5.2 Trigonometry, small angle approximations.
You need to be familiar with the following approximations for small angles θ, which are given in
the formulae booklet. (The angle is measured in radians.) These approximations are very good for
angles smaller than 0.2 radians.
o
sin   
o
cos   1 
o
tan   
2
Note: ≈ means is approximately equal to.
2
Remember that:
180° = π radians
θ° =
q
π radians
180
Examples
Example 1
Use the small angle approximations to find
(a) sin 3°
(b) tan 7°
(a) 3 
3
 p radians = 0.0524
180
sin 3° ≈ 0.0524
(b) 7 
Show that
Use sin   
7
 p radians = 0.122
180
tan 7° ≈ 0.122
Example 2
First change to radians.
(The small angle approximations only
work for angles in radians.)
Use tan   
cos 2 x  1
 2 for small angles x.
x tan x
cos 2 x  1

x tan x

1

(2 x)2
1
2
x x
4 x2
2
2
x
2 x 2
 2  2
x
cos 2 x  1 
(2 x) 2
and tan x  x
2
1  1  0 and (2 x) 2  4 x 2

4 x2
 2 x 2 and cancel the x2
2
Example 3
Show that 1  2cos 2   3  2 2 for small angles θ.
1  2cos2   1  2  cos 
2
 2 
 1  2 1  
2 

2
Use cos   1 

4 
 1  2 1   2  
4 

2
2
.
Square the bracket.
= 3  2 2  12  4
Because θ is small, the θ4 term is very
close to 0 and can be ignored.
 3  2 2
Exercise

1.
Angle θ is in radians and tan  
2.
Show that for a small angle θ, given in radians,
.
45
Use a small angle approximation to find angle θ in degrees.
3  6cos   4cos 2   1   2
cos 4 x  1
when x is both small and in radians.
x sin 3x
3.
Find the value of
4.
When  is small and in radians, find the values of a and b in the following approximation.
cos 6  3 sin 2  a 2  b
5.
For a small angle x which is in radians show that
cos x  1
  14
2 x tan x
6.
(a) Use small angle approximations to show that
3
2
(b) Solve
1
2
 5sin x  cos x  12 x2  5x  12
x2  5x  12  0 . Give your answer to 2 decimal places.
(c) Explain why x = 0.10 is root of
3
2
 5sin x  cos x  0 and why x = 9.9 is not a root.
Answers
1.
2.

45
is a small angle, so tan

45


45
so  

45
radians. Therefore  

45

180

 4
3  6cos  4cos 2   3  6cos  4(cos )2
 2 
 2 
 3  6 1    4 1  
2 
2 


2
 2 

4 
 3  6 1    4 1   2 

2 
4 


 3  6  3 2  4  4 2   4
1 2
3.
4.
cos 4 x  1

x sin 3x
1
(4 x) 2
16 x 2
1 1 
1
8 x 2
8
2
2



2
2
x  3x
3
3x
3x
cos 4 x  1 
(4 x) 2
and
2
sin 3 x  3 x
2
6 

cos 6  3 sin 2  1 
 3  2
2
 1  18 2  6 2  12 2  1
So a = –12 and b =1
x2
 1  1 x2
cos x  1
2

 2 2   14
2 x tan x
2x  x
2x
1
5.
6.
(a)
3
2
x2
)
2
x2
 32  5x  1 
2
1 2
1
 2 x  5x  2
 5sin x  cos x  32  5 x  (1 
(b) x 
(5)  (5) 2  4  12  12
2  12
 5  24  x  0.10 or x  9.90
Use the formula for
quadratic equations.
(c) 0.10 is small, therefore x = 0.10 is a root, but 9.9 is not small and therefore the small angle
approximation is not valid.