a) 24 , - 7 , - 24
2 2 25 25 7
3 1 24 7 24
(b) 2 , 2 , 3 (c) 25 , 25 , 7 (d) 0, -1

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vedanta Excel In Opt. Mathematics - Book 10

9.2 Trigonometric Ratios of Sub-multiple Angles

Let A be any angles. Then A , A , A , ....... etc. are called sub-multiple angles of A. In this
sub-unit, we discuss 2 3 4 ratios of sub-multiple angles $2 , $
the trigonometric 3 , etc.

(I) Trigonometric Ratios of Half Angles :

sinA = 2sin A A 2tan A2 = 2cot A
((a) 2 cos 2 = 2
= A A
1 + tan2 2 1 + cot2 2
A
Here, sinA A + 2 (
2 (
A A A A A A
= sin 2 cos 2 + cos 2 . sin 2 = 2sin 2 cos 2 .

⸫ sinA = 2sin A cos A ............ (i)
2 2
A A
sinA = 2sin 2 cos 2

2sin A . cos A
2 2
= A A
sin2 2 + cos2 2

Dividing numerator and denominator by cos2 A
2
A A
2tan 2 2tan 2

= 1sh+owtanth2 aA2t ⸫ sinA = 1+ tan2 A ........... (ii)
can 2
Similarly we

2cot A
2
sinA = A .......... (iii)
1+ cot2 2

Combing (i), (ii) and (iii), we get,

A A 2tan A 2cot A
2 2 2 2
⸫ sinA = 2sin . cos = A = A
1+ tan2 2 1+ cot2 2

A A 1-tan2 A cot2 A - 1
2 2 2 cot2 2
cosA - sin2
((b) = cos2 = 1+ tan2 A = A +
2 21

Here, cosA = cos A + A
2 2

= cos A . cos A - sin A . sin A
2 2 2 2
A A
= cos2 2 - sin2 2

cosA = cos2 A - sin2 A ........... (i)
2 2

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= cos2 A - 1 + cos2 A
2 2
A
= 2cos2 2 - 1

Also, 2cos2 A = 1 + cosA
2
A A
Again, cosA = cos2 2 - sin2 2

= 1 - 2sin2 A
2
A
2.sin2 2 = 1 - cosA

cosA = cos2 A - sin2 A
2 2
A A
cos2 2 - sin2 2

= sin2 A + cos2 A
2 2
Dividing numerator and denominator by cos2 A .
2
A
1-tan2 2

⸫ cosA = 1+ tan2 A ....... (ii)
2 cot2
A - 1
Similarly, we can show cotA = 2 ............ (iii)
A
cot2 2 + 1

Combining (i), (ii) and (iii), we get

A A 1-tan2 A cot2 A -1
2 2 2 2
⸫ cosA = cos2 - sin2 = A = .
1+ tan2 2 A
cot2 2 + 1

2tan2 A
2
tanA =
((c) 1- tan2 A
2 A A
tan 2 + tan 2
A( + A
Here, tanA = tan 2 ( 2 = 1- tan A . tan A
2 2
2tan A
2
⸫ tanA = A
1- tan2 2

((d) A + A
cotA = cot 2 2

cot A . cot A - 1 cot2 A - 1
2 2 2
Here, cotA = A A = A
cot 2 + cot 2 2cot 2

cot2 A - 1
2
⸫ cotA = A
2cot 2

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II. Trigonometric Ratios of A in terms of A .
3
((a) A 2A ( A A
sinA = sin 3 + 3 = 3sin 3 - 4sin3 3
(
(Here, sinA ( 2A A
= sin 3 + 3

= sin 2A . cos A + cos 2A . sin A
3 3 3 3
(= A A A A ( A
2sin 3 . cos 3 cos 3 + 1- 2sin2 3 . sin 3

(= A 1- sin2 A A A
2sin 3 3 + sin 3 - 2sin3 3

= 2sin A - 2sin3 A + sin A - 2sin3 A
3 3 3 3
A A
? sinA = 3sin 3 - 4 sin3 3

= 4cos3 A - 3cos A
((b) cosA 3 3

Here, cosA = cos (2A+ A
( 3 3

= cos 2A (.cos A - sin 2A . sin A
3 ( 3 3 3
(= A A A A A
2cos2 3 -1 . cos 3 - 2sin 3 . cos 3 . sin 3

(= A - cos A - A 1 - cos2 A (
2cos3 3 3 2cos 3 3

= 2cos3 A - cos A - 2cos A + 2cos3 A = 4cos3 A - 3cos A
3 3 3 3 3 3
A A
? cosA = 4cos3 3 - 3cos 3

((c) tanA = tan A 3tan A - tan2 A
3 3 3
3 . = A
1 - 3tan2 3

(Here, tanA 2A A
= tan 3 + 3

2tan A A
3 3
+ 2tan
tan 2A + tan A 1 - tan2 A
3 3 3
= =
1 - tan 2A . 3tan A A
3 3 1 - 2tan 3 . A
3
1 - tan2 A tan
3
A A A A A
2tan 3 + tan 3 - tan3 3 3tan 3 - tan3 3

= 1 - tan2 A - 2tan2 A = 1 - 3. tan2 A
3 3 3
A A
3tan 3 - tan3 3

? tanA = 1 - 3. tan2 A
3

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((d) cotA = cot A ( cot3 A - 3cot A
3 3 3
3 . = A (For practice do yourself.)
3cot2 3 - 1

Worked out Examples

Example 1. If sin T = 3 find the values of.
Solution: 2 5
(a) sinT (b) cosT (c) tanT
Example 2.
Solution: Here, sin T = 3
2 5

cos T = 1 - sin2 T = 1 - 9
2 2 25

= 25-9 = 16 = 4
25 25 5
T 3
T sin 2 5 3
2 4 4
tan = cos T = 5 =
2
3 4 24
(a) sinT = 2 sin T . cos T = 2. 5 . 5 = 25
2 2
T T 16 9 16-9 7
(b) cosT = cos2 2 - sin2 2 = 25 - 25 = 25 = 25

2tan T 2. 3 3 16 3 16 24
2 4 2 16-9 2 7 7
(c) tanT = = 9 = × = × =
1 - tan2 T 1- 16
2

Alternative 24

tanT sinT = 25 = 24
cosT 7 7
25

If cos45° = 1 , show that cos2212 °= 1 . 2+ 2
2 2

Here, cos45° = 1 2

(or, 45° =1
2 2
cos2 (

or, 224cc25oo°ss22is442255p°°osi-=t1iv=e12[( 1 ( cosT = 2cos2 T - 1)
or, 2 quadrant] 2
of cos +
45° 1
2 lies
( (Value in 1st

? cos 45° = 1 + 1 = 2+1 × 2 (
2 22 2 22 2

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vedanta Excel In Opt. Mathematics - Book 10

= 2+ 2 = 1 . 2+ 2
2 2

? cos 45° = 1 . 2 + 2 Proved.
2 2

Example 3. Prove that
Solution:
(a) tan T = sinT (b) 1 + sinT - cosT = tan T
Example 4. 2 1 + cosT 1 + sinT + cosT 2
Solution: sinT
(a) RHS = 1 + cosT

2sin T . cos T sin T T
2 2 2 2
= = = tan = LHS. Proved.
2cos2 T cos T
2 2
1 + sinT - cosT
1 + sinT + cosT
((b) LHS =

1 + (2sin T . cos T - 1- 2sin2 T
(( 2 2 2
=
( T T T
1 + 2sin 2 . cos 2 + 2cos2 2 -1
(
1 + (2sinT . cos T - 1+ 2sin2 T
(= ( 2 2 2

((2cosT sin T + cos T
2 2 2

2sin T cos T + sin T
(= 2 2 2
= RHS. Proved.
T T T
2cos 2 cos 2 + sin 2

Prove that

((a) tan Sc + A = secA + tanA
4 2
( ((b) cos2 Sc D Sc D D
4 - 4 - sin2 4 - 4 = sin 2

((a) Sc + A
LHS = 4 2

tan Sc + tan A 1+ sin A
4 2 2
= =
Sc $ $
1 - tan 4 . tan 2 1 - tan 2

1 + sin A
2

cos A cos A + sin A cos $
2 2 2 2
= = ×
$ $ A A
1 - sin 2 cos 2 cos 2 + sin 2

cos A
2

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cos $ + sin $ cos $ + sin $
2 2 2 2
= ×
A A A A
cos 2 - sin 2 cos 2 + sin 2

cos2 $ + sin2 $ + 2sin $ .cos $ 1 + sinA
2 2 2 2 cosA
= =
A A
cos2 2 - sin2 2

= 1 + sinA = secA + tanA = RHS. Proved.
cosA cosA
( ((b)
LHS = cos2 Sc - D (- sin2Sc-D
( (= cos2 4 4 (( 4 4
((
Sc - D = cos Sc - D = sin D = RHS Proved.
(Prove that (cosD + cosE)2 + (sinD + sinE)2 = 4cos244 2 2 2

Example 5: D-E
Solution: 2

LHS = (cosD + cosE)2 + (sinD + sinE)2

= cos2D + 2cosD cosE + cos2E + sin2D + 2sinD.cosD + sin2E

= (cos2D + sin2D) + (sin2E + cos2E) + 2(cosD.cosE + sinD.sin2E)

(= 2[1 + cos(D - E)] = 2.2cos2
= 1 + 1 + 2cos (D - E)

(= 4cos2 D-E ((
(( D-E 2 ((
( 2
= RHS. Proved. (
((
Example 6: (Proved that (( ( ( (1+ cos3Sc
1 + cos Sc ((
1+ cos5Sc 1+ cos7Sc = 1
(( 8
Solution: Here,
( ( { ( )}{ ( )}LHS. = (
( ( ( (= 1+ cos Sc 1+ cos3Sc 1 + cos S - 3Sc 1 + cos Sc - Sc
( (= 8 8
( { ( )}=
( (= 1+ cos Sc 1+ cos3Sc 1- cos3Sc 1- cosSc


1- cos2 Sc 1- cos23Sc


1- cos2 Sc 1 - cos2 Sc - Sc
2 8

1- cos2Sc 1- sin2Sc


= sin2 Sc. cos2 Sc

( (= 1
2
2sinSc. cos Sc

( (=1
2
sin2Sc


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( ( (= 1 sin2Sc = 1 1 = 1 = RHS Proved.
(

( (Example 7.
( ( ( (Solution: D 1 1 1 a2 +a12
If cos 2 = 2 a + a , prove that cosD = 2

Here, cos D = 1 a + 1 ,
a

cosD = 2cos2 D - 1
[ ( (]= 2

1 a + 1 2 -1
( (= a

2. 1 a2 + 2.a 1 +a12 -1
( (= 1 a
( (
a2 + 1 +1-1= 1 a2 + 1 = RHS. Proved.
a2 a2

Exercise 9.2

Very Short Questions

1. (a) Define sub-multiple angle with an example.

(b) Express sinT in terms of sin T and cos T .

T T T
(c) Express cosT in terms of cos , sin and tan .

(d) Express tanT in terms of tan T .

T
2. (a) Express sinT in terms of sin .

(b) Express cosT in terms of cos T .

T
(c) Express tanT in terms of tan .

3. (a) If sin T = , find the value of sinT.

T 3
(b) If cos = 2 , find the value of cosT.

(c) If tan T = 3, find the value of tanT.
(d) If = find the value of cosT.
T 4
cos 5 ,

Short Questions

4. (a) If cos30° = 23, find the values sin15°, cos15° and tan15°.
1
(b) If cos45° = 2 , then prove that :

(i) sin2212° = 1 2- 2 (ii) cos2212° = 1 2+ 2 (iii) tan2212° = 3-2 2
2 2

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(c) If cos330° = 3 prove the following :
2
3-1 3-1
(i) cos165° = - 22 (ii) sin165° = 22 (iii) tan165° = - (2 - 3)

5. Prove that:
( ( ( ((a)
1 - tan2 x (b) sinx = 3 sin x - 4sin3 x
2
cosx = x
1 + tan2 2
( ( ( ( ( ( ( (3tan x - tan3 x
cot3 x - 3cot x

1 - 3tan2 x 3cot2 x - 1
( ( ( ((c) tanx =
(d) cotx =

( ( ( (6.

(a) If cos T = 1 p + 1 , prove that cosT = 1 p2 + 1 .
( ( ( ((b) 2 p 2 p2

If cos T = 1 p + 1 , prove that cosT = 1 p3 + 1 .
( ( ( ((c) 2 p 2 p3

If sin T = 1 p+ 1 , prove that cosT = - 1 p2 + 1 .
2 p 2 p2
(7. Prove that:
((b) 1 + sinT =
(a) 1 - sinT = sin T - cos T ((2 sin T +cos T 2
2 2 2 2

(c) 1 + cosT = cot T (d) 1 + secT = cot T
sinT 2 tanT 2
1 + cosT
(e) cosecT - cotT =tan T (f) 1 - cosT = cot2 (2T)
2
A
1 - sinA 1 - tan 2 sin3 T - cos3 T 1
cosA 2 2 2
(g) = A (h) = 1 + sinT
1 + tan 2 sin T - cos T
2 2

1+ secT T 1+ sinT cos T + sin T
secT 2 cosT 2 2
(i) = 2cos2 (j) =
cos T - sin T
(((l) 2 2

((k) (( Sc T 1 - tan2 Sc - T ( T
(4 2 4 4 2
1 - 2sin2 - = sinT ( = sin
Sc T (
1 +tan2 4 – 4

Long Questions

( (8. Prove that:T T 2sinT - sin2T = tan2 T
(a) cos4 2 - sin4 2 = cosT 2sinT + sin2T 2
((b)
(c) 1 sin2T . cosT = tan T (d) 1 +sinT = tan2 Sc + T
+ cos2T 1 + cosT 2 1 - sinT 4 2

cos T 1+sinT T sin T 1+sinT T
2 2 2 2
(e) = tan (f) = cot
T T
sin 2 1+sinT cos 2 1+sinT

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