“Teach A Level Maths”
Vol. 1: AS Core Modules
43: Quadratic Trig
Equations and Use of
Identities
© Christine Crisp
Quadratic Trig Equations
Module C2
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Quadratic Trig Equations
1
e.g. 1 Solve the equation sin2 x 
for the interval
4


 180  x  180
This is the shorthand
1
2
sin x 
Solution:
notation for
4
sin x  sin x
Quadratic equation
1
2
sin x  
Square rooting:
or (sin x )
so 2 solutions!
4
1
sin x  

4
1
1
or sin x  
 sin x  
2
2
The original problem has become 2 simple trig
equations, so we solve in the usual way.
Quadratic Trig Equations
sin x   0  5 or sin x   0  5 for  180  x  180
sin x   0  5 : 1st solution:
x  30
sin x   0  5 : 1st solution:
x  30
y
1
 150
 180
 30
y  0 5
30 
180
360 
x
150
y  0 5
-1
Ans:
 150 ,  30 , 30 , 150
y  sin x
Quadratic Trig Equations
e.g. 2 Solve the equation 3 sin2 x  sin x  0 for the
interval 0  x  360, giving answers to 1 d.p.
Solution: Let
s  sin x. Then,
2
3s  s  0
This is a quadratic equation, so it has 2 solutions.
( Method: Try to factorise; if there are no
factors, use the formula or complete the square. )
Common factor: s(3s  1)  0
s  0 or s   13


sin x  0
or
sin x   13
The original problem has become 2 simple trig
equations, so we again solve in the usual way.
Quadratic Trig Equations
sin x  0
sin x  0 :
or
sin x   13
for
0  x  360
This is easy! We can just use the sketch.
y
1
 180
0
-1
sin x   13 :
Principal value:
180
180
360 x
360 
y  sin x
x  ( 19  5  )
Quadratic Trig Equations
sin x  0
sin x  0 :
or
sin x   13
for
0  x  360
This is easy! We can just use the sketch.
y
1
 180
( 19  5  )
y   13
-1
sin x   13 :
Principal value:
199 5 
180
340 5  x
360 
y  sin x
x  ( 19  5  )
Quadratic Trig Equations
sin x  0
sin x  0 :
or
sin x   13
for
0  x  360
This is easy! We can just use the sketch.
y
1
 180
199 5 
( 19  5  )
0
180
340 5 x
360 
y   13
-1
sin x   13 :
Ans:
Principal value:
y  sin x
x  ( 19  5  )
0  , 180 , 199  5  , 340  5  , 360
Quadratic Trig Equations
e.g. 3 Solve the equation 2 cos   3 cos   2  0 for
the interval 0    2 , giving exact answers.
2
Solution: Let c  cos  . Then,
2c  3c  2  0
Factorising:
(2c  1)(c  2)  0  c 
cos   2
or
cos   1

2
2
The graph of y  cos . . .
shows that cos  always
lies between -1 and +1 so,
cos   2 has
no solutions for  .
1
2
or
c  2
y
1
0
-1


2
y  cos
Quadratic Trig Equations
for 0    2 .
Solving cos   1
2
Principal Solution:

  60 

3
y
1
0
y  0 5

3

5
3
y  cos
-1
Ans:

5
  ,
3
3

2
Quadratic Trig Equations
Exercises
1. Solve the equation 6 cos 2 x  cos x  1  0 for
0  x  360.
Solution: 6c 2  c  1  0  (3c  1)(2c  1)  0
 c   13 or c  12 cos x   13 or cos x  12
Ans: 60 , 109  5  , 250  5  , 300
2. Solve the equation 2 cos 2 x  cos x  0 for
  x   giving the answers as exact
fractions of  .
Solution:
2c 2  c  0  c(2c  1)  0
 c  0 or c  12  cos x  0 or cos x  12

  
Ans:  ,  , ,
2
3 3 2
Use of Trig Identities
We can only solve a trig equation if we can reduce it
to one, or more, of the following:
sin  c , cos  c
or
tan  c
So, if we have an equation with sin and cos 
. . .
e.g. 3 sin2   5 cos   5  0
. . . we need a formula that will change one of these
trig ratios into a function of the other.
The formula we use is sometimes called the
Pythagorean Identity and we will prove it now.
Use of Trig Identities
A
Proof of the Pythagorean Identity.
Consider the right angled triangle ABC.
Using Pythagoras’ theorem:
c
a b  c
Divide by c 2:
a 2 b2 c2



c2 c2 c2
2
2
2
B
2
 a
b
     1
 c
c

 cos  
b
2
2

a
C
a
b
But cos  
and sin 
c
c
 sin   1
2

cos 2   sin2   1
Use of Trig Identities
We have shown that this formula holds for any angle
 in a right angled triangle.
However, because of the symmetries of cos and
sin , it actually holds for any value of  .
A formula like this which is true for any value of the
variable is called an identity.
cos 2   sin2   1
Identity symbol
Identity symbols are normally only used when we
want to stress that we have an identity. In the trig
equations we use an  sign.
Use of Trig Identities
e.g.4 Solve the equation 3 sin2   5 cos   5  0
for 0    360 giving answers correct to 1 d.p.
Method: We use the identity cos 2   sin2   1
to replace sin 2  in the equation.
Solution:
Rearranging: cos 2   sin2   1  sin2   1  cos 2 
Substitute in 3 sin2   5 cos   5  0 We always use the
 3( 1  cos  )  5 cos  5  0
2
identity to
substitute for the
squared term.
Let cos   c and multiply out the brackets:
 3(1  c 2 )  5c  5  0
 3  3c 2  5c  5  0
Use of Trig Identities
3  3c 2  5c  5  0 


c

Principal values: cos 
 3c 2  5c  2  0
2
3
c
 5c  2  0 is easier if
Tip: Factorising
(3c squared
2)(c  1) term
0 is positive.
the
2
or c  1
3
 23    48 2 
cos   1
1
y  23
0
-1
48 2
180
y  cos

311 8 360
Use of Trig Identities
3  3c 2  5c  5  0   3c 2  5c  2  0

3c 2  5c  2  0
 (3c  2)(c  1)  0
c  23 or c  1

Principal values: cos   2    48 2 
3
cos   1 We just look at the graph!
Ans:
1
y
0
-1
48 2
2
3

180
y  cos
  0  , 48  2  ,

311 8 360

311 8  , 360
Use of Trig Identities
A 2nd Trig Identity
A
Consider the right angled triangle ABC.
a
b
cos   ,
sin 
c
c
 c cos   a and c sin  b

B
b
Also, tan  
a
c sin
So, tan  
c cos 
sin
 tan  
cos 
c
b
a
C
Use of Trig Identities
e.g.5 Solve the equation sin   cos  for     
giving exact answers.
Warning! We notice that there are 2 trig ratios but
no squared term. We MUST NOT try to square root
the Pythagorean identity since
cos 2   sin2   1 DOES NOT GIVE cos   sin  1
Method: Divide sin   cos  by cos 
sin
not zero, we can divide by it.
Since cos
is


 1
cos 
sin
We can now use the identity tan  
cos 
 tan  1
We now have one simple trig equation.
Use of Trig Identities
tan   1 for
    

Principal value:   45  
rads.
4

Add

to get 2nd solution:
3
    
4
4


3
Ans:    ,
4
4
Use of Trig Identities
SUMMARY
 With a quadratic equation, if there is only 1 trig
ratio
• Replace the ratio by c, s or t as appropriate.
•
•
Collect the terms with zero on one side of
the equation.
Factorise the quadratic and solve the
resulting 2 trig equations.
 If there are 2 trig ratios, use
cos 2   sin2   1
or
sin
tan  
cos 
to substitute for
cos 2  or sin 2 
if there are no
squared terms.
Use of Trig Identities
Exercises
1. Solve the equation 5 cos 2   3 sin  3 for
 180  x  180
2. Solve the equation sin  2 cos  for
giving the answers correct to 3
0  x  2
significant figures.
Use of Trig Identities
Solutions
1. Solve the equation 5 cos 2   3 sin  3 for
 180    180
Solution: cos 2   sin2   1  cos 2   1  sin2 
Substitute in 5 cos 2   3 sin  3
 5 (1  sin2  )  3 sin  3
5 (1  s 2 )  3 s  3

2
5

5
s
 3s  3

2

0

5
s
3sthe
 2 r.h.s.
We’ll collect the terms on
so that the squared

0  (5term
s  2)(is
s positive.
1)

sin  52
or sin  1
Use of Trig Identities
sin  52 or sin  1 for  180    180
Principal values: sin  2    23 6 
5
sin  1    90
y
1
 180
 90
23 6 
-1
Ans:
y  0 4
180
156 4 
360 
x
y  sin x
 90 , 23  6  , 156  4 
Use of Trig Identities
Solutions
2. Solve the equation sin  2 cos  for 0    2
giving the answers correct to 3 significant figures.
Solution:
Divide by cos  : sin  2 cos  
sin
2
cos 
sin
 tan   tan   2
Substitute using
cos 
Principal value:   1 11 rads.
Add
:
  1  11    4 25
Ans: 1.11c , 4  25 c ( 3 s.f.)
Quadratic Trig Equations
Quadratic Trig Equations
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Quadratic Trig Equations and Use of Identities
1
e.g. 1 Solve the equation sin2 x 
for the interval
4


 180  x  180
This is the shorthand
1
2
sin x 
Solution:
notation for
4
Quadratic equation
sin x  sin x
2
so 2 solutions!
or (sin x )
1
Square rooting:
sin x  
4

sin x  
1
4
1
1
or sin x  
 sin x  
2
2
The original problem has become 2 simple trig
equations, so we solve in the usual way.
Quadratic Trig Equations and Use of Identities
e.g. 2 Solve the equation 3 sin2 x  sin x  0 for the
interval 0  x  360, giving answers to 1 d.p.
Solution: Let
s  sin x. Then,
2
3s  s  0
This is a quadratic equation, so it has 2 solutions.
( Method: Try to factorise; if there are no
factors, use the formula or complete the square. )
Common factor: s(3s  1)  0
s  0 or s   13


sin x  0
or
sin x   13
The original problem has become 2 simple trig
equations, so we solve in the usual way.
Quadratic Trig Equations and Use of Identities
e.g. 3 Solve the equation 2 cos   3 cos   2  0 for
the interval 0    2 , giving exact answers.
2
Solution: Let c  cos  . Then,
2c  3c  2  0
Factorising:
(2c  1)(c  2)  0  c 
cos   2
or
cos   1

2
1
2
or
c  2
2
The graph of y  cos . . .
shows that cos  always
lies between -1 and +1 so,
cos   2 has
no solutions for  .

y  cos
Quadratic Trig Equations and Use of Identities
for 0    2 .
Solving cos   1
2
Principal Solution:

  60 

3
y  0 5


3
y  cos
Ans:

5
  ,
3
3
5
3
Quadratic Trig Equations and Use of Identities
We can only solve a trig equation if we can reduce it
to one, or more, of the following:
sin  c , cos  c
or
tan  c
So, if we have an equation with sin and cos 
. . .
e.g. 3 sin2   5 cos   5  0
. . . we need a formula that will change one of these
trig ratios into a function of the other.
cos 2   sin2   1
A formula like this which is true for any value of the
variable is called an identity.
This formula is sometimes called a Pythagorean
Identity ( since its proof uses Pythagoras’ theorem ).
Quadratic Trig Equations and Use of Identities
e.g.4 Solve the equation 3 sin2   5 cos   5  0
for 0    360 giving answers correct to 1 d.p.
Method: We use the identity cos 2   sin2   1
to replace sin 2  in the equation.
Solution:
Rearranging: cos 2   sin2   1  sin2   1  cos 2 
Substitute in 3 sin2   5 cos   5  0 We always use the
 3( 1  cos  )  5 cos  5  0
2
identity to
substitute for the
squared term.
Let cos   c and multiply out the brackets:
 3(1  c 2 )  5c  5  0
 3  3c 2  5c  5  0
Quadratic Trig Equations and Use of Identities
3  3c 2  5c  5  0   3c 2  5c  2  0

3c 2  5c  2  0
 (3c  2)(c  1)  0
c  23 or c  1

Principal values: cos   2    48 2 
3
cos   1 We just look at the graph!
y
Ans:
2
3
  0  , 48  2  ,

48 2
311 8
y  cos
311 8  , 360
Quadratic Trig Equations and Use of Identities
e.g.5 Solve the equation sin   cos  for     
giving exact answers.
Warning! We notice that there are 2 trig ratios but
no squared term. We MUST NOT try to square root
the Pythagorean identity since
cos 2   sin2   1 DOES NOT GIVE cos   sin  1
Method: Divide sin   cos  by cos 
Since
cos 

is not zero, we can divide by it.
sin
 1
cos 
We can now use the identity

tan  1
sin
tan  
cos 
We now have one simple trig equation.
Quadratic Trig Equations and Use of Identities
SUMMARY
 With a quadratic equation, if there is only 1 trig
ratio
• Replace the ratio by c, s or t as appropriate.
•
•
Collect the terms with zero on one side of
the equation.
Factorise the quadratic and solve the
resulting 2 trig equations.
 If there are 2 trig ratios, use
cos 2   sin2   1
or
sin
tan  
cos 
to substitute for
cos 2  or sin 2 
if there are no
squared terms.