AS Level Trigonometry Revision
Trig ratios for 30, 60, 45

30
2
1
2
2
3
45
60
1
1
1
sin 30  cos 60 
sin 60  cos 30 
tan 60  3
sin 45  cos 45 
1
2
tan 45  1
3
2
tan 30 
1
3
Trig ratios for all angles

1
2
NB the CAST DIAGRAM
For the sign of a trig ratio
S
A
T
C
All positive in first quadrant
Sine (only) in second quadrant
Etc…
Example
Without using a calculator find
(i) cos 150 
(i)
(ii) tan 210

(iii) sin  240 
(ii)
S
A

(iii)
S
A
S
210
150
A
60
30
30
-240
C
T
cos 150    cos 30

3
2
C
T
tan 210   tan 30

1
3
C
T
sin  240    sin 60

3
2

Trig of Scalene triangles
Sine rule
B
a
b

sin A sin B
A
 c 


 sin C 
a
b
C
Given AAS use it to find a second side
Given SSA use it to find a second angle (but take care to choose the angle size appropriately
–it could be acute or obtuse).
Cosine rule
a 2  b 2  c 2  2bc cos A
B
c
A
cos A 
a
b2  c2  a2
2bc
Both formulae with two more sets.
b
C
Given SAS use it to find the third side
Given SSS use it to find an angle (no possible ambiguity here).
Example
Triangle PQR has PR = 3cm, QR = 7cm and QPˆ R  36 
Find (i) QR using the cosine rule and then (ii) PQˆ R using the sine rule.
Q
(i) QR 2  9  49  42 cos 36  24.021...
3
QR  4.901..  4.90
R
(ii)
36
P
7
7
4.901..

sin PQR sin 36
sin PQR 
7 sin 36
 0.8394...
4.901..
 PQR  57.086.. or PQR  122.914..
It can’t be 57.08.. since R would be 86.92.. and would be the largest angle in the triangle,
but R faces the smallest side so is the smallest angle. Hence PQR  122.91

Area “ 12 absin C ” rule given SAS
A
Area of triangle = 12 absin C
b
C
a

B
Circular measures
Remember 2 c  360 o ,
 c  180 o
1.
r
r
 1 
c
1c
r
c
2
c
3
180 

 90 o ,
,1 

c
4
c
180
 45 o ,
c
6
 30 o ,
 60 o , etc…
2. Arc length & Area of a sector

r
r
A

A

r
r
  r [ in radians]
A = area sector – area 
 12 r 2  12 r 2 sin 
A  12 r 2
Essential to learn formulae for arc length and sector area, and that  is in RADIANS! The
formula for segment might be learnt!
In triangles, where angles are given in or are required in radians set your calculator into RAD
mode
Example
5.8
9
0.57c
4.3
x
x 2  5.8 2  4.32  2  5.8  4.3  cos 0.57 c
x  2.54
Graphs of trig functions (all periodic)

1. Graph of y  sin x
y
Period 2
1
sin(2  x)  sin x
sin x  1
0
 2


2
2
3
2
x
-1
2. Graph of y  cos x
y
Period 2
1
cos(2  x)  cos x
cos x  1
0
 2


2
3
2
2
x
-1
3. Graph of y  tan x
Period 
y
tan(  x)  tan x
Vertical asymptotes at
x   2 , x   32 , etc
 2
0

2

3
2
2
Vertical asymptotes
x

Boundary values of trig ratios
Verify these from graphs
S=1
T- C=0
T
S=T=0
C= -1
S=T=0
C=1
T S= -1 T-
C=0

Two important trig identities
sin 
 tan
cos 
Example
sin 2   cos 2   1
Given  is obtuse and sin   178 find the values of cos  and tan .
sin 2   cos 2   1
 cos 2   1  sin 2 
S
64
 1  289

A

225
289
15
 cos    17
sin 
tan 
cos 
 tan 
8
17
15
17

T
C
  158
NB Learn how to rearrange the identities

cos  
cos 2   1  sin 2 
sin 2   1  cos 2 
Complementary angles are those which add up to 90
sin(90   )  cos 

sin 
tan
sin   cos tan
cos(90   )  sin 
Supplementary angles
sin(180   )  sin 
tan(90   )  cot 
are those which add up to 180
cos(180   )   cos 
tan(180   )   tan

Trig equations Remember that from your calculator sin 1 , cos 1 and tan 1 give the
principal value (p.v.)
Example
Solve the equations
(i) tan  1.5 for 0    360
(ii) sin 2  0.5 for  180    180
(iii) 2 cos 2   1  sin  for 0    360
(iv) 2 sin 2   sin  cos  for 0    360
(v) sin   80 
3
for  180    180
2
(i)
A
S
tan  1.5
PV = -56.30
  124 , 304
T
(ii) sin 2  0.5 …..first solve for 2 for  360    360
2  30, 150;  210,  330
  15 , 75 ;  105 ,  165
C
S
A
PV = 30

T
C
(iii) (In this example, use cos 2   1  sin 2  )

2 cos 2   1  sin 

2 1  sin 2   1  sin 
2  2 sin 2   1  sin 
S
A
2 sin 2   sin   1  0
PV = -30
sin   12 sin   1  0

sin   1
  90 
or
sin   
T
1
2
  210  , 330 
   90 , 210  , 330
(iv) Don’t cancel out sin  . Bring to LHS and factorise
2 sin 2   sin  cos 
2 sin 2   sin  cos   0
sin  2 sin   cos    0
C
 sin   0
2 sin   cos 
or
sin  1

cos  2
  0  , 180
1
2
  27  , 207 
   0 , 180 , 27 , 207


(v) sin   80 

3
2
  80  60 ,
PV = 26.56
T
C

solve first for  260    100
 240
  140 ,  160

A
S
tan 
A
S

PV = 60
T
C
In the next example, angles are in radians. The radian sign c is sometimes omitted, but is
implied when the interval contains .
Example
(i)
Solve the following equations
(i)
cos x  0.3 for 0  x  2 , answers correct to 2d.p.
(ii)
tan
x
 3 for  2  x  2 , answers in exact form
2
cos x  0.3 ……put calculator into RAD mode.
x  1.266..., 2  1.266...
A
S
x  1.27, 5.02
PV = 1.2661..
T
C
(ii)
In exact terms means in terms of . The implication is that the angles will be exact
form in degrees. So, work in degrees first and then convert to radians.
tan
x
 3 … solve first for    x  
2
x
 60  ,120 
2
 x  120  ,240 
x 
2
4
, 
3
3
S
A
PV = 60
T
C