Mt18 - Combining sine and cosine into harmonic form | OCR A-Level Mathematics B (MEI)

Choose equivalent sine and cosine forms by expanding them, interpret a combined wave on full and restricted domains, and solve a shifted equation with a non-acute phase.

Matching coefficients in cosine form

Can $3\cos\theta+4\sin\theta$ ever reach $7$ ? Each term has its own maximum, but those maxima occur at different angles. Surprisingly, the combined wave reaches only $5$ . A single shifted cosine will explain why.

You already know

Expand $\cos(\theta-\alpha)=\cos\theta\cos\alpha+\sin\theta\sin\alpha$ . Also $\sin^2\alpha+\cos^2\alpha=1$ . Keep the signs of both coefficients when choosing a quadrant.

The height and horizontal position of the combined wave are different pieces of information.

Amplitude and phase

The amplitude is half the difference between a full wave's maximum and minimum heights. The phase in $R\cos(\theta-\alpha)$ , with $R>0$ , is the argument offset $\alpha$ . For $\cos\theta$ , the amplitude is $1$ and a phase is $0$ ; adding a constant changes the midline, not the amplitude.

Matching the coefficients of $\cos\theta$ and $\sin\theta$ gives $R\cos\alpha=a$ and $R\sin\alpha=b$ . Squaring and adding determines the amplitude.

Cosine harmonic form

If $(a,b)\ne(0,0)$ , then

a\cos\theta+b\sin\theta=R\cos(\theta-\alpha),\qquad R=\sqrt{a^2+b^2}>0,

where $\cos\alpha=a/R$ and $\sin\alpha=b/R$ . Choose $\alpha$ in the requested range. This is derived from compound-angle identities, not supplied as a separate result.

Not in the formula sheet: learn it.

The ratio $\tan\alpha=b/a$ is useful when $a\ne0$ , but it cannot distinguish opposite quadrants.

Worked example 1: positive coefficients

Question. Write $3\cos\theta+4\sin\theta$ as $R\cos(\theta-\alpha)$ , with $R>0$ and $0\le\alpha<2\pi$ .

Plan. Find the amplitude, then match both coefficient signs.

\begin{aligned} R&=\sqrt{3^2+4^2}=5 &&\text{square and add}\\ \cos\alpha&=3/5,\quad\sin\alpha=4/5 &&\text{match coefficients}\\ \alpha&=\arctan(4/3)=0.927295\ldots &&\text{quadrant I} \end{aligned}

Why this step? Both phase components are positive, so the acute inverse-tangent value is appropriate.

Answer. $5\cos(\theta-\alpha)$ , where $\alpha=\arctan(4/3)$ .

Check. At $\theta=0$ , it gives $5\cos\alpha=3$ , matching the expression.

Change only the cosine coefficient's sign. The amplitude survives, but the phase moves quadrants.

Worked example 2: a quadrant-II phase

Question. Write $-3\cos\theta+4\sin\theta$ in the same form and phase range.

Plan. Use the same amplitude and locate the phase from its sine and cosine.

\begin{aligned} R&=5 &&\text{unchanged squares}\\ \cos\alpha&=-3/5,\quad\sin\alpha=4/5 &&\text{signed coefficients}\\ \alpha&=\pi-\arctan(4/3)=2.214297\ldots &&\text{quadrant II} \end{aligned}

Why this step? An acute phase would put a positive cosine coefficient back into the expansion.

Answer. $5\cos(\theta-\alpha)$ , where $\alpha=\pi-\arctan(4/3)$ .

Check. At zero the form gives $-3$ ; at $\theta=\pi/2$ it gives $4$ .

A plus-cosine form has the opposite sine coefficient: $R\cos(\theta+\beta)=R\cos\beta\cos\theta-R\sin\beta\sin\theta$ . Expand the requested form before matching. If $a=0$ , select the phase directly from its sine and cosine; if both coefficients are zero, the function is zero and has no unique phase.

Equivalent sine forms and phase signs

A cosine and a sine can describe the same shifted wave. Their coefficient-to-phase correspondence differs, so do not just replace the function's name.

Sine harmonic form

For $R>0$ ,

R\sin(\theta+\beta)=R\sin\beta\cos\theta+R\cos\beta\sin\theta,

while a minus sign gives cosine coefficient $-R\sin\beta$ . Match each coefficient to the expanded form. The amplitude remains $\sqrt{a^2+b^2}$ for a nonzero combination.

Not in the formula sheet: learn it.

The phase belongs to the form you chose. A complementary acute phase changes a cosine-minus form into a sine-plus form.

Worked example 3: a sine-plus form

Question. Express $5\sin\theta+12\cos\theta$ as $R\sin(\theta+\beta)$ , with $0<\beta<\pi/2$ .

Plan. Expand the requested sine form and match coefficients.

\begin{aligned} R&=13 &&\text{amplitude}\\ R\cos\beta&=5,\quad R\sin\beta=12 &&\text{match expansion}\\ \beta&=\arctan(12/5) &&\text{acute phase} \end{aligned}

Why this step? In the sine-plus form the cosine coefficient matches $R\sin\beta$ , not $R\cos\beta$ .

Answer. $13\sin(\theta+\arctan(12/5))$ .

Check. At zero, the form gives $13\sin\beta=12$ . Also $\arctan(5/12)+\arctan(12/5)=\pi/2$ , linking a cosine form to this sine form.

Change only the cosine coefficient's sign; a sine-minus form now keeps the phase acute.

Worked example 4: a sine-minus form

Question. Express $5\sin\theta-12\cos\theta$ as $R\sin(\theta-\gamma)$ , with $0<\gamma<\pi/2$ .

Plan. Match $R\cos\gamma$ and $-R\sin\gamma$ .

\begin{aligned} R&=13 &&\text{same amplitude}\\ \cos\gamma&=5/13,\quad\sin\gamma=12/13 &&\text{match both signs}\\ \gamma&=\arctan(12/5) &&\text{acute range} \end{aligned}

Why this step? The minus in the angle produces the required negative cosine coefficient.

Answer. $13\sin(\theta-\arctan(12/5))$ .

Check. At $\theta=0$ the answer gives $-12$ , and at $\theta=\pi/2$ it gives $5$ .

Pure sine and cosine are boundary cases, too: $4\sin\theta=4\cos(\theta-\pi/2)$ . Equivalent phases differ by full turns; the question's phase range selects the representative to report.

Watch out

$R\sin(\theta-\gamma)$ has cosine coefficient $-R\sin\gamma$ . Writing $R\sin\gamma=-12$ in the last example forgets that leading minus and sends the phase to the wrong quadrant.

Use new coefficients to make that sign choice yourself.

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Mt18 - Combining sine and cosine into harmonic form

Matching coefficients in cosine form

Equivalent sine forms and phase signs

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