Double-angle Formulae | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Make the two input angles equal

A double angle is twice an input angle. Set $B=A$ in the compound-angle formulae.

\begin{gathered}\sin(2A)\equiv2\sin A\cos A\\ \cos(2A)\equiv\cos^2A-\sin^2A\\ \tan(2A)\equiv\frac{2\tan A}{1-\tan^2A}\end{gathered}

Do not confuse $\sin(2A)$ , the sine of twice the angle, with $\sin^2A=(\sin A)^2$ , the square of the sine value. At $A=30^\circ$ , these are $\sqrt3/2$ and $1/4$ .

The tangent quotient needs $\cos A\ne0$ and $1-\tan^2A\ne0$ . For example, $\tan180^\circ=0$ , but using $A=90^\circ$ in the quotient would be invalid. Use the original angle instead.

Your turnPractice

Write $2\sin(3x)\cos(3x)$ as a single ratio. Explain why the input is not $3x$ .

Show worked answer

Answer

\sin(6x)

The angle playing the role of $A$ is the whole $3x$ ; it doubles to $6x$ .

Choose the cosine form that leaves one ratio

Substitute $\cos^2A=1-\sin^2A$ or $\sin^2A=1-\cos^2A$ into the cosine identity.

\begin{gathered}\cos2A\equiv1-2\sin^2A\\ \cos2A\equiv2\cos^2A-1\end{gathered}

Use the sine-only form when the other terms contain sine, and the cosine-only form when they contain cosine. Use $\cos^2A-\sin^2A$ when both values are already known. All three forms are supplied in MF19.

Worked example

Given tan A = 3/4 and A acute, find sin 2A, cos 2A and tan 2A.

A right triangle with opposite side 3 and adjacent side 4 has hypotenuse $\sqrt{3^2+4^2}=5$ . Thus $\sin A=3/5$ and $\cos A=4/5$ .

\begin{gathered}\sin2A=2\frac35\frac45\\=\frac{24}{25}\\\cos2A=\frac{16}{25}-\frac9{25}\\=\frac7{25}\\\tan2A=\frac{24/25}{7/25}\\=\frac{24}{7}\end{gathered}

Your turnPractice

Rewrite $\cos2x+2\sin x$ using sine only, and $\cos2x-3\cos x$ using cosine only.

Show worked answer

Answer

\begin{gathered}1-2\sin^2x+2\sin x\\2\cos^2x-1-3\cos x\end{gathered}

Make the two input angles equal

A double angle is twice an input angle. Set $B=A$ in the compound-angle formulae.

\begin{gathered}\sin(2A)\equiv2\sin A\cos A\\ \cos(2A)\equiv\cos^2A-\sin^2A\\ \tan(2A)\equiv\frac{2\tan A}{1-\tan^2A}\end{gathered}

Do not confuse $\sin(2A)$ , the sine of twice the angle, with $\sin^2A=(\sin A)^2$ , the square of the sine value. At $A=30^\circ$ , these are $\sqrt3/2$ and $1/4$ .

The tangent quotient needs $\cos A\ne0$ and $1-\tan^2A\ne0$ . For example, $\tan180^\circ=0$ , but using $A=90^\circ$ in the quotient would be invalid. Use the original angle instead.

Your turnPractice

Write $2\sin(3x)\cos(3x)$ as a single ratio. Explain why the input is not $3x$ .

Show worked answer

Answer

\sin(6x)

The angle playing the role of $A$ is the whole $3x$ ; it doubles to $6x$ .

Choose the cosine form that leaves one ratio

Substitute $\cos^2A=1-\sin^2A$ or $\sin^2A=1-\cos^2A$ into the cosine identity.

\begin{gathered}\cos2A\equiv1-2\sin^2A\\ \cos2A\equiv2\cos^2A-1\end{gathered}

Worked example

Given tan A = 3/4 and A acute, find sin 2A, cos 2A and tan 2A.

A right triangle with opposite side 3 and adjacent side 4 has hypotenuse $\sqrt{3^2+4^2}=5$ . Thus $\sin A=3/5$ and $\cos A=4/5$ .

\begin{gathered}\sin2A=2\frac35\frac45\\=\frac{24}{25}\\\cos2A=\frac{16}{25}-\frac9{25}\\=\frac7{25}\\\tan2A=\frac{24/25}{7/25}\\=\frac{24}{7}\end{gathered}

Your turnPractice

Rewrite $\cos2x+2\sin x$ using sine only, and $\cos2x-3\cos x$ using cosine only.

Show worked answer

Answer

\begin{gathered}1-2\sin^2x+2\sin x\\2\cos^2x-1-3\cos x\end{gathered}