# Trigonometry A level ```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
o
sin   
o
cos   1 
o
tan   
2
Note: ≈ means is approximately equal to.
2
Remember that:
θ&deg; =
q
180
Examples
Example 1
Use the small angle approximations to find
(a) sin 3&deg;
(b) tan 7&deg;
(a) 3 
3
180
sin 3&deg; ≈ 0.0524
(b) 7 
Show that
Use sin   
7
180
tan 7&deg; ≈ 0.122
Example 2
(The small angle approximations only
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.
1.
2.

45
is a small angle, so tan

45


45
so  

45

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