Multiple-angle Identities | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Split a larger angle into angles you can expand

You do not need a separate memorised triple-angle formula. Write $3x=2x+x$ , expand, and use the double-angle identities. To express the result in sine only, replace cosine squared as well.

\begin{gathered}\sin3x\\=\sin2x\cos x+\cos2x\sin x\\=2\sin x\cos^2x\\{}+(1-2\sin^2x)\sin x\\=2\sin x(1-\sin^2x)\\{}+\sin x-2\sin^3x\\=3\sin x-4\sin^3x\end{gathered}

The exponent 3 in $\sin^3x$ means cube the sine value. At $x=30^\circ$ , the expression gives $3/2-4(1/8)=1$ , agreeing with $\sin90^\circ$ .

Your turnPractice

By expanding $\cos(2x+x)$ , express $\cos3x$ in terms of $\cos x$ only.

Show worked answer

\begin{gathered}\cos3x\\=(2\cos^2x-1)\cos x\\{}-2\sin^2x\cos x\\=2\cos^3x-\cos x\\{}-2(1-\cos^2x)\cos x\end{gathered}

Answer

4\cos^3x-3\cos x

Add or subtract expansions to cancel matching terms

When a question suggests two angle splits, expand both. Adding often cancels the sine products; subtracting often cancels the cosine products.

Worked example

Express cos 3x + cos x as a product.

\begin{gathered}\cos3x=\cos(2x+x)\\=\cos2x\cos x\\{}-\sin2x\sin x\\\cos x=\cos(2x-x)\\=\cos2x\cos x\\{}+\sin2x\sin x\\\cos3x+\cos x\\=2\cos2x\cos x\end{gathered}

Your turnPractice

Use $4x=3x+x$ and $2x=3x-x$ to express $\cos2x-\cos4x$ as a product.

Show worked answer

Answer

2\sin3x\sin x

The two cosine products cancel on subtraction; the sine products add.

A sine-to-cosine quotient gives a tangent identity

Before division, the triple-angle expansions can be kept in both ratios:

\begin{gathered}\sin3x\\=3\sin x\cos^2x-\sin^3x\\\cos3x\\=\cos^3x-3\sin^2x\cos x\end{gathered}

Divide the first by the second, then divide numerator and denominator by $\cos^3x$ . This gives:

\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}

Require $\cos x\ne0$ and $\cos3x\ne0$ . This route avoids introducing $\tan2x$ , which may be undefined even when the final identity has a value.

Your turnPractice

Check the identity at $x=45^\circ$ . Why would an intermediate $\tan2x$ be unsuitable here?

Show worked answer

The quotient is $(3-1)/(1-3)=-1=\tan135^\circ$ . But $\tan90^\circ$ is undefined.

Split a larger angle into angles you can expand

You do not need a separate memorised triple-angle formula. Write $3x=2x+x$ , expand, and use the double-angle identities. To express the result in sine only, replace cosine squared as well.

\begin{gathered}\sin3x\\=\sin2x\cos x+\cos2x\sin x\\=2\sin x\cos^2x\\{}+(1-2\sin^2x)\sin x\\=2\sin x(1-\sin^2x)\\{}+\sin x-2\sin^3x\\=3\sin x-4\sin^3x\end{gathered}

The exponent 3 in $\sin^3x$ means cube the sine value. At $x=30^\circ$ , the expression gives $3/2-4(1/8)=1$ , agreeing with $\sin90^\circ$ .

Your turnPractice

By expanding $\cos(2x+x)$ , express $\cos3x$ in terms of $\cos x$ only.

Show worked answer

\begin{gathered}\cos3x\\=(2\cos^2x-1)\cos x\\{}-2\sin^2x\cos x\\=2\cos^3x-\cos x\\{}-2(1-\cos^2x)\cos x\end{gathered}

Answer

4\cos^3x-3\cos x

Add or subtract expansions to cancel matching terms

When a question suggests two angle splits, expand both. Adding often cancels the sine products; subtracting often cancels the cosine products.

Worked example

Express cos 3x + cos x as a product.

\begin{gathered}\cos3x=\cos(2x+x)\\=\cos2x\cos x\\{}-\sin2x\sin x\\\cos x=\cos(2x-x)\\=\cos2x\cos x\\{}+\sin2x\sin x\\\cos3x+\cos x\\=2\cos2x\cos x\end{gathered}

Your turnPractice

Use $4x=3x+x$ and $2x=3x-x$ to express $\cos2x-\cos4x$ as a product.

Show worked answer

Answer

2\sin3x\sin x

The two cosine products cancel on subtraction; the sine products add.

A sine-to-cosine quotient gives a tangent identity

Before division, the triple-angle expansions can be kept in both ratios:

\begin{gathered}\sin3x\\=3\sin x\cos^2x-\sin^3x\\\cos3x\\=\cos^3x-3\sin^2x\cos x\end{gathered}

Divide the first by the second, then divide numerator and denominator by $\cos^3x$ . This gives:

\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}

Require $\cos x\ne0$ and $\cos3x\ne0$ . This route avoids introducing $\tan2x$ , which may be undefined even when the final identity has a value.

Your turnPractice

Check the identity at $x=45^\circ$ . Why would an intermediate $\tan2x$ be unsuitable here?

Show worked answer

The quotient is $(3-1)/(1-3)=-1=\tan135^\circ$ . But $\tan90^\circ$ is undefined.