Hidden R-form | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Expand a product before combining the wave

An expression may need compound- and double-angle identities before R-form becomes possible. Expand first, then look for squared ratios and products with the same input.

Worked example

Show that 4 sin θ sin(θ + 60°) = 1 + 2 sin(2θ − 30°).

(9709/22/O/N/24 Q7(a)) Expand the shifted sine, then use $2\sin^2\theta=1-\cos2\theta$ and $2\sin\theta\cos\theta=\sin2\theta$ .

\begin{gathered}4\sin\theta\sin(\theta+60^\circ)\\=2\sin^2\theta\\{}+2\sqrt3\sin\theta\cos\theta\\=1-\cos2\theta+\sqrt3\sin2\theta\\=1+2\sin(2\theta-30^\circ)\end{gathered}

In the last line, expand $2\sin(2\theta-30^\circ)$ to check the coefficients $\sqrt3$ and −1. The constant 1 stays outside.

Your turnPractice

Express $7\sin^2x+9\cos^2x+4\sin x\cos x$ as $c+R\cos(2x-\alpha)$ with $R>0$ and acute $\alpha$ . Then find its range for all real $x$ .

Show worked answer

\begin{gathered}\frac72(1-\cos2x)\\{}+\frac92(1+\cos2x)\\{}+2\sin2x\\=8+\cos2x+2\sin2x\\=8+\sqrt5\cos(2x-\alpha)\\\alpha=\tan^{-1}2\end{gathered}

Answer

8-\sqrt5\le y\le8+\sqrt5

Both extremes occur because the unrestricted input covers complete periods.

For the smallest positive solution, check the preceding branch

Do not discard a negative shifted angle automatically: it can give a positive original angle after the shift is undone. Generate the nearby branches, undo the shift, then compare positive answers.

Worked example

Find the smallest positive θ satisfying 1/5 + 4 sin θ sin(θ + 60°) = 0.

(9709/22/O/N/24 Q7(b)) Use the preceding identity.

\begin{gathered}\frac65+2\sin(2\theta-30^\circ)=0\\u=2\theta-30^\circ\\\sin u=-\frac35\\\theta>0\ \Longleftrightarrow\ u>-30^\circ\end{gathered}

The principal angle is approximately $-36.9^\circ$ , below −30°, so it would give a negative $\theta$ . The next sine branch is $u=180^\circ+\sin^{-1}(3/5)$ . It precedes $360^\circ-\sin^{-1}(3/5)$ .

\theta=\frac{210^\circ+\sin^{-1}(3/5)}2

Answer

\theta\approx123.4^\circ

Substitution into the original equation checks the value; comparing branches establishes that it is the smallest positive one.

Your turnPractice

Find the smallest positive $x$ satisfying $\sqrt3\sin(2x)-\cos(2x)=-1/2$ , in degrees.

Show worked answer

\begin{gathered}2\sin(2x-30^\circ)=-\frac12\\ u=2x-30^\circ>-30^\circ\\ u=-\sin^{-1}(1/4)\end{gathered}

Answer

\begin{gathered}x=\frac{30^\circ-\sin^{-1}(1/4)}2\\\approx7.8^\circ\end{gathered}

This time the negative shifted angle is above −30°, so it gives the first positive original angle.

Expand a product before combining the wave

An expression may need compound- and double-angle identities before R-form becomes possible. Expand first, then look for squared ratios and products with the same input.

Worked example

Show that 4 sin θ sin(θ + 60°) = 1 + 2 sin(2θ − 30°).

(9709/22/O/N/24 Q7(a)) Expand the shifted sine, then use $2\sin^2\theta=1-\cos2\theta$ and $2\sin\theta\cos\theta=\sin2\theta$ .

\begin{gathered}4\sin\theta\sin(\theta+60^\circ)\\=2\sin^2\theta\\{}+2\sqrt3\sin\theta\cos\theta\\=1-\cos2\theta+\sqrt3\sin2\theta\\=1+2\sin(2\theta-30^\circ)\end{gathered}

In the last line, expand $2\sin(2\theta-30^\circ)$ to check the coefficients $\sqrt3$ and −1. The constant 1 stays outside.

Your turnPractice

Express $7\sin^2x+9\cos^2x+4\sin x\cos x$ as $c+R\cos(2x-\alpha)$ with $R>0$ and acute $\alpha$ . Then find its range for all real $x$ .

Show worked answer

\begin{gathered}\frac72(1-\cos2x)\\{}+\frac92(1+\cos2x)\\{}+2\sin2x\\=8+\cos2x+2\sin2x\\=8+\sqrt5\cos(2x-\alpha)\\\alpha=\tan^{-1}2\end{gathered}

Answer

8-\sqrt5\le y\le8+\sqrt5

Both extremes occur because the unrestricted input covers complete periods.

For the smallest positive solution, check the preceding branch

Do not discard a negative shifted angle automatically: it can give a positive original angle after the shift is undone. Generate the nearby branches, undo the shift, then compare positive answers.

Worked example

Find the smallest positive θ satisfying 1/5 + 4 sin θ sin(θ + 60°) = 0.

(9709/22/O/N/24 Q7(b)) Use the preceding identity.

\begin{gathered}\frac65+2\sin(2\theta-30^\circ)=0\\u=2\theta-30^\circ\\\sin u=-\frac35\\\theta>0\ \Longleftrightarrow\ u>-30^\circ\end{gathered}

\theta=\frac{210^\circ+\sin^{-1}(3/5)}2

Answer

\theta\approx123.4^\circ

Substitution into the original equation checks the value; comparing branches establishes that it is the smallest positive one.

Your turnPractice

Find the smallest positive $x$ satisfying $\sqrt3\sin(2x)-\cos(2x)=-1/2$ , in degrees.

Show worked answer

\begin{gathered}2\sin(2x-30^\circ)=-\frac12\\ u=2x-30^\circ>-30^\circ\\ u=-\sin^{-1}(1/4)\end{gathered}

Answer

\begin{gathered}x=\frac{30^\circ-\sin^{-1}(1/4)}2\\\approx7.8^\circ\end{gathered}

This time the negative shifted angle is above −30°, so it gives the first positive original angle.