Choosing Sine or Cosine R-form | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Expand the form the question asks for

The same sum can be written with a shifted sine or cosine. The coefficient matches change with the chosen form. For positive $R$ and acute $\alpha$ , the four expansions are:

\begin{gathered}R\sin(\theta+\alpha)\\=(R\cos\alpha)\sin\theta\\{}+(R\sin\alpha)\cos\theta\\[6pt]R\sin(\theta-\alpha)\\=(R\cos\alpha)\sin\theta\\{}-(R\sin\alpha)\cos\theta\\[6pt]R\cos(\theta-\alpha)\\=(R\cos\alpha)\cos\theta\\{}+(R\sin\alpha)\sin\theta\\[6pt]R\cos(\theta+\alpha)\\=(R\cos\alpha)\cos\theta\\{}-(R\sin\alpha)\sin\theta\end{gathered}

Worked example

Write 4 cos θ − 3 sin θ as R cos(θ + α).

\begin{gathered}R\cos\alpha=4\\R\sin\alpha=3\\R=5\\\alpha=\tan^{-1}(3/4)\end{gathered}

The minus sign already comes from the cosine addition formula. There is no need to make $R$ negative.

Answer

\begin{gathered}5\cos(\theta+\alpha)\\\alpha\approx36.9^\circ\end{gathered}

Your turnPractice

Write $\sqrt3\cos x+\sin x$ in the form $R\cos(x-\alpha)$ , giving exact constants.

Show worked answer

Answer

2\cos(x-30^\circ)

Matching gives $R\cos\alpha=\sqrt3$ and $R\sin\alpha=1$ . Hence $R=2$ , $\cos\alpha=\sqrt3/2$ , $\sin\alpha=1/2$ .

Use both coefficient signs to choose the quadrant

When a required form has negative coefficients, an acute phase may not work. For $-3\sin x+4\cos x=R\sin(x+\alpha)$ , matching gives $\cos\alpha=-3/5$ and $\sin\alpha=4/5$ . Therefore $\alpha$ lies in quadrant II.

\begin{gathered}R=5\\\alpha=180^\circ-\tan^{-1}(4/3)\end{gathered}

The negative principal value from $\tan^{-1}(-4/3)$ alone would choose the wrong quadrant. Check sine and cosine separately. Angles differing by a full turn describe the same wave; use the interval requested.

If one coefficient is zero, match the other directly: $4\cos x=4\sin(x+90^\circ)$ . If both are zero, the expression is identically zero; its amplitude is 0 and no unique phase can be determined.

Your turnPractice

Express $-5\sin x$ as $R\sin(x+\alpha)$ with $R>0$ and $0^\circ\le\alpha<360^\circ$ .

Show worked answer

Answer

5\sin(x+180^\circ)

Matching needs $\cos\alpha=-1$ and $\sin\alpha=0$ .

Expand the form the question asks for

The same sum can be written with a shifted sine or cosine. The coefficient matches change with the chosen form. For positive $R$ and acute $\alpha$ , the four expansions are:

\begin{gathered}R\sin(\theta+\alpha)\\=(R\cos\alpha)\sin\theta\\{}+(R\sin\alpha)\cos\theta\\[6pt]R\sin(\theta-\alpha)\\=(R\cos\alpha)\sin\theta\\{}-(R\sin\alpha)\cos\theta\\[6pt]R\cos(\theta-\alpha)\\=(R\cos\alpha)\cos\theta\\{}+(R\sin\alpha)\sin\theta\\[6pt]R\cos(\theta+\alpha)\\=(R\cos\alpha)\cos\theta\\{}-(R\sin\alpha)\sin\theta\end{gathered}

Worked example

Write 4 cos θ − 3 sin θ as R cos(θ + α).

\begin{gathered}R\cos\alpha=4\\R\sin\alpha=3\\R=5\\\alpha=\tan^{-1}(3/4)\end{gathered}

The minus sign already comes from the cosine addition formula. There is no need to make $R$ negative.

Answer

\begin{gathered}5\cos(\theta+\alpha)\\\alpha\approx36.9^\circ\end{gathered}

Your turnPractice

Write $\sqrt3\cos x+\sin x$ in the form $R\cos(x-\alpha)$ , giving exact constants.

Show worked answer

Answer

2\cos(x-30^\circ)

Matching gives $R\cos\alpha=\sqrt3$ and $R\sin\alpha=1$ . Hence $R=2$ , $\cos\alpha=\sqrt3/2$ , $\sin\alpha=1/2$ .

Use both coefficient signs to choose the quadrant

\begin{gathered}R=5\\\alpha=180^\circ-\tan^{-1}(4/3)\end{gathered}

If one coefficient is zero, match the other directly: $4\cos x=4\sin(x+90^\circ)$ . If both are zero, the expression is identically zero; its amplitude is 0 and no unique phase can be determined.

Your turnPractice

Express $-5\sin x$ as $R\sin(x+\alpha)$ with $R>0$ and $0^\circ\le\alpha<360^\circ$ .

Show worked answer

Answer

5\sin(x+180^\circ)

Matching needs $\cos\alpha=-1$ and $\sin\alpha=0$ .