Trig for M2 © Christine Crisp

We sometimes find it useful to remember the trig. ratios for the angles 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.

2 Consider an equilateral triangle. Trig ratios don’t depend on the size of the triangle, so we can let the sides be any convenient length. 2 ( You’ll see why 2 is useful in a minute ). Divide the triangle into 2 equal right angled triangles.

2 1 Consider an equilateral triangle. Trig ratios don’t depend on the size of the triangle, so we can let the sides be any convenient length. 2 ( You’ll see why 2 is useful in a minute ). 1 Divide the triangle into 2 equal right angled triangles. We now consider just one of the triangles.

2 1 Consider an equilateral triangle. Trig ratios don’t depend on the size of the triangle, so we can let the sides be any convenient length. 2 ( You’ll see why 2 is useful in a minute ). 1 Divide the triangle into 2 equal right angled triangles. We now consider just one of the triangles.

( Choosing 2 for the original side means we don’t have a fraction for the 2nd side ) Pythagoras’ theorem gives the 3rd side. 2 From the triangle, we can now write down the trig ratios for 1

600is opposite the 3 as 3 is larger than 1 Memory Aid 1, 2, 3 not 3 but 3 600is opposite the 3 as 3 is larger than 1 2 1

1 For we again need a right angled triangle. By making the triangle isosceles, there are 2 angles each of . 1 We 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

Memory Aid 1, 1, 2 2 is the longest side 1 1

SUMMARY The trig. ratios for are:

c a b A B C Proof of the Pythagorean Identity. Consider the right angled triangle ABC. Using Pythagoras’ theorem: Divide by : But and

We have shown that this formula holds for any angle in a right angled triangle. However, because of the symmetries of and , it actually holds for any value of . A formula like this which is true for any value of the variable is called an identity. Identity symbol Identity symbols are normally only used when we want to stress that we have an identity. In the trig equations we use an = sign.

c a b A B C A 2nd Trig Identity Consider the right angled triangle ABC. Also, So,

c s c s t We’ve been using 3 trig ratios: but we need another 3. These are the 3 reciprocal ratios: c s c s t Tip: I remember which is which by noticing that: cot and tan are the only ones with t in them

s s c t c t We’ve been using 3 trig ratios: but we need another 3. These are the 3 reciprocal ratios: s c t s c t Tip: Remember which is which by noticing that: cot and tan are the only ones with t in them cosec and sec go with sin and cos respectively, 3rd letters match up. Also, since , we get

SUMMARY The 3 reciprocal ratios are: Also,

There are 2 identities involving the reciprocal ratios which we will prove. We start with the identity we met in AS Dividing by : But, and So,

There are 2 identities involving the reciprocal ratios which we will prove. We start with the identity we met in AS Dividing by : But, and So,

There are 2 identities involving the reciprocal ratios which we will prove. We start with the identity we met in AS Dividing by : But, and So,

There are 2 identities involving the reciprocal ratios which we will prove. We start with the identity we met in AS Dividing by : But, and So,

There are 2 identities involving the reciprocal ratios which we will prove. We start with the identity we met in AS Dividing by : But, and So,

Exercise Starting with find an identity linking and Solution: Dividing by : But, and So,

Never try to square root these identities. SUMMARY There are 3 quadratic trig identities: Never try to square root these identities.

LHS is either tan2q or cot2q SUMMARY There are 3 quadratic trig identities: Memory aid LHS is either tan2q or cot2q RHS is either sec2q or cosec2q If the LHS has a Co so does the RHS

The trig identities are used in 2 ways: to solve some quadratic trig equations to prove other identities

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).

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