Twitter: @Owen134866 · acute angle Cos(-120)º = -Cos 60º Draw a sketch of the graph Mark on -120º Using the fact that the graph has symmetry, find an acute value of θwhich has

Twitter: @Owen134866

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Prior Knowledge Check

1a) Sketch the graph of 𝑦 = 𝑠𝑖𝑛𝑥for 0 ≤ 𝑥 ≤ 540°

b) How many solutions are there to the equation 𝑠𝑖𝑛𝑥 = 0.6 in the range 0 ≤ 𝑥 ≤ 540°?

c) Given that 𝑠𝑖𝑛−1 0.6 = 36. 9° (to 3sf), write down three other solutions to the equation 𝑠𝑖𝑛𝑥 = 0.6

2) Solve the equation 𝑠𝑖𝑛𝑥 = −0.7

3) Solve these equations

a) 𝑥2 − 4𝑥 + 3 = 0

b) 𝑥2 + 8𝑥 − 9 = 0

c) 2𝑥2 − 3𝑥 − 7 = 0

4

143.1, 396.9, 503.1

-44.4

1, 3

1, -9

3 ± 65

4

Trig Identities and Equations

You need to understand how to use the trig graphs to find other

values of sine, cosine or tan

10A

You need to be able to work out larger values of sin, cos and tan as

acute angles (0º - 90º)

Write sin 130º as sine of an acute angle

(sometimes asked as a ‘trigonometric ratio’)

Sin 130º = Sin 50º

y = sinθ

1

90º 180º 270º 360º

y

θ0

-1

13050

-40-40

Draw a sketch of the graph

Mark on 130º

Using the fact that the graph has symmetry, find an acute value of θ which has the same value as sin 130


You need to understand how to use the trig graphs to find other

values of sine, cosine or tan

10A

+30

+30You need to be able to work out larger values of sin, cos and tan as


Write cos (-120)º as cos of an acute angle

Cos(-120)º = -Cos 60º Draw a sketch of the graph

Mark on -120º

Using the fact that the graph has symmetry, find an acute value of θ which has the same numerical value as cos (-120)

y = cosθ

-270º

y

1

0

-1

θ90º 180º 270º-180º

-120

-60 60

-90º

+60 +60

The value you find here will have the same digits in it, but will be multiplied by -1

Graphs of Trigonometric Functions

You need to be able to recognise the graphs of sinθ, cosθ and

tanθ

You need to be able to work out larger values of sin, cos and tan as


Write tan190 as tan of an acute angle

Tan190 = Tan10

10A

Draw a sketch of the graph

Mark on 190

Using the fact that the graph has symmetry, find an acute value of θ which has the same numerical value as tan190

y = tanθ

1

0

-1

θ

36027018090

19010

10+10+

You need to be able to find the exact values of some Trigonometrical Ratios

Some values of Sin, Cos or Tan can be written using fractions, surds, or combinations of both…

We can use an Equilateral Triangle with sides of length 2 to show this.

Using Pythagoras, the missing side in the right angled triangle is √3 (Square root of 22-12)

60˚ 60˚

60˚

2

2 2

2

1

√3

60˚

30˚

OppHyp

Opp

Hyp

Sinθ =12Sin30 =

√32Sin60 =

Opp


10B





60˚ 60˚

60˚

2

2 2

2

1

√3

60˚

30˚

AdjHyp

Adj

Hyp

Cosθ =√32Cos30 =

12Cos60 =

Adj


10B





60˚ 60˚

60˚

2

2 2

2

1

√3

60˚

30˚

OppAdj

AdjOpp

Tanθ =1

√3Tan30 =

√3Tan60 =

OppAdj√33

=


10B



We can also do a similar demonstration with a right-angled Isosceles triangle, with the equal sides being of length 1 unit.

Using Pythagoras’ Theorem, the hypotenuse will be of length √2 (Square root of 12 + 12)

45˚

1

1

√2

OppHypSinθ =

1√2Sin45 =

Opp

Hyp

√22

=


10B





45˚

1

1

√2

AdjHypCosθ =

1√2Cos45 =

Adj

Hyp

√22

=


10B





45˚

1

1

√2

OppAdjTanθ =

11Tan45 =

Opp

Adj

1=


10B


You can use sin, cos and tan along with Pythagoras’ Theorem to find

two useful identities.

To the right is a ‘unit circle’, a circle with a radius of 1 and a

centre at (0,0)

We can draw a right-angled triangle and label sides using GCSE

trigonometry

10C

1Sinθ

𝜃

Cosθ

H

O

A

OT A

AC H

OS H

𝑂 = 𝐻𝑆𝑖𝑛𝜃 𝐴 = 𝐻𝐶𝑜𝑠𝜃

1

1

-1

-1




Given that:

We can see from the triangle in the unit circle that:

10C

1Sinθ

𝜃

Cosθ

OT A

AC H

OS H

1

1

-1

-1

𝑇𝑎𝑛𝜃 =𝑂

𝐴

H

O

A

𝑇𝑎𝑛𝜃 =𝑆𝑖𝑛𝜃

𝐶𝑜𝑠𝜃




Given that:

We can see from the triangle in the unit circle that:

10C

1Sinθ

𝜃

Cosθ

OT A

AC H

OS H

1

1

-1

-1

H

O

A

𝑇𝑎𝑛𝜃 ≡𝑆𝑖𝑛𝜃

𝐶𝑜𝑠𝜃

𝑎2 + 𝑏2 = 𝑐2

(𝑆𝑖𝑛𝜃)2+(𝐶𝑜𝑠𝜃)2= 12

𝑆𝑖𝑛2𝜃 + 𝐶𝑜𝑠2𝜃 = 1




Simplify the following expression:

10C


𝐶𝑜𝑠𝜃𝑆𝑖𝑛2𝜃 + 𝐶𝑜𝑠2𝜃 ≡ 1

𝑠𝑖𝑛23𝜃 + 𝑐𝑜𝑠23𝜃

𝑠𝑖𝑛23𝜃 + 𝑐𝑜𝑠23𝜃

= 1

Use the identity above (it will always work as long as the

amounts of 𝜃 are the same)





10C



5 − 5𝑠𝑖𝑛2𝜃

5 − 5𝑠𝑖𝑛2𝜃

= 5(1 − 𝑠𝑖𝑛2𝜃)Factorise

= 5𝑐𝑜𝑠2𝜃

𝑆𝑖𝑛2𝜃 + 𝐶𝑜𝑠2𝜃 ≡ 1

𝑆𝑖𝑛2𝜃 ≡ 1 − 𝐶𝑜𝑠2𝜃 𝐶𝑜𝑠2𝜃 ≡ 1 − 𝑆𝑖𝑛2𝜃

Subtract 𝐶𝑜𝑠2𝜃 Subtract 𝑆𝑖𝑛2𝜃

You can use a rearranged version of the identity above

to replace the bracket





10C



𝑠𝑖𝑛2𝜃

1 − 𝑠𝑖𝑛22𝜃

𝑠𝑖𝑛2𝜃

1 − 𝑠𝑖𝑛22𝜃

=𝑠𝑖𝑛2𝜃

𝑐𝑜𝑠22𝜃

=𝑠𝑖𝑛2𝜃

𝑐𝑜𝑠2𝜃

= 𝑡𝑎𝑛2𝜃

Replace the rooted part as in the previous

example

You can work out the square root

This is equal to the other identity!




Prove that:

10C



𝑐𝑜𝑠4𝜃 − 𝑠𝑖𝑛4𝜃

𝑐𝑜𝑠2𝜃≡ 1 − 𝑡𝑎𝑛2𝜃

𝑐𝑜𝑠4𝜃 − 𝑠𝑖𝑛4𝜃

𝑐𝑜𝑠2𝜃

You should choose one side to begin with…

=(𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃)(𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛2𝜃)

𝑐𝑜𝑠2𝜃

=(𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛2𝜃)

𝑐𝑜𝑠2𝜃

=𝑐𝑜𝑠2𝜃

𝑐𝑜𝑠2𝜃−𝑠𝑖𝑛2𝜃

𝑐𝑜𝑠2𝜃

= 1 − 𝑡𝑎𝑛2𝜃

You can factorise the top (similar to the

difference of 2 squares)

The left of the brackets is equal to 1

You can write this as 2 separate fractions

These fractions can then be simplified

You need to be able to use the Trigonometrical identities

90 180 270 360y = Sinθ

y = Cosθ

y = Tanθ


10C



Look at the 3 graphs. What can we say about the values for each when the angle is:

a) Acute?b) Obtuse?c) Reflex?


You also need to be able to work out exact values of Sinθ, Cosθ or Tanθ, having been given one of

the others.

You will also need to use whether θ is Acute, Obtuse, or Reflex…

Example Question

Given that Cosθ is -3/5 and θ is reflex, find the value of Sinθ

θ

OT A

AC H

OS H

ACos

H

3

5Cos

5

3

4

You were effectively told A and H in the question. IGNORE the negative for now…

The other side should be worked out using Pythagoras’ Theorem…

90 180 270 360y = Sinθ

y = Cosθ

y = Tanθ

OSin

H

4

5Sin

4

5Sin

Put in the values from the Triangle

Consider the region on the

diagram

Draw a Right Angled

Triangle


10C



90


You also need to be able to work out exact values of Sinθ, Cosθ or Tanθ, having been given one of

the others.

You will also need to use whether θ is Acute, Obtuse, or Reflex…

Example Question

Given that Sinθ is 2/5 and θ is obtuse, find the value of Cosθ

θ

OT A

AC H

OS H

OSin

H

2

5Sin

52

√21

You were effectively told O and H in the question.

The other side should be worked out using Pythagoras’ Theorem…

180 270 360y = Sinθ

y = Cosθ

y = Tanθ

ACos

H

21

5Cos

21

5Cos

Put in the values from the Triangle

Consider the region on the

diagram

Draw a Right Angled

Triangle


10C




Given that 𝑝 = 3𝑐𝑜𝑠𝜃 and that 𝑞 = 2𝑠𝑖𝑛𝜃, show that

4𝑝2 + 9𝑞2 = 36


10C



𝑝 = 3𝑐𝑜𝑠𝜃 𝑞 = 2𝑠𝑖𝑛𝜃

𝑝

3= 𝑐𝑜𝑠𝜃

𝑞

2= 𝑠𝑖𝑛𝜃

Divide by 3

Divide by 2

𝑆𝑖𝑛2𝜃 + 𝐶𝑜𝑠2𝜃 = 1

𝑞

2

2

+𝑝

3

2

= 1

𝑞2

4+𝑝2

9= 1

9𝑞2 + 4𝑞2 = 36

Replace sin and cos using the above relationships

Square each fraction

Multiply by 36

You will need to start with one of the identities, and replace the cos

and sin parts!


You need to be able to solve trigonometrical equations of the

form 𝒔𝒊𝒏𝜽 = 𝒌 and 𝒄𝒐𝒔𝜽 = 𝒌(where −𝟏 ≤ 𝒌 ≤ 𝟏) and 𝒕𝒂𝒏𝜽 = 𝒑

(where 𝒑 ∈ ℝ)

This will involve finding an answer using your calculator, and then

checking whether there are other possible solutions in the given range

It is very important that you check the range you are given for

θ!!!!

10D

Solve the equation 0.5Sin in the interval 0 360

0.5Sin

10.5Sin

30

Use Sin-1

This will give you one answer

90 180 270 360

30 150

0.5

30 or 150

y = Sinθ

𝑠𝑖𝑛𝜃 = 𝑠𝑖𝑛(180 − 𝜃) 𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑠(360 − 𝜃) 𝑡𝑎𝑛𝜃 = 𝑡𝑎𝑛(𝜃 ± 180)








θ!!!!

10D


Solve the equation 5 2Sin in the interval 0 360

5 2Sin Divide by 5

Use Sin-1

90 180 270 360

336.4

-0.4

203.6 or 336.4

y = Sinθ

0.4Sin

23.6 Not within the range. You can add 360° to obtain an equivalent

value336.4

203.6








θ!!!!

10D


Solve the equation 2Sin Cos in the interval 0 360

2Sin Cos Divide by

Cosθ

Use Trig Identities

90 180 270 360

243.4

2

63.4 or 243.4

y = Tanθ

Use Tan-1

63.4

2Sin

Cos

2Tan

63.4


You need to be able to solve equations of the form 𝒔𝒊𝒏 𝒏𝜽 = 𝒌,

𝐜𝐨𝐬 𝒏𝜽 = 𝒌 and 𝒕𝒂𝒏 𝒏𝜽 = 𝒑.

10E

Example Question

Solve the equation 2 1Cos in the interval 0 360

0 360

0 2 720

2 1Cos

2 180

90

180

y = Cosθ

-1270 360180

2 180 , 540 , 900

90 , 270

1) Work out the acceptable interval for 2θ

2) Work out one possible answer as before. Find all values in the

standard 0 – 360 range

3) Add/Subtract 360 to these values until you have all the answers within the 2θ range

4) These answers are for 2θ. Undo them to find values for θ itself

Multiply by 2

Solve using Cos-1

Adding 360 to the value we worked

out (staying within the range)Divide

by 2


You need to be able to solve equations of the form 𝒔𝒊𝒏 𝒏𝜽 = 𝒌,

𝐜𝐨𝐬 𝒏𝜽 = 𝒌 and 𝒕𝒂𝒏 𝒏𝜽 = 𝒑.

10E

Example Question

Solve the equation Sin(2 35) 1 in the interval -180 180

180 180

395 2 35 325

(2 35) 1Sin

(2 35) 270

90

270

y = Sinθ

-1270 360180

2 35 270 , 630, -90

27.5 , 152.5

1) Work out the acceptable interval for (2θ – 35)

2) Work out one possible answer as before. Find all values in the

standard 0 – 360 range

3) Add/Subtract 360 to these values until you have all the

answers within the (2θ - 35) range

4) These answers are for (2θ – 35). Undo this to find values for θ

itself

Multiply by 2.

Subtract 35

Solve using Sin-1

Adding/Subtracting 360 to the value we worked out (staying within the range)

Add 35, Divide by 2

, -450

2 35 90 , 270


You need to be able to solve quadratic equations given in sin,

cos or tan.

10F

Solve the following equation

2 3 2 0x x

( 2)( 1) 0x x

2 or 1x x

Factorise

Work out what value would make either bracket 0

Solve the following Equation

2 3 2 0Sin Sin

( 2)( 1) 0Sin Sin

2 or 1Sin Sin

Factorise


90

90

y = Sinθ

1

270 360180

2

Sinθ = 2 has no solutions

Sinθ = 1 has 1 solution

90





cos or tan.

10F

Solve the following equation

22 1 0Cos Cos

(2 1)( 1) 0Cos Cos

0.5 or Cos 1Cos

Factorise


90

0

y = Cosθ-0.5 270 360180

1 Cosθ = 1 has 2 solutions

Cosθ = -0.5 has 2 solutions

0 , 120 , 240 , 360

360

120 240





cos or tan.

10F

Solve the following equation in the range 0 ≤ θ ≤ 360

2 12

( 30)Sin Square root both sides. On fractions root top

and bottom separately. Can be positive or

negative.

90

1/√2y = Sinθ

270 360180

45

( 30) 45 , 135 , 225 , 315

1

2( 30)Sin

0 360

30 30 330

Work out the acceptable

range. Subtract 30

1

2( 30)Sin

( 30) 45 -1/√2

135

225 315

1

2( 30)Sin

( 30) 45

( 30) 315

75 , 165 , 255 , 345 360 added to get a value in

the range





cos or tan.

10F

Solve the following equation in the range -180 ≤ x ≤ 180

2𝑐𝑜𝑠2𝑥 + 9𝑠𝑖𝑛𝑥 = 3𝑠𝑖𝑛2𝑥



2(1 − 𝑠𝑖𝑛2𝑥) + 9𝑠𝑖𝑛𝑥 = 3𝑠𝑖𝑛2𝑥

2 − 2𝑠𝑖𝑛2𝑥 + 9𝑠𝑖𝑛𝑥 = 3𝑠𝑖𝑛2𝑥

5𝑠𝑖𝑛2𝑥 − 9𝑠𝑖𝑛𝑥 − 2 = 0

(5𝑠𝑖𝑛𝑥 + 1)(𝑠𝑖𝑛𝑥 − 2) = 0

𝑠𝑖𝑛𝑥 = −1

5

90-1/5 180

-11.5-90-180

-168.5

𝑥 = −11.5 , 168.5

Replace cosx using the identity above

Multiply the bracket

Rearrange

Factorise

Solve (sinx = 2 has no solutions)

You can use a sketch to find other values in

the given range

Documents

Twitter: @Owen134866 · acute angle Cos(-120)º = -Cos 60º Draw a sketch of the graph Mark on -120º Using the fact that the graph has symmetry, find an acute value of θwhich has