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| − | This unit looks at trigonometric formulae known as the double angle formulae. They are called
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| − | this because they involve trigonometric functions of double angles, i.e. sin 2A, cos 2A and tan2A.
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| − | In order to master the techniques explained here it is vital that you undertake the practice
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| − | exercises provided.
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| − | After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
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| − | • derive the double angle formulae from the addition formulae
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| − | • write the formula for cos 2A in alternative forms
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| − | • use the formulae to write trigonometric expressions in different forms
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| − | • use the formulae in the solution of trigonometric equations
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| − | Contents
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| − | 1. Introduction 2
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| − | 2. The double angle formulae for sin 2A, cos 2A and tan2A 2
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| − | 3. The formula cos2A = cos2 A − sin2 A 3
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| − | 4. Finding sin3x in terms of sin x 3
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| − | 5. Using the formulae to solve an equation 4
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| − | 1. Introduction
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| − | This unit looks at trigonometric formulae known as the double angle formulae. They are called
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| − | this because they involve trigonometric functions of double angles, i.e. sin 2A, cos 2A and tan2A.
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| − | 2. The double angle formulae for sin 2A, cos 2A and tan 2A
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| − | We start by recalling the addition formulae which have already been described in the unit of the
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| − | same name.
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| − | sin(A + B) = sin A cos B + cos A sin B
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| − | cos(A + B) = cos A cos B − sin A sin B
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| − | tan(A + B) = tan A + tan B
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| − | 1 − tan A tan B
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| − | We consider what happens if we let B equal to A. Then the first of these formulae becomes:
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| − | sin(A + A) = sin A cos A + cos A sin A
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| − | so that
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| − | sin 2A = 2 sin A cos A
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| − | This is our first double-angle formula, so called because we are doubling the angle (as in 2A).
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| − | Similarly, if we put B equal to A in the second addition formula we have
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| − | cos(A + A) = cos A cos A − sin A sin A
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| − | so that
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| − | cos 2A = cos2 A − sin2 A
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| − | and this is our second double angle formula.
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| − | Similarly
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| − | tan(A + A) = tan A + tan A
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| − | 1 − tan A tan A
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| − | so that
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| − | tan 2A =
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| − | 2 tan A
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| − | 1 − tan2 A
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| − | These three double angle formulae should be learnt.
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| − | 3. The formula cos 2A = cos2 A − sin2 A
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| − | We now examine this formula more closely.
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| − | We know from an important trigonometric identity that
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| − | cos2 A + sin2 A = 1
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| − | so that by rearrangement
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| − | sin2 A = 1 − cos2 A.
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| − | So using this result we can replace the term sin2 A in the double angle formula. This gives
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| − | cos 2A = cos2 A − sin2 A
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| − | = cos2 A − (1 − cos2 A)
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| − | = 2 cos2 A − 1
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| − | This is another double angle formula for cos 2A.
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| − | Alternatively we could replace the term cos2 A by 1 − sin2 A which gives rise to:
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| − | cos 2A = cos2 A − sin2 A
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| − | = (1 − sin2 A) − sin2 A
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| − | = 1 − 2 sin2 A
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| − | which is yet a third form.
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| − | 4. Finding sin 3x in terms of sin x
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| − | Example
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| − | Consider the expression sin 3x. We will use the addition formulae and double angle formulae to
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| − | write this in a different form using only terms involving sin x and its powers.
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| − | We begin by thinking of 3x as 2x + x and then using an addition formula:
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| − | sin 3x = sin(2x + x)
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| − | = sin 2x cos x + cos 2x sin x using the first addition formula
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| − | = (2 sin x cos x) cos x + (1 − 2 sin2 x) sin x using the double angle formula
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| − | cos 2x = 1 − 2 sin2 x
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| − | = 2 sin x cos2 x + sin x − 2 sin3 x
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| − | = 2 sin x(1 − sin2 x) + sin x − 2 sin3 x from the identity cos2 x + sin2 x = 1
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| − | = 2 sin x − 2 sin3 x + sin x − 2 sin3 x
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| − | = 3 sin x − 4 sin3 x
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| − | We have derived another identity
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| − | sin 3x = 3 sin x − 4 sin3 x
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| − | Note that by using these formulae we have written sin 3x in terms of sin x (and its powers). You
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| − | could carry out a similar exercise to write cos 3x in terms of cos x.
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| − | 5. Using the formulae to solve an equation
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| − | Example
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| − | Suppose we wish to solve the equation cos 2x = sin x, for values of x in the interval −π ≤ x < π.
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| − | We would like to try to write this equation so that it involves just one trigonometric function, in
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| − | this case sin x. To do this we will use the double angle formula
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| − | cos 2x = 1 − 2 sin2 x
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| − | The given equation becomes
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| − | 1 − 2 sin2
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| − | x = sin x
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| − | which can be rewritten as
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| − | 0 = 2 sin2 x + sin x − 1
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| − | This is a quadratic equation in the variable sin x. It factorises as follows:
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| − | 0 = (2 sin x − 1)(sin x + 1)
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| − | It follows that one or both of these brackets must be zero:
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| − | 2 sin x − 1 = 0 or sin x + 1 = 0
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| − | so that
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| − | sin x =
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| − | 1
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| − | 2
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| − | or sin x = −1
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