Download Trigonometry

CHAPTER-23
Trigonometry
IMPORTANT FORMULAE
Trigonometric Ratios :
C
Hypotenous
Perpendicular

A
Base
B
2
2
(Hypotenous) = (Base) + (Perpendicular)
(i) sin =
Perpendicular
Hypotenuse
(ii) cos =
Base
OM

Hypotenuse OP
(iii) tan =
Perpendicular PM

,
Base
OM
Hypotenuse
 Perpendicular
(iv) cosec 
(v) sec 

2
Hypotenuse

Base
vicot 
Base
Perpendicular
Trigonometric Identies : Following results hold for all values of These results are
called Trigonometric identities.
(i) sin2   cos2   1
(ii) 1 tan2   sec2 
2
2
(iii) 1 + cot = cosec 
sin 
Remark : We have (i) tan   cos  

(ii) cot  
cos 
sin  
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


iiitan × cot = 1

Complementary angles : The angles  and (90-) are called complementary
angles.
Results on complementary angles:
0
(i) sin (90 - ) = cos 
0
(ii) cos (90 - ) = sin 
0
(iii) tan (90 - ) = cot 
0
(iv) cosec (90 - ) = sec 
0
(v) sec (90 - ) = cosec 
0
(vi) cot (90 - ) = tan 
Value of trigonometric ratios -
Some fomulas:
1. Sin(A + B) = sin A cos B + cos A sin B
2. sin (A-B) = sin A cos B - cos A sin B
3. cos (A + B) = cos A cos B - sin A sin B
4. cos (A-B) = cos A cos B + sin A sin B
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5. tan (A-B)
6. tan (A+B)
tan A  tanB
1  tan A tanB
tan A  tanB

1  tan A tanB

7. sin2 A = 2sin A cos A
2
2
8. cos2 A = cos A -sin A
2
9. cos2 A = 1-2sin A
2
10. cos2 A = 2cos A-1
Circular Measure:- The unit of measurement of angles in this system is a
c
radian (or 1 ).
A radian is defined as the angle subtended at the centre of a circle by an
c
arc whose length is equal to the radius of the circle and it is denoted by 1 .
In the figure PO = OQ = r, (radius of the circle) and PQ
P
O

r
Q
r 
Length of arc (L) = 180o
Maxima and Minima Minima
–1
Maxima
+1
sin , sin 2,sin3,.......
sin n
 cos , cos 2,cos 3,.......
–1
+1
cos n
2
2
2
 sin , sin 2,sin 3,.......
0
+1
2
sin n
2
2
2
 cos , cos 2,cos 3,......
0
+1
2
cos n
 Important formula for Maxima and Minima (a sin  + b cos ) maxima value = a2  b2
(a sin  + b cos ) minima value =  a2  b2
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(a tan  + b cot ) minima value = ab
0
0
Example: Sin30 cos 45
tan 600
?
1 1

sin30 cos 45
1
2
2 

tan600
3
2 2
0
Answer:
0

 3 

1
2 6

6
6
6
12

Example:  sin2 300 cos2 450  4 tan2 300  2 sin2 900 – 2cos2 900   ?
1


2
0
2
0
2
0
2
0
2
0
Answer: sin 30 cos 45 + 4 tan 30 + sin 90 –2 cos 90 .
2
2
2
=
 1  1
2
2
 1  1 
   1 – 2   0 
  4
 2 
   2
 3 2

1 1
1 1
1 4 1
  4  – 0   
4 2
3 2
8 3 2
=
3  32  12 47

24
24
0
0
0
0
0
Example:.tan 35 tan 40 tan 45 tan 50 tan 55 = ?
0
0
0
0
0
0
0
0
Answer: tan 35 tan 40 tan 45 tan 50 tan 55
0
0
0
= tan 35 tan 40 ×1 × tan (90–40 )× tan (90–35 )
0
0
= tan 35 tan 40 × cot 40 ×cot 35
0
0
0
0
= (tan 35 cot 35 )×(tan 40 × cot 40 )
= 1×1 { tan  × cot  = 1} = 1
Example: If tan  = 34 and is acute angle, then cosec 
Answer:
A
5
C
3

B
4
tan =
3
4
In, ABC
AC = AB2  BC2
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 32  42
=5
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AC
5
 cosec  = AB  3

Example:The least value of  4 sec 2   9 cos ec 2   is4sec  + 9 cosec 
2
Answer:
2
= 4(1+tan ) + 9(1+cot )
2
2
= 13+(4tan  +9cot )
Now,
AM > GM
2

2
4 tan2   9cot 2 
 4 tan2 .9cot 2 
2
 4tan  + 9cot  > 2×
2
2
36
> 12
 Minimum value of 4sec  + 9 cosec 
= 13 + 12 = 25
Example:The value of sin 39  2tan 11 tan 31 tan 45 tan 59
2
2
0
0
0
0
0
cos 510
isAnswer:
=
sin 39o
cos 51o
o
o
o
o
2
o
2
o
+ 2 tan 11 tan 31 tan 45 . tan 59 tan 79 - 3 (sin 21 + sin 69 )
sin (90o  51o )
cos 51o
sin 51o
cos 51o
o
tan 790 -3(sin2 210  sin2 690 )
o
0
o
o
2
o
2
0
0
o
2
o
2
o
+ 2 tan 11 tan (90 - 31 ) - 3(sin 21 + cos 21 )
o
o
o
o
= 1 + 2 (tan 11 cot 11 ) tan 31 cot 31 ) - 3
=1+2××1-3
=1+2-3
=3-3
=0
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o
+ 2 tan 11 . tan 79 . tan 31 tan 59 - [sin 21 + sin (90 - 21 )]
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