Super Trig PowerPoint

Introduction- Finding lengths

Sin, cos or tan to find lengths?

Finding missing lengths worksheet

Finding the Area of Triangles and

Segments

The Sine rule The Cosine RuleThe Sine and

Cosine Rule (quiz and worksheet)

Finding lengths and angles (more

practise)

Finding Missing Angles

The Sine rule proof

The Cosine Rule proof

Trig graphs Combining the Rules

Trig graphs- matching cards

a

b

c

d

e

f

g h

i10cm30˚

40˚50˚

35˚

45˚ 42˚

27˚

38˚

51˚

Use sin, cos and tan to find the missing lengths, round them to 1 d.p, and use that

answer to work out the next length.Side Length (rounded to 1 dp)

a

b

c

d

e

f

g

h

i

a

b

c

d

e

f

g h

i10cm30˚

40˚50˚

35˚

45˚ 42˚

27˚

38˚

51˚

Use sin, cos and tan to find the missing lengths, round them to 1 d.p, and use that

answer to work out the next length.Side Length (rounded to 1 dp)

a

b

c

d

e

f

g

h

i

Side Length (rounded to 1 dp)

a 5

b 4.2

c 3.2

d 2.2

e 3.1

f 3.4

g 1.7

h 2.8

i 9.5

Trigonometry 1

Warm upSolve the following

equations:

1) 20=

2) 15=

3) 8=

4) 7=

5) 16=

X2

X3

32 X

21 X

64 X

Trigonometry• We can use trigonometry to find missing

angles and lengths of triangles.• Trigonometry uses three functions, these are

called:– Sine (shortened to Sin and pronounced “sign”)– Cosine (shortened to Cos)– Tangent (shortened to Tan)

• We will start working with right angled triangles

Labelling the sides

Hypotenuse

The longest side, the one opposite the right angle is called the hypotenuse

Before we can use Sin, Cos and Tan we need to be able to label the sides of a right angled triangle

Labelling the sidesO

ppos

iteWhat we call the other two sides will

change depending on which angle we are working with, for example..

ϴAdjacent

If we are given (or need to work out) this angle, we label the other

sides like this..But if we are working

with this angle, we label the sides like

this...

Opposite

Adja

cent

Labelling Right Angle Triangle

10 multiple choice questions

OppositeAdjacent

Hypotenuse

A) B)

C)

ϴX

What is the side marked with an X?

Adjacent

Hypotenuse OppositeA) B)

C)

ϴ

X


Hypotenuse Opposite

Adjacent

A) B)

C)

ϴ

X


Opposite Adjacent

Hypotenuse

A) B)

C)

ϴ

X


Hypotenuse

Adjacent OppositeA) B)

C)

ϴX


AdjacentOpposite

Hypotenuse

A) B)

C)

ϴX


AdjacentOpposite

Hypotenuse

A) B)

C)

ϴ

X


Opposite Hypotenuse

Adjacent

A) B)

C)

ϴX


Opposite

Hypotenuse AdjacentA) B)

C)

ϴ

X


Hypotenuse Opposite

Adjacent

A) B)

C)

ϴ

X


Sine (sin)

30˚

5cm10cm We use Sine when we have the

Opposite length and the Hypotenuse

Try entering sin30 in your calculator, it should give the same answer as 5 ÷ 10

Sin30= 510

The rule we use is:

Sinϴ= OppositeHypotenuse

Sin Example 1

42˚

O 7cm We can use Sin as the question involves the Opposite length and the Hypotenuse

Sin42= O7

The rule we use is:


7 x Sin42= O

4.68 cm (2dp)= O

Sin Example 2

17˚

10cm H We can use Sin as the question involves the

Opposite length and the Hypotenuse

Sin17= 10H

The rule we use is:


H x Sin17= 10

H= 10 Sin17 H= 34.2 cm (1dp)

Cosine (cos)

50˚

Adjacent

Hypotenuse

We use cosine when we have the Adjacent length and the Hypotenuse

The rule we use is:

Cosϴ= AdjacentHypotenuse

Cos Example 1

53˚

A

9cm We can use Cos as the question involves the Adjacent length and the Hypotenuse

Cos53= A9

The rule we use is:


9 x Cos53= A

5.42 cm (2dp)= A

Cos Example 217˚

9cm H We can use Cos as the question involves

the Adjacent length and the Hypotenuse

Cos17= 9H

The rule we use is:


H x Cos17= 10

H= 9 Cos17 H= 9.41 cm (2dp)

Tangent (tan)

50˚

6.4cm (1dp)

10cm

We use tangent when we have the Opposite and Adjacent lengths.

The rule we use is:

Tanϴ= OppositeAdjacent

Tan Example 1

53˚

O

11cm

We can use Tan as the question involves the Adjacent and Opposite lengths

Tan53= O11

The rule we use is:


11 x Tan53= O

14.6 cm (1dp)= O

Tan Example 2

35˚

A

21cm

We can use Tan as the question involves the Adjacent and Opposite lengths

Tan35= 21A

The rule we use is:


A x Tan35= 21

A= 21 Tan35

A= 29.99 cm (2dp)

The three rulesSo we have:




Sinϴ= Tanϴ= OA

Cosϴ= AH

OH

SOHCAHTOAThere are a few ways to remember this

Silly Old Horses Can’t Always Hear The Other Animals

SOHCAHTOA

WHAT?

Practise1. Use Sine to find the missing lengths on these triangles:

2. Use Cosine to find the missing lengths on these triangles:

3. Use Tangent to find the missing lengths on these triangles:

O15cm

50˚17cm

H60˚

A

22cm

38˚

25cmH60˚

O

15cm42˚

11cm

A60˚ Tanϴ=Opposite

Adjacent


Sinϴ=Opposite

Hypotenuse

Trigonometry 2

Skiers On Holiday Can Always Have The Occasional Accident

SOHCAHTOA




http://www.google.co.uk/url?sa=i&rct=j&q=skiing+accident&source=images&cd=&cad=rja&docid=OrgF1n7GTrkOEM&tbnid=FPRRPU0dCxQCwM:&ved=0CAUQjRw&url=http://www.maineinjurylawyerblog.com/2012/01/man-dies-on-way-to-hospital-after-bangor-skiing-accident.html&ei=juJRUbiFEO_u0gXl_oEQ&bvm=bv.44342787,d.d2k&psig=AFQjCNHKCvhJdRswodP95TXfZBHw3p7n0g&ust=1364407102160647

Our aim today• We have looked at the three rules and have

practised labelling triangles.• Today we will have to decide whether we are

using Sin, Cos or Tan when answering questions.

SOH CAH TOA

7cmX

35˚oppo

site

Hypotenuse

This question will use Sine

Sin35= X7

Sinϴ= OH

SOH CAH TOA

X

8cm

17˚

opposite

Adja

cent

This question will use Tan

Tan17= 8X

Tanϴ= OA

SOH CAH TOA

X

8cm

43˚

opposite

Hypotenuse

This question will use Sin

Sin43= 8X

Sinϴ= OH

SOH CAH TOA

X

8cm26˚

Hypotenuse

Adja

cent

This question will use Cosine

cos26= X8

cosϴ= OA

Sin, Cos or Tan?


SinCos

Tan

A) B)

C)

35˚X

Will you use Sin, Cos or Tan with this question?

11cm

Cos

Sin TanA) B)

C)

14˚X


15cm

Sin Cos

Tan

A) B)

C)

40˚ XWill you use Sin, Cos or Tan with this question?

17cm

Tan Sin

Cos

A) B)

C)

50˚

X


5cm

Sin

Cos TanA) B)

C)

51˚X


6cm

TanSin

Cos

A) B)

C)

16˚X


8cm

CosSin

Tan

A) B)

C)

42˚14cm


X

Tan Cos

Sin

A) B)

C)

35˚X


4cm

Sin

Cos TanA) B)

C)

63˚

X


3.4cm

Sin Tan

Cos

A) B)

C)

71˚

X


5mm

Answers:1) 3.1cm2) 6.1cm3) 5.1cm4) 17.1cm5) 4.5cm6) 8.6cm7) 20.5cm8) 31.1cm9) 117.6cm10)1.5cm11)4.1cm12)108.9cm

Practise

The Sine Rule

a

b c

C

A

B

SinA SinB SinC==a b c

This is the Sine rule

We can also write it like this

We can use it to find out missing sides and angles of non right angle triangles

7cm

5cm 50˚

B

Because we’re looking for an angle, I'm going to use the version of the rule which has Sin on top

SinA SinB=a b

Sin50 SinB=7 5

(multiply both sides by 5)

5xSin50 SinB=7

SinB=0.54717...

Sin-10.54717... = 33.2˚ (1dp)

11cm

8cm38˚

C

Because we’re looking for an angle, I'm going to use the version of the rule which has Sin on top

SinA SinB=a b

Sin38 SinC=11 8

(multiply both sides by 58)

8xSin38 SinC=11

SinC=0.447753..

Sin-10.447753... =26.6˚ (1dp)

a

5cm 50˚

35˚

Because we’re looking for a length, I'm going to use the version of the rule which has the length on top

a b=sinA sinB

Sin35 Sin50=5 a (multiply both sides by

Sin50)

5xSin50 a=sin35

a=6.7cm(1dp)

c

9cm

64˚

21˚

Because we’re looking for a length, I'm going to use the version of the rule which has the length on top

a c=SinA SinC

Sin64 Sin21=9 c (multiply both sides by

Sin21)

9xSin21 c=sin64

c=3.6cm(1dp)

The Sine Rule Quiz


Yes NoA) B)

Can you use the sine rule to find the missing value?

10cm

12cm

a

20˚

NoYesA) B)


9cm

15cm

a

30˚

Yes NoA) B)


10cm

X20˚

85˚

NoYesA) B)


10cm 12cm

a

15cm

Yes NoA) B)


10cm

12cm

38 ˚

20˚

X

NoYesA) B)


a

20˚

72˚ 88˚

NoYesA) B)


10cm

12cm

Yes NoA) B)


10cm

x35˚

But you could also useSin35= 10 ÷ x

Yes NoA) B)


9.8cm

21cm

a

34˚

Yes NoA) B)


5cm

7cm

a

6cm

Practise Questions

37˚

45˚

61˚45˚

23˚

37˚

80˚

a

b

19˚

110˚c

11cm

BC 8cm

25cm

C

5.6cm

7.2cm

59cm

47cm57cm

34cm

Answers:1) 9.4cm2) 9.9cm3) 38.5cm4) 50˚5) 15˚6) 34.1˚

The Cosine Rule

Warm up

7cm

x

x23cm

19cm29cm 7.1cm

13.4cm

35˚ 41˚

ϴ

ϴ

x= 8.5cmx= 15.1cm

ϴ=49.1˚

ϴ=27.9˚

Find the missing sides and angles

a

Abc

a2=b2 + c2 – 2bccosA

The length here, has to be the length opposite this angle

Which side is b and which is c doesn’t matter

a

95˚5cm8cm

a2=52 + 82 – 2x5x11xcos95

The length here, has to be the length opposite this anglea2=25 + 64 – 110xcos95a2=89– 110xcos95a2=89– -9.58.....a2=98.58...

a=9.9291..a=9.9cm (1dp)

a

81˚7cm12cm

a2=72 + 122 – 2x7x12xcos81

The length here, has to be the length opposite this anglea2=49 + 144 – 168xcos81a2=193– 168xcos81a2=193– 26.280...a2=166.7190...

a=12.911..a=12.9cm (1dp)

11cm

A5cm8cm

112=52 + 82 – 2x5x11xcosA


112=52 + 82 – 2x5x11xcosA121=89– 2x5x11xcosA32=-110xcosA32 ÷ -110=cosA

-0.2909090..=cosACos-1-0.209090..=A106.9˚ (1dp)=A

20cm

A13cm8cm

202=132 + 82 – 2x13x11xcosA


202=132 + 82 – 2x13x8xcosA400=233– 208cosA167=-208xcosA167 ÷ -208=cosA

-0.80288.....cosACos-1-0.80288...=A143.4˚ (1dp)=A

Practise Questions

12cm

45˚ 61˚

15cm

28cm

37˚

3.4cm

a

b

51cm

23cm

c

11cm

xx 8cm

25cm

C

5.6cm

7.2cm

59cm

47cm57cm

48cm

Answers:1) 1.8.9cm2) 13.1cm3) 17.1cm4) 49.1˚5) 49.9˚6) 55.8 ˚

The Sine Rule proof

What would you like to do?

Be shown the proof Try to prove it yourself

Why does the Sine Rule Work?A

B

bc

aB

Using the rule that:Sinϴ=O/AWe can show that the red line is: Red line=c x sinB

b

C

Using the rule that:Sinϴ=O/AWe can show that the red line is: Red line=b x sinC

The red line is b x sinC and c x sinBSo b x sinC= c x sinB

We can rearrange to make:

bsinB =

csinC

SinBb =

SinCc

or

Why does the Sine Rule Work?A

B

bc

aB

Using the rule that:Sinϴ=O/AWe can show that the red line is: Red line=a x sinB

b

C

Using the rule that:Sinϴ=O/AWe can show that the red line is: Red line=b x sinA

The red line is b x sinA and a x sinBSo b x sinA= a x sinB

We can rearrange to make:

bsinB =

asinA

A

a

SinBb =

SinAa

or

Why does the Sine Rule Work?

So far we have shown that:

bsinB =

asinA

bsinB =

csinCand

Therefore:a

sinA =b

sinB =c

sinC

(We could have also used the version with the angles on top)

The Sine Rule



Prove the Sine Rule!

Hint 1Start with a diagram like this:

CA

B

b

ac

Hint 2

Hint 3

Split it into 2 right angled triangles

Can you find 2 different ways to find the length of the line you drew?

Hint 4Can you split the triangle any other way?

Show me the proof Home

Finding the Areas of Triangles

40˚

10cm

8cm h8cm h

40˚Can we find the height of this triangle using trigonometry?

Sin40=h÷8

8 x Sin40=hSo the height of the triangle is 8 x sin40, we know that the area of a triangle is half base times height so..

Area= ½ base x height

Area= ½ x 10 x 8 x sin40

Can we find the area of this triangle?We need the base and the height (area= half base times height)

a

hb h

CCan we find the height of this triangle using trigonometry?

SinC=h÷b

b x SinC=hSo the height of the triangle is b x sinC, we know that the area of a triangle is half base times height so..

Area= ½ base x height

Area= ½ x a x b x sinC

Can we find the area of this triangle?This is what we’d call

The formula for the area of a triangle is:

Area= ½ absinC

The formula for the area of a triangle is:

Area= ½ absinCWe can use this formula to find the area of non right angled triangles when we haven’t been given the perpendicular height

All we need to know is:The length of two sides and the size of the angle between them

Practise Questions Challenge Questions

Area=30cm2

Area=30cm2

Area=25m2

Area=75cm2

Area=52cm2

Find the areas of the following triangles Find the missing lengths/angles of the following triangles

10cm

40˚

8cm 12cm

9cm

12˚ 72˚

7cm

6.5cm

19cm 9.1cm

ϴ

20cm

13cm

35˚

ϴ15cm

12cm

14cm11cm

38˚7cm a

a

10cm28˚

28˚

13cm

bAnswers:1) 28.9cm2

2) 21.6cm2

3) 10cm2

4) 49.6cm2

5) 70.8cm2

Answers:1) 13.9cm2) 21.3cm3) 9.8cm4) 56.4˚5) 42.5 ˚

Area of Segments

Here we will look at finding the area of sectors

You will need to be able to do two things:

Area of Segments

1) Find the area of a sector using the formula-

2) Find the area of a triangle using the formula-

Area= ½ absinC

Area of sector= Angle of Sector x πr2

360

C

b

a

Example-find the area of the blue segment

10cm10cm100°

Step 1- find the area of the whole sectorArea= 100/360 x π x r2

= 100/360 x π x 102

=100/360 x π x 100 =87.3cm2

Step 2- find the area of the triangleArea= ½ absinC =1/2 x 10 x 10 x sin100 = 49.2cm2

Step 3- take the area of the triangle from the area of the segment

87.3 – 49.3 = 38 cm2

Example-find the area of the blue segment

12cm12cm120°

Step 1- find the area of the whole sectorArea= 120/360 x π x r2

= 120/360 x π x 122

=120/360 x π x 144 =150.8cm2

Step 2- find the area of the triangleArea= ½ absinC =1/2 x 12 x 12 x sin120 = 62.4cm2

Step 3- take the area of the triangle from the area of the segment

150.8 – 62.4 = 88.4 cm2

Questions

10cm

130°11cm85°

12cm170°

5cm95°6.5cm

Find the area of the blue segments, to 1 decimal place

1 2 3

654

17cm65°160°

Answers:1) 75.1cm2

2) 29.5cm2

3) 201.1cm2

4) 8.3cm2

5) 51.8cm2

6) 33cm2

Trigonometry 3Finding missing angles

Some Old Hairy Camels Are Hairier Than Other Animals

SOHCAHTOA




• Alan presses SIN then an angle, he gets the answer 0.5, what angle did he enter?

Sinϴ=0.5We can use the inverse of sin to find out

Sin-10.5= ϴ30˚= ϴ

Warm up

Warm upUse the inverse of Sin, Cos and Tan to find the missing

angles, rounded to 1dp:1. sinϴ=0.7 (type sin-10.7)

2. sinϴ=0.33. cosϴ=0.54. cosϴ=0.95. tanϴ=0.36. tanϴ=0.25

Answers:1)44.4˚2)17.5˚3)60˚4)25.8˚5)16.7˚6)14˚

SOH CAH TOA

7cm3cm

ϴoppo

site

Hypotenuse

This question will use Sin

Sinϴ= 37

Sinϴ= OH

Find the missing angle

Sinϴ=0.42857...What angle would give us this

answer?

Sin-10.42857...= ϴ25.4˚ (1dp)= ϴ

You could use the ANS button on your calculator

Sin-1ANS= ϴ

SOH CAH TOA

8cm

6cm ϴ

Adja

cent

Opposite

This question will use Tan

Tanϴ= 86

Tanϴ= OA


Tanϴ=1.25What angle would give us this

answer?

Tan-11.25= ϴ51.3˚ (1dp)= ϴ

SOH CAH TOA

12cm9cm ϴ

Adja

cent

Hypotenuse

This question will use Cos

Cosϴ= 912

Cosϴ= AH


Cosϴ=0.75What angle would give us this

answer?

Cos-10.75= ϴ41.4˚ (1dp)= ϴ

You could use the ANS button on your calculator

Cos-1ANS= ϴ

Practise Questions

ϴ

ϴ

ϴ

ϴ

ϴ

ϴ

ϴϴ

10cm 11

cm

17cm

11cm

20cm

12cm 18cm

5cm 15cm

23cm 35cm

22cm

10cm

12cm

6cm

13cmAnswers:1) 50.2˚2) 28.6˚3) 59.1˚4) 52.3˚5) 63.4˚6) 65.4˚7) 28.6˚8) 40.9˚

The Cosine Rule



Prove the Cosine Rule!

Hint 1Start with a diagram like this:

A

b

ac

Hint 2

Hint 3

Split it into 2 right angled triangles

You could use pythagoras to find the length of a in the right angled triangle on the right if you had the other two lengths

Hint 4Expand and tidy up, using factorising (don’t forget the identity at the top!)

Show me the proof Home

For this proof you need to know that- (SinA)2 + (CosA)2 = 1

Find the missing lengths and angles

X16cm

30˚

1

14cm

X

51˚

2

15cm

17cm

ϴ3

8cm12cm

ϴ

4

19cm36cm

ϴ5

X18cm

63˚

6

9cm

8.3cm

ϴ7

11.2cm X

35˚

8

X

15cm

43˚

9

X

23cm

50˚10

40cm53cmϴ

X

46cm

28˚

16cm32cmϴ

X

36cm

18˚

61cm 74cm

ϴ 81cm

106cmϴ

11 12

16151413

Answers:1) 8cm2) 22.2cm3) 48.6 ˚4) 41.8 ˚5) 58.1 ˚6) 16cm7) 42.7 ˚8) 19.5cm9) 11cm10)19.3˚11)41 ˚12)21.6cm13)60 ˚14)11.7cm15)55.5 ˚16)49.8 ˚

Why does the Cosine rule work?

a

a

c

A

The cosine rule has “a” as it’s subject, so we need to think of how we could find a in this triangleIf we knew the lengths of the red and purple lines, we could use Pythagoras’ theorem to find aThe red is cSinA

The purple will be the length of b takeaway the green length

The green will be cCosA so the purple is b-cCosA

b-cCosA

cSinAa

From Pythagoras we know that:

a2= (cSinA)2 + (b-cCosA)2

a2= c2(SinA)2 + b2+c2(CosA)2 –bcCosA-bcCosA

a2= b2+c2(SinA)2 +c2(CosA)2 –2bcCosA

a2= b2+c2((SinA)2 + (CosA)2) –2bcCosA

a2= b2+c2(1)–2bcCosA

a2= b2+c2–2bcCosA

So.....

(SinA)2 + (CosA)2 = 1This is something that will come up whilst you are doing your A levels

QED

QED stands for the latin phrase “quod erat demonstrandum” which mean, “which had to be demonstrated” it’s a way for us to say we have finished our proof

Sine or Cosine Rule?


Sine Rule Cosine RuleA) B)

How would you find the missing value?

X

11cm

120˚

11˚

Cosine RuleSine RuleA) B)


X

62cm40cm

110cm



X

11cm

86˚25˚



X

16cm40cm

32cm



X

6cm22˚ 110˚



X

6cm

4cm110˚



X

3.7cm

4.2cm 81˚



X

6cm32˚

115˚



X

13cm

27cm

114˚



X6cm

4cm

7cm

Find the missing lengths and angles(give your answers to 1dp)

13cm

9.2cm9cm

ϴ

ϴ

8cm

7cm

5cm

11cm

12cm

7cm

ϴ

13cm

11cm35˚

ϴ

ϴ

17cm

22cm

41˚

45˚

17˚ 25cm

X

25cm X77˚

˚32˚

X

14cm11cm 43˚

X14cm

23cm31˚

Answers:1) 91.2˚2)29 ˚3)10.3cm4)9.6cm5)60 ˚6)30.5 ˚7)13.2cm8)27.9cm9)35.1 ˚

For each question say what method you would use to find the missing

value

Warm up

Combining the Rules

Pythagoras’ Theorem

Sine Rule

Cosine Rule

SOHCAHTOAA) B)

C) D)

How would you find the missing length in this triangle?

7cm

6cm

x

SOHCAHTOASine Rule

Cosine RulePythagoras’ Theorem

A) B)

C) D)


12cm x

30˚

Cosine Rule

Sine Rule SOHCAHTOA


A) B)

C) D)


13cmx

46˚ 19cm

SOHCAHTOASine Rule


A) B)

C) D)


13cm

x40˚

Sine Rule SOHCAHTOA


A) B)

C) D)


X

11cm

120˚

11˚

Cosine Rule

Sine Rule SOHCAHTOA


A) B)

C) D)


X

6cm

4cm110˚

SOHCAHTOASine Rule


A) B)

C) D)


12cm

x25˚

Cosine Rule

Sine Rule SOHCAHTOA


A) B)

C) D)

How would you find the missing length in this triangle?X

3.7cm

4.2cm 81˚


Sine Rule

Cosine Rule

SOHCAHTOAA) B)

C) D)

8cm16cm

x

Sine Rule SOHCAHTOA


A) B)

C) D)


X

6cm22˚ 110˚

Work out the missing lengths (give answers to 1dp)

X 14cm

23cm

71˚

9cm

32˚130˚

c

a

b

d

Circumference=100cm

81˚

X

4cm

15cm

15cm

23˚

X

28˚

Area of blue segment- 200cm2

X

120˚

146˚

9cm12cm

X

Answers:1) 23.4cm2)

a) 4.8cmb) 7.6cmc) 14.3cmd) 11.3cm

3) 20.7cm4) 24.7cm5) 18.4cm6) 9.6cm

Graphs of Trig Functions

The Sine Curve

The Cosine Curve

The Tan Curve

Key ValuesAngle SinX CosX

0 0 190 1 0

180 -0 -1270 -1 0360 0 1

As you can see both sine and cosine follow the same pattern.

The difference is that sine starts at 0, whereas cosine starts at 1

Key ValuesAngle TanX

0 090 ∞

180 0270 ∞360 0

Reading off values

If sinx=0.5 what is x?A calculator would tell us that it is 30˚, but there are other values that would give 0.5

30 150

The solutions for sinx=0.5 between 0 and 360 are x=30 and x=150

-0.5

Reading off values

If sinx=-0.3 what is x?

197.5 342.5

The solutions for sinx=-0.3 between 0 and 360 are x=197.5 and x=342.5

-0.3

Transformation of trig graphs• Here instead we will use f(x) instead of sin, cos or tan, as what we will

describe works not only with these three functions but all other functions.• y=f(x) + k Translation through

• y=f(x+k) Translation through

• y=kf(x) Stretch by factor k parallel to the y axis

• y=f(kx) Stretch by factor 1/k parallel to the x axis

• y=-f(x) Reflection in the x axis

• y=f(-x) Reflection in the y axis

-k0 )(

0k )(

y=sinxy=2sinx (stretches the curve vertically)y=0.5sinx (squashes vertically)

y=sinxy=sin0.5x (stretches horizontally)

y=sin2x (squashes horizontally)

y=sinxy=sinx + 1 (moves curve upwards)y=sin(x+90) (moves curve to the right)

Transformations of Trig Graphs


Y=sinx

Y=cosx

Y=cosx-1

Y=tanxA) B)

C) D)

Y=cosxY=tanx

Y=cos2xY=2cosx

A) B)

C) D)

Y=tanx

Y=tan2x Y=0.5tanx

Y=-tanx

A) B)

C) D)

Y=2cosx Y=cos2x

Y=2sinxY=sin2x

A) B)

C) D)

Y=0.5sinx

Y=sin2x y=sin0.5x

Y=0.5cosx

A) B)

C) D)

Y=-cosxY=-sinx

Y=sinx-1Y=cos(-x)

A) B)

C) D)

Y=sin2xY=0.5sinx

Y=2sinxY=sin0.5x

A) B)

C) D)

Y=tanx-1

Y=tanx+1 Y=tan(x-1)

Y=tanx

A) B)

C) D)

Y=-tanx

Y=tan(x+90)

Y=tan2x

Y=2tanxA) B)

C) D)

Y=cosx-1 Y=-cosx

Y=cos(-x)Y=sinx-1

A) B)

C) D)

Practise Questions

1. Sketch the graph y=cosxa) Use your graph to find all of the solutions for cosx=0.7 between 0 and

360b) Use your graph to find all of the solutions for cosx=-0.5 between 0 and

360

2. Sketch the graph y=tanxa) Use your graph to find all the solutions for tanx=1.5 between 0 and 360

3. Sketch the following graphs:b) y=3sinxc) y=sin3xd) y=cosx + 5e) y=cos(x-45)f) Y=tan(x+10)

y=sinxy=cosxy=tanx

y=2cosx

y=sin2xy=-cosx

y=sin(x-90)y=sin(x+90)

y=0.5sinx y=tanx-1y=tan(-x) y=cosx-1y=-tanx y=cos0.5x

Cut out the cards and match them to the graphs- (some graphs may have more than one label!)

Super Trig PowerPoint

Documents

Transcript of Super Trig PowerPoint