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© Christine Crisp Trig for M2

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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.

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

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

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Consider an equilateral triangle. Divide the triangle into 2 equal right angled triangles. Trig 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 ).

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

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

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1 1 For we again need a right angled triangle. By making the triangle isosceles, there are 2 angles each of. We let the equal sides have length 1. Using Pythagoras’ theorem, the 3 rd side is From the triangle, we can now write down the trig ratios for

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

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SUMMARYThe trig. ratios for are:

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

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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. We have shown that this formula holds for any angle in a right angled triangle. 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.

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

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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: c s c s t t We’ve been using 3 trig ratios:

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

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SUMMARY Also, The 3 reciprocal ratios are:

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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,

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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,

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

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

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

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

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

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If the LHS has a Co so does the RHS SUMMARY There are 3 quadratic trig identities: Memory aid LHS is either tan 2 or cot 2 RHS is either sec 2 or cosec 2

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

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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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So,

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