© Christine Crisp

Trig for M2Trig for M2

We sometimes find it useful to remember the trig. ratios for the angles 6045,30 and

These are easy to find using triangles.

In order to use the basic trig. ratios we need right angled triangles which also contain the required angles.

Consider an equilateral triangle.

Divide the triangle into 2 equal right angled triangles.

60 60

60 Trig ratios don’t depend on the size of the triangle, so we can let the sides be any convenient length.

22

2

( You’ll see why 2 is useful in a minute ).

Consider an equilateral triangle.

Divide the triangle into 2 equal right angled triangles.

60 60

30Trig ratios don’t depend on the size of the triangle, so we can let the sides be any convenient length.

1

22

2

We now consider just one of the triangles.

( You’ll see why 2 is useful in a minute ).

Consider an equilateral triangle.

Divide the triangle into 2 equal right angled triangles.

60

30Trig ratios don’t depend on the size of the triangle, so we can let the sides be any convenient length.

1

2

We now consider just one of the triangles.

( You’ll see why 2 is useful in a minute ).

1

2

From the triangle, we can now write down the trig ratios for 6030 and

Pythagoras’ theorem gives the 3rd side.3

60sin2

3 60cos 60tan2

1 3

30sin 30cos 30tan2

32

1

3

1

312 22

( Choosing 2 for the original side means we don’t have a fraction for the 2nd side )

60

30

1

23

60

30Memory Aid

1, 2, 3 not 3 but 3

600is opposite the 3 as 3 is larger than 1

1

1

45

45

211 22

2

45cos45sin 45tan2

11

For we again need a right angled triangle.

45

By making the triangle isosceles, there are 2 angles each of .45We let the equal sides have length 1.Using Pythagoras’ theorem, the 3rd side is

From the triangle, we can now write down the trig ratios for 45

Memory Aid

1, 1, 2

2 is the longest side

45

452

1

1

2

1 45cos45sin 45tan 1

SUMMARY

60sin2

3 60cos 60tan2

13

30sin 30cos 30tan2

32

1

3

1

The trig. ratios for are: 6045,30 and

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

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.

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

but we need another 3.

Tip: I remember which is which by noticing that:• cot and tan are the only ones with t in

them

These are the 3 reciprocal ratios:

in

1osec c

s

os

1ec

cs t

t

an

1co

We’ve been using 3 trig ratios: ,sin tanand cos

t

but we need another 3.These are the 3 reciprocal ratios: 1

co ecθ in

1

os

se

Tip: Remember which is which by noticing that:• cot and tan are the only ones with t in

them

scs

ct

an

1co

We’ve been using 3 trig ratios: ,sin tanand cos

• cosec and sec go with sin and cos respectively, 3rd letters match up.

Also, since ,

cos

sintan we

get

sin

coscot

sin

1cosec

cos

1sec

SUMMARY

tan

1cot

sin

coscot

cos

sintan

Also,

The 3 reciprocal ratios are:

There are 2 identities involving the reciprocal ratios which we will prove.

We start with the identity we met in AS

1sincos 22

Dividing by :

2cos

But,

cos

sin tan and cos

1 sec

2

2

cos

cos

2cos

1

2

2

cos

sin

So,

There are 2 identities involving the reciprocal ratios which we will prove.

We start with the identity we met in AS

1sincos 22

Dividing by :

2cos2cos

1

2

2

cos

sin

2

2

cos

cos

But,

cos

sin tan and cos

1 sec

1So,

There are 2 identities involving the reciprocal ratios which we will prove.

We start with the identity we met in AS

1sincos 22

Dividing by :

2cos

1So, 2tan

2

2

cos

sin

2

2

cos

cos

2cos

1

But,

cos

sin tan and cos

1 sec

There are 2 identities involving the reciprocal ratios which we will prove.

We start with the identity we met in AS

1sincos 22

Dividing by :

2cos

1So, 2tan 2sec

2cos

1

2

2

cos

cos

2

2

cos

sin

But,

cos

sin tan and cos

1 sec

There are 2 identities involving the reciprocal ratios which we will prove.

We start with the identity we met in AS

1sincos 22

Dividing by :

2cos

1So, 2tan 2sec

2cos

1

2

2

cos

cos

2

2

cos

sin

But,

cos

sin tan and cos

1 sec

Exercise

Starting with find an identity linking and

1sincos 22 cosec cot

Solution:

Dividing by :

2sin

22

2

2

2

sin

1

sin

sin

sin

cos

But,

sin

cos cot and sin

1 cosec

1So, 2cot 2cosec

1sincos 22

SUMMARY

221 coseccot

221 sectan

There are 3 quadratic trig identities:

122 sincos

Never try to square root these identities.

If the LHS has a Co so does the RHS

SUMMARY

221 coseccot

221 sectan

There are 3 quadratic trig identities:

122 sincos

Memory aidLHS is either tan2 or cot2RHS is either sec2 or cosec2

The trig identities are used in 2 ways:

• to solve some quadratic trig equations

• to prove other identities

BABABA sincoscossin)sin( )1(

The double angle formulae are used to express an angle such as 2A in terms of A.

We derive the formulae from 3 of the addition formulae.

What do we need to do to obtain formulae for sin 2A

ANS: Replace B by A in (1), (3) and (5).

BABABA sincoscossin)sin( So,

AAAAA sincoscossin2sin )1(AAA cossin22sin