E6 - Combining sine and cosine

E6 - Combining sine and cosine

Replace a sum of sine and cosine by one shifted wave. Match coefficients to choose its amplitude and phase, then use that form to find extreme values and solve equations over a complete interval.

Matching coefficients in a cosine form

How can 3cos⁡x+4sin⁡x3\cos x+4\sin x reach 55, when neither coefficient is 55? At x=0x=0 the expression is 33, and at x=π/2x=\pi/2 it is 44. Its largest value occurs between those angles, when neither separate term is at its own maximum.

You already know

The addition formulae expand cos⁡(x−α)\cos(x-\alpha) into cos⁡xcos⁡α+sin⁡xsin⁡α\cos x\cos\alpha+\sin x\sin\alpha. They are supplied in the formula booklet. Pythagoras gives cos⁡2α+sin⁡2α=1\cos^2\alpha+\sin^2\alpha=1.

Try 5cos⁡(x−α)5\cos(x-\alpha) with cos⁡α=3/5\cos\alpha=3/5 and sin⁡α=4/5\sin\alpha=4/5. Expansion gives exactly 3cos⁡x+4sin⁡x3\cos x+4\sin x. The positive amplitude 55 comes from 32+42\sqrt{3^2+4^2}, rather than adding the coefficients.

Harmonic form

A harmonic form here means a single shifted sine or cosine representing acos⁡x+bsin⁡xa\cos x+b\sin x, with the same argument xx in both original terms. The coefficients a,ba,b are constants. For example, 3cos⁡x+4sin⁡x3\cos x+4\sin x can be combined this way; 3cos⁡x+4sin⁡2x3\cos x+4\sin2x cannot generally be combined into a single wave of this form.

For the general case, expand first and match the coefficients of cos⁡x\cos x and sin⁡x\sin x. Squaring and adding the coefficient equations gives R2=a2+b2R^2=a^2+b^2.

Cosine-minus form

For (a,b)≠(0,0)(a,b)\ne(0,0), choose R>0R>0:

acos⁡x+bsin⁡x≡Rcos⁡(x−α) a\cos x+b\sin x\equiv R\cos(x-\alpha) R=a2+b2Rcos⁡α=aRsin⁡α=b R=\sqrt{a^2+b^2}\qquad R\cos\alpha=a\qquad R\sin\alpha=b

Select a phase interval, such as −π<α≤π-\pi<\alpha\le\pi. Phases differing by 2π2\pi describe the same wave. If both coefficients are zero, the expression is zero for every input and no unique phase is determined.

Not in the formula booklet: learn it.

This result is derived from the supplied addition formulae. Writing the coefficient equations also guards against taking the wrong tangent ratio.

Worked example 1: positive coefficients

Question. Express 3cos⁡x+4sin⁡x3\cos x+4\sin x as Rcos⁡(x−α)R\cos(x-\alpha), where R>0R>0 and 0<α<π/20<\alpha<\pi/2. Give α\alpha to three decimal places in radians.

Plan. Expand the proposed form, match coefficients and choose the acute phase.

Rcos⁡α=3Rsin⁡α=4match coefficientsR=9+16=5square and addtan⁡α=43divide the equationsα=arctan⁡(4/3)≈0.927both signs positive\begin{aligned} R\cos\alpha&=3\qquad R\sin\alpha=4 &&\text{match coefficients}\\ R&=\sqrt{9+16}=5 &&\text{square and add}\\ \tan\alpha&=\frac43 &&\text{divide the equations}\\ \alpha&=\arctan(4/3)\approx0.927 &&\text{both signs positive} \end{aligned}

Why this step? Matching before dividing tells you which coefficient belongs above the tangent ratio.

Answer. 5cos⁡(x−α)5\cos(x-\alpha), with α=arctan⁡(4/3)≈0.927\alpha=\arctan(4/3)\approx0.927 radians.

Check. 5(3/5)=35(3/5)=3 and 5(4/5)=45(4/5)=4 recover the original coefficients.

Change only the sign of the sine coefficient. Use a cosine-plus form so that the phase can remain acute.

Worked example 2: a negative sine coefficient

Question. Express 3cos⁡x−4sin⁡x3\cos x-4\sin x as Rcos⁡(x+α)R\cos(x+\alpha), where R>0R>0 and 0<α<π/20<\alpha<\pi/2. Give α\alpha to three decimal places in radians.

Plan. Expand the plus form, whose sine term has a minus sign.

Rcos⁡(x+α)=Rcos⁡αcos⁡x−Rsin⁡αsin⁡xexpandRcos⁡α=3Rsin⁡α=4match signed termsR=5α=arctan⁡(4/3)≈0.927solve as before\begin{aligned} R\cos(x+\alpha)&=R\cos\alpha\cos x-R\sin\alpha\sin x &&\text{expand}\\ R\cos\alpha&=3\qquad R\sin\alpha=4 &&\text{match signed terms}\\ R&=5\qquad\alpha=\arctan(4/3)\approx0.927 &&\text{solve as before} \end{aligned}

Why this step? The negative sine coefficient is already created by the plus sign inside cosine; it does not require a negative acute phase.

Answer. 5cos⁡(x+α)5\cos(x+\alpha), with α≈0.927\alpha\approx0.927 radians.

Check. At x=π/2x=\pi/2, the new form gives −5sin⁡α=−4-5\sin\alpha=-4, as the original does.

The general amplitude is positive even when either coefficient is negative.

Watch out

Wrong: R=3+4=7R=3+4=7. The maxima of sine and cosine occur at different inputs. Right: R=32+42=5R=\sqrt{3^2+4^2}=5, obtained by squaring and adding the coefficient equations.

Keep the negative sine coefficient and change the coefficient pair for your turn.

Phase quadrants and sine conventions

A tangent ratio does not determine a quadrant: adding π\pi leaves tangent unchanged but reverses both sine and cosine. Use both coefficient signs to select a phase.

The sine form has different coefficient roles. Expanding it gives

Rsin⁡(x+β)=Rsin⁡βcos⁡x+Rcos⁡βsin⁡xR\sin(x+\beta)=R\sin\beta\cos x+R\cos\beta\sin x

Thus for acos⁡x+bsin⁡xa\cos x+b\sin x, match Rsin⁡β=aR\sin\beta=a and Rcos⁡β=bR\cos\beta=b. For Rsin⁡(x−β)R\sin(x-\beta), the cosine coefficient is instead −Rsin⁡β-R\sin\beta. These derived matching relations are not separately in the formula booklet. The same amplitude R=a2+b2R=\sqrt{a^2+b^2} applies.

If a coefficient is zero, use the sine and cosine equations directly. For instance, 4sin⁡x=4cos⁡(x−π/2)4\sin x=4\cos(x-\pi/2). Dividing to form a tangent ratio would be unnecessary and may involve division by zero.

Worked example 3: an exact sine phase

Question. Express −3cos⁡x+3 sin⁡x-3\cos x+\sqrt3\,\sin x as Rsin⁡(x+β)R\sin(x+\beta), with R>0R>0 and −π<β≤π-\pi<\beta\le\pi.

Plan. Match the sine-form coefficients and use their signs to choose the phase.

R=9+3=23amplitudesin⁡β=−323=−32cosine coefficientcos⁡β=323=12sine coefficientβ=−π3negative sine; positive cosine\begin{aligned} R&=\sqrt{9+3}=2\sqrt3 &&\text{amplitude}\\ \sin\beta&=-\frac3{2\sqrt3}=-\frac{\sqrt3}2 &&\text{cosine coefficient}\\ \cos\beta&=\frac{\sqrt3}{2\sqrt3}=\frac12 &&\text{sine coefficient}\\ \beta&=-\frac\pi3 &&\text{negative sine; positive cosine} \end{aligned}

Why this step? The sign pair places the phase in quadrant IV, represented by a negative angle in the chosen interval.

Answer. 23 sin⁡(x−π/3)2\sqrt3\,\sin(x-\pi/3).

Check. Expanding gives cosine coefficient −3-3 and sine coefficient 3\sqrt3.

Change only the sign of the original sine coefficient. The amplitude stays fixed, but the phase moves into a different quadrant.

Worked example 4: selecting the other quadrant

Question. Express −3cos⁡x−3 sin⁡x-3\cos x-\sqrt3\,\sin x as Rsin⁡(x+β)R\sin(x+\beta), with R>0R>0 and −π<β≤π-\pi<\beta\le\pi.

Plan. Keep the amplitude and match both phase signs again.

R=23same squared coefficientssin⁡β=−32cos⁡β=−12match coefficientsβ=−2π3both signs negative\begin{aligned} R&=2\sqrt3 &&\text{same squared coefficients}\\ \sin\beta&=-\frac{\sqrt3}2\qquad\cos\beta=-\frac12 &&\text{match coefficients}\\ \beta&=-\frac{2\pi}3 &&\text{both signs negative} \end{aligned}

Why this step? The principal inverse tangent gives π/3\pi/3, but that has both signs positive and would reverse the whole expression.

Answer. 23 sin⁡(x−2π/3)2\sqrt3\,\sin(x-2\pi/3).

Check. 23 (−3/2)=−32\sqrt3\,(-\sqrt3/2)=-3 and 23 (−1/2)=−32\sqrt3\,(-1/2)=-\sqrt3.

You can avoid memorising a separate tangent ratio for every convention: expand the requested form instead.

Watch out

Wrong: for Rsin⁡(x+β)R\sin(x+\beta), write Rcos⁡β=aR\cos\beta=a and Rsin⁡β=bR\sin\beta=b. That swaps the coefficients. Right: the cosine coefficient is Rsin⁡βR\sin\beta, and the sine coefficient is Rcos⁡βR\cos\beta.

Keep the sine form and change the coefficient magnitudes while keeping both signs negative.

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