Entry Test Series Mathematics (Book-I)

Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

UNIT-10
Distance Formula
Let P(x1, y1) and Q (x2, y2) be two points. If “d” denotes the distance between them,
then = d = |PQ| = (x1 + x2)2 + (y1  y2)2
= (x2  x1)2 + (y2  y2)2
i.e. square root of the sum of square the difference of x-coordinates and square the
difference of y-coordinates.
Example:
Find the distance between the following points:
i. A (3, 8) , B(5, 6)
ii. P(cos x, cos y), Q(sin x, sin y)
Solution:
i. Distance = |AB| = (3  5)2 + (8  6)2 = 4 + 4 = 8 = 2 2
= (5  3)2 + (6  8)2 = 4 + 4 = 8 = 2 2
ii. Distance = (cos x  sin x)2 + (cos y  sin y)2
= cos2 x + sin2 x  2cos x sin x + cos2 y + cos2 y + sin2 y  2cos y sin y
= 2  2cos x sin x  2cos y sin y
= 2  (cos x sin x + cos y sin y)
The Fundamental Law of Trigonometry
 Forany two angles  and  (real numbers)
cos(   )  cos  cos   sin  sin  is called fundamental law of trigonometry.

Addition Formulas:

sin(   )  sin  cos   cos  sin  sin(   )  sin  cos   cos  sin 
cos(   )  cos  cos   sin  sin  cos(   )  cos  cos   sin  sin 
tan   tan  cot  cot   1
tan(   )  cot(   ) 
1  tan  tan  cot   cot 
tan   tan  cot  cot   1
tan(   )  cot(   ) 
1  tan  tan  cot   cot 

Double Angle Formulas

sin 2  2sin  cos 

cos 2  cos2   sin 2   1  2sin 2   2cos 2  1
2 tan 
tan 2 
1  tan 2 
Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

Half Angle Formulas

 1  cos 2  1  cos 2  1  cos 2

sin  tan  cos 
2 2 2 1  cos 2 2 2

Multiple Angle Formulas

sin 3  3sin   4sin 3  cos3  4cos3   3cos 

sin 4  4sin  cos   8sin 3  cos  cos 4  8cos 4   8cos 2   1
3 tan   tan 3  4 tan   4 tan 3 
tan 3  tan 4 
1  3 tan 2  1  6 tan 2   tan 4 

Powers of Trigonometric Functions

1  cos 2 1  cos 2
sin 2   cos 2  
2 2
3 1 3 1
sin 3   sin   sin 3 cos3   cos   cos 3
4 4 4 4

3 1 1 3 1 1
sin 4    cos 2  cos 4 cos 4    cos 2  cos 4
8 2 8 8 2 8
(i).
Sum/Differences to Products Formulas
               
sin   sin   2sin   cos   sin   sin   2 cos   sin  
 2   2   2   2 
               
cos   cos   2 cos   cos   cos   cos   2sin   sin  
 2   2   2   2 
Products to Sum/Differences Formulas

1
sin  sin   cos(   )  cos(   )
2
1
cos  cos   cos(   )  cos(   )
2
1
sin  cos   sin(   )  sin(   )
2
cos A sin B = (1/2) [ sin (A + B) - sin (A - B) ]
Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

Trigonometric Ratios of Allied Angles

The angles associated with basic angles of measure  to a right angle or its multiple are
called allied angles. So, the angles of measure 90o  , 180o  , 270o  , 360o  , are known as
allied angles.

Sine Cosine Tangent

     
sin      cos  cos      sin  tan      cot 
2  2  2 
     
sin      cos  cos       sin  tan       cot 
2  2  2 
sin      sin  cos       cos  tan       tan 
sin       sin  cos       cos  tan      tan 
 3   3   3 
sin       cos  cos       sin  tan      cot 
 2   2   2 
 3   3   3 
sin       cos  cos      sin  tan       cot 
 2   2   2 
sin  2      sin  cos  2     cos  tan  2      tan 
sin  2     sin  cos  2     cos  tan  2     tan 
Note:
The above results also apply to the reciprocals of sine, cosine and tangent. These result are
to be applied frequently in the study of trigonometry, and they can be remembered by using
the following device.
Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

1. If  added to or subtracted from odd multiple of right angle, the trigonometric ratios
change into co-ratios and vice versa.
i. e., sin 
 cos tan 
, sec 
 cosec
  3 
e. g. sin 2   = cos  and cos  2   = sin 
   

2. If  is added to or subtracted from an even multiple of 2 , the trigonometric ratios shall
remain the same.
3. So far as the sign of the result is concerned, it is determined by the quadrant in which the
terminal arm of the angle lies.
e. g. sin (  ) = sin , tan ( + ) = tan  cos (2  ) = cos 

Note:
sin( n   )  (1) n sin  cos(n   )  (1) n cos
 n 1
 n 1
n  (1) 2 cos , n is odd n  (1) 2 sin  , n is odd
sin( )   n cos(   )   n
2 (1) 2 sin  , n is even 2 (1) 2 cos , n is even
 

Note:
If  ,  and  are the angle of triangle ABC then
sin(   )  sin  cos(   )   cos
   tan   tan   tan   tan  tan  tan 
cos( )  sin
2 2
tan(   )  tan   0 1
cos 36   cos 72  
2
      1
tan tan  tan tan  tan tan 1 cos 36 . cos 72  
2 2 2 2 2 2 4

      tan  . tan( 60    ). tan( 60    )  tan 3

cot  cot  cot  cot cot cot
2 2 2 2 2 2
cot cot   cot  cot   cot  cot  1 1
cos . cos(60    ). cos(60    )  cos3
4

1 1
sin . sin( 60    ). sin( 60    )  sin 3 sin  .sin 2 .sin 4  sin 3
4 4
1 3
1 sin 20.sin 40.sin 80  sin 60 
cos  .cos 2 .cos 4  cos 3 4 8
4 Example
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

sin 20.sin 40.sin 60.sin 80

 sin 60(sin 20.sin 40.sin 80)
3 1 3 1 3 3
 . sin 60  . . 
2 4 2 4 2 16
Greatest and least value of a sin   b cos or a cos  b sin  is  a 2  b 2

1 cos 12 = . Solution :

(a) 3 cos3 4  4 cos 4 (b) 4 cos3 4  3 cos 4 Cos12  Cos 4  3 
(c) 3 sin 4  4 sin 4
3
(d) 4 sin 4  3 sin 43

Use cos3  4cos3   3cos 

2 cos 12 + sin 12 Solution:

= .
cos 12  sin 12
1  cot 45cot12
(a) tan 33 (b) tan 33 Cot  45  12  
cot 45  cot12
(c) cot 33 (d) cot 33 cos12
cot 33= 1
1  cot12 sin12 
 
1  cot 45 1  cos12
sin12
Solution:
3 1 + sin 
= .
1 – sin  1 + sin 
L.H.S. =
    1 – sin 
tan 2 + cos 2 sin 2  cos 2
(a) (b)
     sin2 = 2sin cos
tan 2  cos 2 sin 2 + cos 2
sin = 2sin /2 cos /2
   
sin 2 + cos 2 tan 2  cos 2 =
(c) (d)
       
sin 2  cos 2 tan 2 + cos 2 sin2 2 + cos2 2 + 2 sin 2 cos 2
   
sin2 2 + cos2 2 – 2 sin 2 cos 2
2
 sin  + cos  
 2 2
= 2
 sin  – cos  
 2 2
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

 
sin 2 + cos 2
=
 
sin 2 – cos 2

Use allied angle

4   3 
cos 2 +  . cos  2   cosec2  = .
     
1 cos 2 +  =- sin 
(a) 1 (b)  
(c) 0 (d) 2
3 
cos  2   = - sin 
 

1
cosec2 
sin 2 

5 sin ( + ) sin (  ) = . Use

(a) cos2   cos2  (b) cos2  + cos2  sin(   )  sin  cos   cos  sin 
(c) cos   cos 
2 2
(d) sin   sin 
2 2

sin(   )  sin  cos   cos  sin 

6 4 5 
If cos ( + ) = 5 , sin (  ) = 13 , 0 < ,  < 4

then tan 2 =
33 33
(a) 56 (b) 48
56
(c) 33 (d) None of these

7 If cos 34 = x then sin 17 = . Hint : cos 34 = x then sin 17
x
(a) 2x 1  x2 (b)  1  cos 2
1  x2 sin 
2 2
1x 1x
(c)  (d)
2 2

8 cos 15  sin 15 = . Hint : cos 15  sin 15 =cos(90-75) – sin15=
1 1
(a)  (b) Sin75-sin15=
2 2
(c)  2 (d) 2
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

USE

       
sin   sin   2 cos   sin  
 2   2 

9 If  is an acute angle then 90  ; 180  , 270   and Definition

360   are called:
(a) Quadrantal angles (b) Half angles
(c) Allied angles (d) Double angles

10 If  lies in quad II then terminal arm of angles 2   lies Suppose angle is 170
in  So 360-170 =190 lies in III
(a) Quad III (b) Quad IV
(c) Quad I (d) Quad II
Formula
11 tan 3 = .
3 tan  + tan3  3 tan   3 tan2 
(a) (b)
1  3 tan2  1 + 3 tan2 
3 tan   tan3  tan  + 3 tan2 
(c) (d)
1  3 tan2  1 + 3 tan2 

12 Hint :
cot 54 tan 20
The value of + = .
tan 36 cot 70 If 1   2   90 then
(a) 0 (b) 2
tan1 tan  2  1
(c) 3 (d) 1
cot 1 cot  2  1

13 If cot ( + ) = 0 then sin ( + 2) = .

(a) sin  (b) cos 
(c) sin  (d) cos 2

14 1  cos 2A 1  cos 2A 2sin 2 A

1 + cos 2A = . 1 + cos 2A = = tan A
2 cos 2 A
(a) cos A (b) tan A
(c) sin A (d) cot A
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

15 If , ,  are angles of a ABC then sin ( + ) = For a triangle ABC we know that
(a) sin  (b) sin   +  +  = 180o
(c) cos  (d) cos   +  = 180o – 
sin ( + ) = sin (180o – )
= sin  ( sin ( – ) = sin )
Hint : ALLIED ANGLE
16 3 
cot  2   = .
 
(a) tan  (b) cot 
(c) tan  (d) cot 

Hint:
17 cos ( + )  cos (  ) = .
(a) 2 cos  cos  (b) 2 sin  sin  cos(   )  cos  cos   sin  sin 
(c) 2 sin  sin  (d) 2 sin  cos  cos(   )  cos  cos   sin  sin 

Hint :
18 sin 5 cos 2 = .
sin 7 + sin 3 2 sin  cos  = sin ( + ) + sin ( – )
(a) sin 7 + sin 3 (b)
2
1
2 (cos   sin )
(c) (d) None of these

Hint :
19 cos (2x + 30) cos (2x  30) = .

(a)
1
(cos 4x + cos 60) (b)
1 2 cos  cos  = cos ( + ) + cos ( – )
2 2 (sin 4x + sin
60)
(c) cos 4x + sin 60 (d) sin 4x + sin 60

20 12 5 12
If sin  = 13 and cos  =  13 then  lies in: Hint : sin  = 13 +ve in I and II and

(a) I quadrant (b) III quadrant 5

(c) II quadrant (d) IV quadrant cos  =  13 -ve in II and III

21 Hint: Value of
The coordinates of a point whose terminal ray makes
2 1 2 3 3
an angle of 60 and 4 units from origin is: cos 60   and sin60= 
4 2 4 2
(a) ( 3 3) (b) (1, 1)
4
(c) (0, 0) (d) (2 2 3)
2 3
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

2
Formula
22 1  cos 
 is equal to:
2
 
(a) cos 2 (b) sin 2


(c) tan 2 (d) sin 

Hint: Multiple of 360 is odd so convert it into cot

23 If n is an odd integer, then cot (n . 360 + ) =
(a) tan  (b) tan  And lies in First Quadrant.
(c) cot  (d) cot 

24 sin 2 = . 2 tan 

Solution : =
1 + tan2 
2 tan  1  tan2 
(a)
1 + tan2 
(b)
1 + tan2  2sin 
cos   2sin  cos 
1 + tan2 
(c) (d) None of these sin 2 
2 tan  1
cos 2 
 sin 2

25 sec (300) = . Solution : sec (300)= sec (300) and

(a) 1 (b) 1 sec (360- 60)=sec60=2
(c) 2 (d) 2

26 sin 20 sin 40 sin 80 = .

3 1 1
(a) (b) Hint : sin  .sin 2 .sin 4  sin 3
8 8 4
1 3
(c) (d)
16 16

27 Solution :
Which of the following is rational?
(a) sin 15 (b) cos 15
(c) sin 15 cos 15 (d) sin 15 cos 75
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

28 cos 20 + cos 100 + cos 140 = .

(a) 0 (b) 1
(c) 1 (d) None of these

Hint :
29 
cos 12 = .

 15 and
31 3+1 12
(a) (b)
2 2 2 2 cos(60  45)  cos 60cos 45  sin 60sin 45
1 3 31
(c) (d)
2 2 2 2

30 sin P  Cos Q = 0 if: Solution: sin P  Cos Q=0

(a) P + Q = 90 (b) P + Q = 180 =SinP=CosQ
(c) P + Q = 45 (d) P + Q = 270
=SinP=Sin(90-Q)

Implies that P=90-Q

31 Definition
Co-ratio of cosec is:
(a) sin (b) sec
(c) cos (d) cosec

32 The sign of tan 2140 is: Tan 2140 =Tan (6.360 - 20)
(a) +ve (b) ve Fourth Quadrant
(c) may be any (d) None of these

33 19 19
cot( )? cot( )?
3 3
1 19  19 18  
A) B) 3 tan  tan     tan  3
3 3  3 3  3
3
C)  3 D)
2

34 cot 59  If 1   2   90 then


tan 31 tan1 tan  2  1
A)0 B)1 cot 1 cot  2  1
 Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com

1
C) D) 2
2

35 sin 19 cos11  sin 71 sin 11  Hint:

sin(   )  sin  cos   cos  sin 
3  3
A) B)
2 2
sin 71  sin  90  11  cos11
1 1
C) D)  1
2 3 sin(30) 
2
Entry Test Series Mathematics (Book-I)
Muhammad Asim Ali -Lecturer Punjab Group of Colleges- Lahore, masimali99@gmail.com