Expand expressions using double and half angle identities

Where the argument of a trigonometric function is a multiple of some variable, say $2x$2x or $3x$3x or $\frac{x}{2}$x2​, we may wish to express the function in terms of the variable $x$x itself. In other situations, we may wish to work in the opposite direction write a complicated expression more simply in terms of a multiple argument.

Example 1

The expression $\cos3x$cos3x can be expressed in terms of $\cos x$cosx.

One way to do this is to think of the argument $3x$3x in the form $2x+x$2x+x. Then, $\cos3x=\cos\left(2x+x\right)$cos3x=cos(2x+x)$=\cos2x\cos x-\sin2x\sin x$=cos2xcosx−sin2xsinx

We further expand $\cos2x$cos2x and $\sin2x$sin2x to obtain

$\cos3x=\left(\cos^2x-\sin^2x\right)\cos x-2\sin^2x\cos x$cos3x=(cos2x−sin2x)cosx−2sin2xcosx

Because the required end result is to be in terms of the cosine function, we must replace $\sin^2x$sin2x with $1-\cos^2x$1−cos2x.

Then,

$\cos3x=\left(2\cos^2x-1\right)\cos x-2\left(1-\cos^2x\right)\cos x$cos3x=(2cos2x−1)cosx−2(1−cos2x)cosx and finally, on collecting like terms, we find $\cos3x\equiv4\cos^3x-3\cos x$cos3x≡4cos3x−3cosx

 

Example 2

The expression $\sin^4x-\cos^4x$sin4x−cos4x can be written more simply in terms of a multiple angle.

The expression is the difference of two squares.

It can be factorised to $\left(\sin^2x+\cos^2x\right)\left(\sin^2x-\cos^2x\right)$(sin2x+cos2x)(sin2x−cos2x).

The first bracket is just $1$1, and the second bracket is the negative of the expansion of $\cos2x$cos2x.

So, we have $\sin^4x-\cos^4x\equiv-\cos2x$sin4x−cos4x≡−cos2x

 

Example 3

It is possible to deduce exact values for the trigonometric functions of angles that are multiples of $\frac{\pi}{10}$π10​ or $18^\circ$18°. Such angles are related to angles found in the regular pentagon.

Suppose we look for an angle $\theta$θ such that $\cos3\theta=\sin2\theta$cos3θ=sin2θ. We found in Example 1 above, that $\cos3\theta=4\cos^3\theta-3\cos\theta$cos3θ=4cos3θ−3cosθ. We also know that $\sin2\theta=2\sin\theta\cos\theta$sin2θ=2sinθcosθ. Thus, we can write

$4\cos^3\theta-3\cos\theta=2\sin\theta\cos\theta$4cos3θ−3cosθ=2sinθcosθ. On cancelling the common factor $\cos\theta$cosθ from each term and replacing $\cos^2\theta$cos2θ with $1-\sin^2\theta$1−sin2θ, we have a quadratic in $\sin\theta$sinθ.

$4\sin^2\theta+2\sin\theta-1=0$4sin2θ+2sinθ−1=0

This has a positive solution $\sin\theta=\frac{1}{4}\left(\sqrt{5}-1\right)$sinθ=14​(√5−1)

But what is the angle $\theta$θ? We began with the equation $\cos3\theta=\sin2\theta$cos3θ=sin2θ. In the first quadrant, where $\sin$sin and $\cos$cos are both positive,  $\sin x\equiv\cos\left(\frac{\pi}{2}-x\right)$sinx≡cos(π2​−x). So, 

$\cos3\theta=\cos\left(\frac{\pi}{2}-2\theta\right)$cos3θ=cos(π2​−2θ). It must be the case that $3\theta=\frac{\pi}{2}-2\theta$3θ=π2​−2θ and hence, 

$\theta=\frac{\pi}{10}$θ=π10​

Thus,

$\sin\frac{\pi}{10}=\frac{1}{4}\left(\sqrt{5}-1\right)$sinπ10​=14​(√5−1). 

With the Pythagorean identity and with the definition of the tangent function, we can deduce that

$\cos\frac{\pi}{10}=\frac{1}{4}\sqrt{10+2\sqrt{5}}$cosπ10​=14​√10+2√5

and

$\tan\frac{\pi}{10}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$tanπ10​=√5−1√10+2√5​

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3