Trigonometric Identities

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Expand expressions using double and half angle identities

Lesson

Where the argument of a trigonometric function is a multiple of some variable, say $2x$2`x` or $3x$3`x` or $\frac{x}{2}$`x`2, 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.

The expression $\cos3x$`c``o``s`3`x` can be expressed in terms of $\cos x$`c``o``s``x`.

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

We further expand $\cos2x$`c``o``s`2`x` and $\sin2x$`s``i``n`2`x` to obtain

$\cos3x=\left(\cos^2x-\sin^2x\right)\cos x-2\sin^2x\cos x$`c``o``s`3`x`=(`c``o``s`2`x`−`s``i``n`2`x`)`c``o``s``x`−2`s``i``n`2`x``c``o``s``x`

Because the required end result is to be in terms of the cosine function, we must replace $\sin^2x$`s``i``n`2`x` with $1-\cos^2x$1−`c``o``s`2`x`.

Then,

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

The expression $\sin^4x-\cos^4x$`s``i``n`4`x`−`c``o``s`4`x` can be written more simply in terms of a multiple angle.

The expression is the difference of two squares.

It can be factorised to $\left(\sin^2x+\cos^2x\right)\left(\sin^2x-\cos^2x\right)$(`s``i``n`2`x`+`c``o``s`2`x`)(`s``i``n`2`x`−`c``o``s`2`x`).

The first bracket is just $1$1, and the second bracket is the negative of the expansion of $\cos2x$`c``o``s`2`x`.

So, we have $\sin^4x-\cos^4x\equiv-\cos2x$`s``i``n`4`x`−`c``o``s`4`x`≡−`c``o``s`2`x`

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 $\cos3\theta=\sin2\theta$`c``o``s`3`θ`=`s``i``n`2`θ`. We found in Example 1 above, that $\cos3\theta=4\cos^3\theta-3\cos\theta$`c``o``s`3`θ`=4`c``o``s`3`θ`−3`c``o``s``θ`. We also know that $\sin2\theta=2\sin\theta\cos\theta$`s``i``n`2`θ`=2`s``i``n``θ``c``o``s``θ`. Thus, we can write

$4\cos^3\theta-3\cos\theta=2\sin\theta\cos\theta$4`c``o``s`3`θ`−3`c``o``s``θ`=2`s``i``n``θ``c``o``s``θ`. On cancelling the common factor $\cos\theta$`c``o``s``θ` from each term and replacing $\cos^2\theta$`c``o``s`2`θ` with $1-\sin^2\theta$1−`s``i``n`2`θ`, we have a quadratic in $\sin\theta$`s``i``n``θ`.

$4\sin^2\theta+2\sin\theta-1=0$4`s``i``n`2`θ`+2`s``i``n``θ`−1=0

This has a positive solution $\sin\theta=\frac{1}{4}\left(\sqrt{5}-1\right)$`s``i``n``θ`=14(√5−1)

But what is the angle $\theta$`θ`? We began with the equation $\cos3\theta=\sin2\theta$`c``o``s`3`θ`=`s``i``n`2`θ`. In the first quadrant, where $\sin$`s``i``n` and $\cos$`c``o``s` are both positive, $\sin x\equiv\cos\left(\frac{\pi}{2}-x\right)$`s``i``n``x`≡`c``o``s`(π2−`x`). So,

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

$\theta=\frac{\pi}{10}$`θ`=π10

Thus,

$\sin\frac{\pi}{10}=\frac{1}{4}\left(\sqrt{5}-1\right)$`s``i``n`π10=14(√5−1).

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

$\cos\frac{\pi}{10}=\frac{1}{4}\sqrt{10+2\sqrt{5}}$`c``o``s`π10=14√10+2√5

and

$\tan\frac{\pi}{10}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$`t``a``n`π10=√5−1√10+2√5

Fully simplify $\sin x\cos^3\left(x\right)-\sin^3\left(x\right)\cos x$`s``i``n``x``c``o``s`3(`x`)−`s``i``n`3(`x`)`c``o``s``x`, using trigonometric identities where necessary.

Write $\sin3x$`s``i``n`3`x` in terms of $\sin x$`s``i``n``x`.

Leave your answer in expanded form.

Using a double angle identity, simplify the expression $\sin5x\cos5x$`s``i``n`5`x``c``o``s`5`x`.

Express your answer as a single trigonometric function.

Manipulate trigonometric expressions

Apply trigonometric methods in solving problems