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