 New Zealand
Level 8 - NCEA Level 3

Apply sum and difference and double angle identities

Lesson

In a previous chapter it was shown how the expression $\cos3x$cos3x could be expanded into a form containing trigonometric functions of $x$x only. This can be done with any of the trigonometric functions of an integer multiple of the variable quantity.

There is a tidy way of obtaining these so-called multiple angle formulas through a connection between trigonometry and complex numbers, using a theorem discovered by the 18th-century French mathematician Abraham deMoivre. But, for the moment, we will use the trigonometric identities already discussed to achieve the required expansions.

We might perhaps wish to expand $\sin5\theta$sin5θ. A workable strategy would be to write this as $\sin\left(\theta+4\theta\right)$sin(θ+4θ). This expands to $\sin\theta\cos4\theta+\sin4\theta\cos\theta$sinθcos4θ+sin4θcosθ. Next, the functions of $4\theta$4θ could be expanded as functions of $\theta$θ and $3\theta$3θ. Continuing in this way, we reach after four steps an expression in $\theta$θ only. The iterative process has involved thinking of $5\theta$5θ as the sum $$A slightly quicker process, involving just three steps and quite a lot of tidying up, would be to decompose 5\theta5θ into the sum \theta+\left(2\theta+2\theta\right)θ+(2θ+2θ), as follows:  \sin5\thetasin5θ == \sin\theta\cos4\theta+\cos\theta\sin4\thetasinθcos4θ+cosθsin4θ == \sin\theta\left(\cos^22\theta-\sin^22\theta\right)+\cos\theta\left(2\sin2\theta\cos2\theta\right)sinθ(cos22θ−sin22θ)+cosθ(2sin2θcos2θ) == \sin\theta\left(\left(\cos^2\theta-\sin^2\theta\right)^2-\left(2\sin\theta\cos\theta\right)^2\right)+4\cos^2\theta\sin\theta\left(\cos^2\theta-\sin^2\theta\right)sinθ((cos2θ−sin2θ)2−(2sinθcosθ)2)+4cos2θsinθ(cos2θ−sin2θ) After collecting like terms, this can be written \sin5\theta=\sin\theta\left(5\cos^4\theta-10\sin^2\theta\cos^2\theta+\sin^4\theta\right)sin5θ=sinθ(5cos4θ10sin2θcos2θ+sin4θ) We could use the Pythagorean identity \cos^2\theta=1-\sin^2\thetacos2θ=1sin2θ to obtain from this an expression using just the sine function: \sin5\theta=\sin\theta\left(5-20\sin^2\theta+16\sin^4\theta\right)sin5θ=sinθ(520sin2θ+16sin4θ) It is interesting to observe that \sin5\thetasin5θ has been expressed as a quintic in \sin\theta.sinθ.We might imagine that it would be possible to rearrange this so that \sin\thetasinθ is expressed in terms of \sin5\thetasin5θ. Thus, we might write \sin\frac{\theta}{5}sinθ5 in terms of \sin\thetasinθ and so obtain some further 'exact values'. It turns out that this is not possible due to the fact, discovered by the Norwegian mathematician Niels Abel, that quintic equations are in general not solvable algebraically. Some more techniques and ideas that are useful in dealing with trigonometric functions are demonstrated in the following examples. Example 1 Odd and even functions. An even function possesses the property that its graph is unchanged on reflection in the vertical axis. An odd function possesses rotational symmetry through the angle \piπ about the origin. These characteristics are also expressed by observing that for an even function ff, f\left(-x\right)=f\left(x\right)f(x)=f(x), and for an odd function, f\left(-x\right)=-f\left(x\right)f(x)=f(x). Thus \tan\left(x\right)tan(x) is an odd function because \tan\left(-x\right)=\tan\left(2\pi-x\right)=\frac{\tan\left(2\pi\right)-\tan\left(x\right)}{1+\tan\left(2\pi\right)\tan\left(x\right)}=-\tan\left(x\right)tan(x)=tan(2πx)=tan(2π)tan(x)1+tan(2π)tan(x)=tan(x). Example 2 Express \csc\left(-\alpha\right)csc(α) as the cosecant of a positive argument. We write \csc\left(-\alpha\right)=\csc\left(2\pi-\alpha\right)csc(α)=csc(2πα). The argument will now be positive provided 0<\alpha<2\pi0<α<2π. Example 3 Establish as an identity: \frac{1}{\tan\theta+\cot\theta}=\frac{1}{2}\sin2\theta1tanθ+cotθ=12sin2θ. We manipulate the simplest looking side of the equation to show that it is identical to the other side. In this case the left side is  \frac{1}{\tan\theta+\cot\theta}1tanθ+cotθ​ ==$$ $=$=  $=$= $\sin\theta.\cos\theta$sinθ.cosθ $=$= $\frac{1}{2}\sin2\theta$12​sin2θ

and since the left hand side is the same as the right, we conclude that the statement is an identity.

More Worked Examples

QUESTION 1

Write $\tan3x$tan3x in terms of $\tan x$tanx.

QUESTION 2

Write $\cos4x$cos4x in terms of $\cos x$cosx.

QUESTION 3

Verify that $\frac{1-\tan^2\left(x\right)}{1+\tan^2\left(x\right)}=\cos2x$1tan2(x)1+tan2(x)=cos2x is an identity.

Outcomes

M8-6

Manipulate trigonometric expressions

91575

Apply trigonometric methods in solving problems