The double angle formulae

1 The double angleformulaemc-TY- doubleangle -2009-1 This unit looks at trigonometric formulae known as thedouble angle formulae . They are calledthis because they involve trigonometric functions of double angles, 2A,cos 2 Aandtan order to master the techniques explained here it is vital that you undertake the practiceexercises reading this text, and/or viewing the video tutorial on this topic, you should be able to: derive the double angle formulae from the addition formulae write the formula forcos 2 Ain alternative forms use the formulae to write trigonometric expressions in different forms use the formulae in the solution of trigonometric double angle formulae forsin 2A,cos 2 Aandtan formulacos 2A= cos2A 3xin terms the formulae to solve an mathcentre 20091. IntroductionThis unit looks at trigonometric formulae known as thedouble angle formulae . They are calledthis because they involve trigonometric functions of double angles, 2A,cos 2 Aandtan The double angle formulae forsin 2A,cos 2 Aandtan 2 AWe start by recalling the addition formulae which have already been described in the unit of thesame (A+B) = sinAcosB+ cosAsinBcos(A+B) = cosAcosB sinAsinBtan(A+B) =tanA+ tanB1 tanAtanBWe consider what happens if we letBequal toA.

2 Then the first of these formulae becomes:sin(A+A) = sinAcosA+ cosAsinAso thatsin 2A= 2 sinAcosAThis is our firstdouble- angle formula, so called because we are doubling the angle (as in 2A).Similarly, if we putBequal toAin the second addition formula we havecos(A+A) = cosAcosA sinAsinAso thatcos 2A= cos2A sin2 Aand this is our second double angle (A+A) =tanA+ tanA1 tanAtanAso thattan 2A=2 tanA1 tan2 AThese three double angle formulae should be Pointsin 2A= 2 sinAcosAcos 2A= cos2A sin2 Atan 2A=2 tanA1 mathcentre 20093. The formulacos 2A= cos2A sin2 AWe now examine this formula more know from an important trigonometric identity thatcos2A+ sin2A= 1so that by rearrangementsin2A= 1 using this result we can replace the termsin2 Ain the double angle formula. This givescos 2A= cos2A sin2A= cos2A (1 cos2A)= 2 cos2A 1 This is another double angle formula forcos we could replace the termcos2 Aby1 sin2 Awhich gives rise to:cos 2A= cos2A sin2A= (1 sin2A) sin2A= 1 2 sin2 Awhich is yet a third Pointcos 2A= cos2A sin2A= 2 cos2A 1= 1 2 sin2A4.

3 Findingsin 3xin terms ofsinxExampleConsider the expressionsin 3x. We will use the addition formulae and double angle formulaetowrite this in a different form using only terms involvingsinxand its begin by thinking of3xas2x+xand then using an addition mathcentre 2009sin 3x=sin(2x+x)=sin 2xcosx+ cos 2xsinxusing the first addition formula=(2 sinxcosx) cosx+ (1 2 sin2x) sinxusing the double angle formulacos 2x= 1 2 sin2x=2 sinxcos2x+ sinx 2 sin3x=2 sinx(1 sin2x) + sinx 2 sin3xfrom the identitycos2x+ sin2x= 1=2 sinx 2 sin3x+ sinx 2 sin3x=3 sinx 4 sin3xWe have derived another identitysin 3x= 3 sinx 4 sin3xNote that by using these formulae we have writtensin 3xin terms ofsinx(and its powers). Youcould carry out a similar exercise to writecos 3xin terms Using the formulae to solve an equationExampleSuppose we wish to solve the equationcos 2x= sinx, for values ofxin the interval x < .We would like to try to write this equation so that it involvesjust one trigonometric function, inthis casesinx.

4 To do this we will use the double angle formulacos 2x= 1 2 sin2xThe given equation becomes1 2 sin2x= sinxwhich can be rewritten as0 = 2 sin2x+ sinx 1 This is a quadratic equation in the variablesinx. It factorises as follows:0 = (2 sinx 1)(sinx+ 1)It follows that one or both of these brackets must be zero:2 sinx 1 = 0orsinx+ 1 = 0so thatsinx=12orsinx= 1We can solve these two equations by referring to the graph ofsinxover the interval x < which is shown in Figure 1. 1 1sinxx 2 212 6 65 -- -Figure 1. A graph ofsinxover the interval x < . mathcentre 2009 From the graph we see that the angle whose sine is 1is 2. The angle whose sine is12is astandard result, namely 6, or30 . Using the graph, and making use of symmetry we note thereis another solution atx=5 6. So, in summary, the solutions arex= 6,5 6and 2 ExampleSuppose we wish to solve the equationsin 2x= sinx x < In this case we will use the double angle formulaesin 2x= 2 gives2 sinxcosx= sinxWe rearrange this and factorise as follows:2 sinxcosx sinx= 0sinx(2 cosx 1) = 0from whichsinx= 0or2 cosx 1 = 0We have reduced the given equation to two simpler deal first withsinx= 0.

5 Byreferring to the graph ofsinxin Figure 1 we see that the two required solutions arex= andx= 0. The potential solution atx= is excluded because it is outside the interval specified inthe original equation2 cosx 1 = 0givescosx=12. The angle whose cosine is12is60 or 3, anotherstandard result. By referring to the graph ofcosxshown in Figure 2 we deduce that the solutionsarex= 3andx= 3. 1 1cosxx 3 3-- - Figure 2. A graph ofcosxover the interval x < .Exercises1. Verify the three double angle formulae (forsin 2A,cos 2A,tan 2A) for the casesA= 30oandA= By writingcos(3x) = cos(2x+x)determine a formula forcos(3x)in terms mathcentre 20093. Determine a formula forcos(4x)in terms Solve the equationsin 2x= cosxfor x < .5. Solve the equationcos 2x= cosxfor0 x < cos3x 3 cos4x 8 cos2x+ 14. 2, 6, 2,5 65. 0 and2 mathcentre 2009