43: Quadratic Trig 43: Quadratic Trig Equations and Use of Equations and Use of
IdentitiesIdentities
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Quadratic Trig Equations
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Module C2
Quadratic Trig Equations
4
1sin 2 xSolutio
n:xsinSquare
rooting: 4
1
2
1sin x
4
1sin x
2
1sin x
e.g. 1 Solve the equation for the interval 180180 x
4
1sin 2 x
or
xx sinsin
This is the shorthand notation for
2)(sin xor Quadratic
equation so 2 solutions!
The original problem has become 2 simple trig equations, so we solve in the usual
way.
Quadratic Trig Equations
y
1
-1
180 360x
xy sin
180
1st solution:
:50sin x 30x
:50sin x 1st solution:
30x
50y
50y
30
150150
50sin x 50sin xor 180180 xfor
30
Ans:
150,30,30,150
Quadratic Trig Equations
e.g. 2 Solve the equation for the interval , giving answers to 1 d.p.3600 x
0sinsin3 2 xx
0sin x31sin x or
The original problem has become 2 simple trig equations, so we again solve in the
usual way.
Solution: Let . Then,xs sin03 2 ss
This is a quadratic equation, so it has 2 solutions.
Common factor:
0)13( ss
( Method: Try to factorise; if there are no factors, use the formula or complete the square. )
310 ss or
Quadratic Trig Equations
y
1
-1 xy sin
x180 360
180
:0sin x
:sin31x Principal value: )519( x
0sin x31sin xor 3600 xfor
0 180 360
This is easy! We can just use the sketch.
Quadratic Trig Equations
y
1
-1 xy sin
x180180 360
31y
53405199
:0sin x
0sin x31sin xor 3600 xfor
)519(
This is easy! We can just use the sketch.
:sin31x Principal value: )519( x
Quadratic Trig Equations
y
1
-1 xy sin
x180
31y
5340
0sin x31sin xor 3600 xfor
)519(
Ans:
360,5340,5199,180,0
51990
360180
:0sin x This is easy! We can just use the sketch.
:sin31x Principal value: )519( x
Quadratic Trig Equations
e.g. 3 Solve the equation for the interval , giving exact answers. 20
02cos3cos2 2
21cos 2cos or
0232 2 ccFactorising:
0)2)(12( cc 221 cc or
The graph of . . .cosy
Solution: Let . Then,cosc
shows that always lies between -1 and +1 so, has no solutions for .
cos
2cos
0 2
1
y
cosy-1
Quadratic Trig Equations
cosy-1
0 2
1
y
Principal Solution: 3
60
Solving for .21cos 20
50y
3
3
5
Ans: 3
5,
3
Quadratic Trig Equations
Exercises
giving the answers as exact fractions of .
2. Solve the equation for0coscos2 2 xx x
1. Solve the equation for .
01coscos6 2 xx3600 x
Solution:
016 2 cc 0)12)(13( cc
21
31 cc or
Ans: 300,5250,5109,60
Solution:
02 2 cc 0)12( cc
210 cc or
Ans: 2
,3
,3
,2
21
31 coscos xx or
21cos0cos xx or
Use of Trig Identities
05cos5sin3 2 e.g.
The formula we use is sometimes called the Pythagorean Identity and we will
prove it now.
We can only solve a trig equation if we can reduce it to one, or more, of the following:
,sin c ccos ctanor
So, if we have an equation with and . . .
sin cos
. . . we need a formula that will change one of these trig ratios into a function of the other.
Use of Trig Identities
Proof of the Pythagorean Identity.
Using Pythagoras’ theorem: 222 cba Divide by :2c
122
c
b
c
a
1sincos 22 1sincos 22
2
2
2
2
2
2
c
c
c
b
c
a
Consider the right angled triangle ABC. c
a
b
A
B C
c
acosBut and
c
bsin
Use of Trig Identities
1sincos 22
However, because of the symmetries of and , it actually holds for any value of .
cossin
A formula like this which is true for any value of the variable is called an identity.
We have shown that this formula holds for any angle in a right angled triangle.
Identity symbolIdentity 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
Let and multiply out the brackets:
ccos
Solution: 2222 cos1sin1sincos Rearranging:
05cos5sin3 2 Substitute in
05cos5)(3 2cos1
055)1(3 2 cc
05533 2 cc
05cos5sin3 2 e.g.4 Solve the equationfor giving answers correct to 1 d.p.3600
We always use the identity to substitute for the squared term.
Method: We use the identity to replace in the equation.
1sincos 22 2sin
Use of Trig Identities
0
cosy
1
-1
180 360
05533 2 cc
8311
Tip: Factorising is easier if the squared term is positive.
0253 2 cc0253 2 cc
32c 1c o
r Principal values:
24832cos
1cos
0)1)(23( cc
32y
248
Use of Trig Identities
cosy
1
-1
180
05533 2 cc
8311
0253 2 cc0253 2 cc
32c 1c o
r Principal values:
24832cos
1cos
Ans: ,248,0 360,8311
0)1)(23( cc
We just look at the graph!
32y
2480 360
Use of Trig Identities
A 2nd Trig Identity
Consider the right angled triangle ABC. c
a
b
A
B C
,cosc
a
c
bsin
Also, a
btan
bcac sincos and
So,
cos
sintan
c
c
cos
sintan
Use of Trig Identities
Method: Divide by
cossin cos
1cos
sin
e.g.5 Solve the equation for cossin 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
1sincos 22 DOES NOT GIVE 1sincos
We can now use the identity
cos
sintan
Since is not zero, we can divide by it.
cos
1tan We now have one simple trig
equation.
Use of Trig Identities
1tan for
Principal value: 454
rads
.Add to get 2nd solution:
4 4
3
Ans: 4
3,
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
1sincos 22
or
cos
sintan if there are no
squared terms.
to substitute for
or 2cos 2sin
Use of Trig Identities
2. Solve the equation for giving the answers correct to 3 significant figures.
cos2sin 20 x
1. Solve the equation for3sin3cos5 2 180180 x
Exercises
Use of Trig IdentitiesSolution
s1. Solve the equation for
180180 Solution:
2222 sin1cos1sincos
33)1(5 2 ss3355 2 ss
2350 2 ss )1)(25(0 ss
3sin3cos5 2
Substitute in
3sin3cos5 2
52sin 1sin o
r
3sin3)sin1(5 2
We’ll collect the terms on the r.h.s. so that the squared term is positive.
Use of Trig Identities
y
1
-1 xy sin
180 x
180 36090
40y
4156623
Ans: 4156,623,90
6231sin 9052sin Principal
values:
for 180180 52sin 1sin o
r
Use of Trig Identities
Solution:
2. Solve the equation forgiving the answers correct to 3 significant figures.
cos2sin 20
cosDivide by : cos2sin 2cos
sin
2tan
tancos
sin Substitute using
Principal value: rads. 111
111 254Add :cc 254,11.1 Ans: ( 3 s.f.)
Solutions
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
4
1sin 2 xSolutio
n:
xsinSquare rooting: 4
1
2
1sin x
4
1sin x
2
1sin x
e.g. 1 Solve the equation for the interval 180180 x
4
1sin 2 x
or
xx sinsin
This is the shorthand notation for
2)(sin xor Quadratic
equation so 2 solutions!
The original problem has become 2 simple trig equations, so we solve in the usual
way.
Quadratic Trig Equations and Use of Identities
0sin x31sin x or
The original problem has become 2 simple trig equations, so we solve in the usual
way.
Solution: Let . Then,xs sin03 2 ss
This is a quadratic equation, so it has 2 solutions.
Common factor:
0)13( ss
( Method: Try to factorise; if there are no factors, use the formula or complete the square. )
310 ss or
e.g. 2 Solve the equation for the interval , giving answers to 1 d.p.3600 x
0sinsin3 2 xx
Quadratic Trig Equations and Use of Identities
e.g. 3 Solve the equation for the interval , giving exact answers. 20
02cos3cos2 2
21cos 2cos or
0232 2 ccFactorising:
0)2)(12( cc 221 cc or
The graph of . . .cosy
Solution: Let . Then,cosc
shows that always lies between -1 and +1 so, has no solutions for .
cos
2cos
cosy
Quadratic Trig Equations and Use of Identities
Principal Solution:
Solving for .
cosy
360
21cos 20
50y
3
3
5
Ans: 3
5,
3
Quadratic Trig Equations and Use of Identities
05cos5sin3 2 e.g.
This formula is sometimes called a Pythagorean Identity ( since its proof uses Pythagoras’ theorem ).
We can only solve a trig equation if we can reduce it to one, or more, of the following:
,sin c ccos ctanor
So, if we have an equation with and . . .
sin cos
. . . we need a formula that will change one of these trig ratios into a function of the other.
1sincos 22
A formula like this which is true for any value of the variable is called an identity.
Quadratic Trig Equations and Use of Identities
Let and multiply out the brackets:
ccos
Solution: 2222 cos1sin1sincos Rearranging:
05cos5sin3 2 Substitute in
05cos5)(3 2cos1
055)1(3 2 cc
05533 2 cc
05cos5sin3 2 e.g.4 Solve the equationfor giving answers correct to 1 d.p.3600
We always use the identity to substitute for the squared term.
Method: We use the identity to replace in the equation.
1sincos 22 2sin
Quadratic Trig Equations and Use of Identities
cosy
05533 2 cc
8311
0253 2 cc0253 2 cc
32c 1c o
r Principal values:
24832cos
1cos
Ans: ,248,0 360,8311
0)1)(23( cc
We just look at the graph!
32y
248
Quadratic Trig Equations and Use of Identities
Method: Divide by
cossin cos
1cos
sin
e.g.5 Solve the equation for cossin 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
1sincos 22 DOES NOT GIVE 1sincos
We can now use the identity
cos
sintan
Since is not zero, we can divide by it.
cos
1tan 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
1sincos 22
or
cos
sintan if there are no
squared terms.
to substitute for
or 2cos 2sin
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