P3 2.3 - Combining sine and cosine

P3 2.3 - Combining sine and cosine

Two waves of the same frequency can combine into one shifted wave. In P3 2.3 you will rewrite a sine–cosine combination with a positive amplitude, choose the phase angle from both coefficient signs, and use the result to solve equations and find greatest and least values.

Converting to a shifted sine or cosine

How large can 3cos⁡x+4sin⁡x3\cos x+4\sin x become? Adding 3+43+4 suggests 77, but sine and cosine cannot both equal 11 at the same angle. A compound-angle expansion combines the two terms correctly.

Expand Rcos⁡(x−α)R\cos(x-\alpha):

Rcos⁡(x−α)=Rcos⁡αcos⁡x+Rsin⁡αsin⁡x.R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x.

To match acos⁡x+bsin⁡xa\cos x+b\sin x, we need Rcos⁡α=aR\cos\alpha=a and Rsin⁡α=bR\sin\alpha=b. Squaring and adding gives R2=a2+b2R^2=a^2+b^2. Choose R>0R>0 so it is the wave's amplitude; if a=b=0a=b=0, the expression is simply zero and no phase angle is needed.

Combining equal-frequency sine and cosine

For (a,b)≠(0,0)(a,b)\ne(0,0),

acos⁡x+bsin⁡x=Rcos⁡(x−α),R=a2+b2,cos⁡α=aR,sin⁡α=bR.a\cos x+b\sin x=R\cos(x-\alpha),\qquad R=\sqrt{a^2+b^2},\quad \cos\alpha=\frac aR,\quad\sin\alpha=\frac bR.

Use both signs to locate α\alpha, not just tan⁡−1(b/a)\tan^{-1}(b/a). This form follows from the supplied compound-angle formula.

Other requested forms use different sign matches:

FormCoefficient of cos⁡x\cos xCoefficient of sin⁡x\sin x
Rcos⁡(x−α)R\cos(x-\alpha)Rcos⁡αR\cos\alphaRsin⁡αR\sin\alpha
Rcos⁡(x+α)R\cos(x+\alpha)Rcos⁡αR\cos\alpha−Rsin⁡α-R\sin\alpha
Rsin⁡(x+α)R\sin(x+\alpha)Rsin⁡αR\sin\alphaRcos⁡αR\cos\alpha
Rsin⁡(x−α)R\sin(x-\alpha)−Rsin⁡α-R\sin\alphaRcos⁡αR\cos\alpha

Expand the requested form before matching. This avoids remembering a phase angle for the wrong form. When a coefficient is zero, match sine and cosine directly; division to form a tangent ratio may be unavailable.

Worked example 1

Write 3cos⁡x+4sin⁡x3\cos x+4\sin x as Rcos⁡(x−α)R\cos(x-\alpha), with R>0R>0 and 0∘≤α<360∘0^\circ\le\alpha<360^\circ.

Plan. Find the amplitude, then choose the phase angle whose cosine and sine have the two required signs.

R=32+42=5,cos⁡α=35,sin⁡α=45.R=\sqrt{3^2+4^2}=5, \qquad\cos\alpha=\frac35,\quad\sin\alpha=\frac45.

Both are positive, so α\alpha is in quadrant I:

α=tan⁡−1(4/3)=53.1301…∘.\alpha=\tan^{-1}(4/3)=53.1301\ldots^\circ.

Answer. 5cos⁡(x−53.1301…∘)5\cos(x-53.1301\ldots^\circ).

Check. At x=0x=0, the result is 5(3/5)=35(3/5)=3; at x=90∘x=90^\circ, it is 5(4/5)=45(4/5)=4, matching the original expression at two useful points. Expansion proves the equality at every angle.

Change only the sign of the cosine coefficient. The amplitude and tangent ratio's magnitude stay the same, but the phase angle moves to quadrant II.

Worked example 2

Write −3cos⁡x+4sin⁡x-3\cos x+4\sin x as Rcos⁡(x−β)R\cos(x-\beta), with R>0R>0 and 0∘≤β<360∘0^\circ\le\beta<360^\circ.

Plan. Match the negative cosine and positive sine instead of accepting a calculator's negative principal arctangent.

R=5,cos⁡β=−35,sin⁡β=45.R=5,\qquad\cos\beta=-\frac35,\quad\sin\beta=\frac45.

Thus

β=180∘−tan⁡−1(4/3)=126.8698…∘.\beta=180^\circ-\tan^{-1}(4/3)=126.8698\ldots^\circ.

Answer. 5cos⁡(x−126.8698…∘)5\cos(x-126.8698\ldots^\circ).

Check. At x=0x=0 the value is −3-3, while at x=90∘x=90^\circ it is 44. Choosing −53.1301…∘-53.1301\ldots^\circ would give the wrong signs at both checkpoints.

For comparison, the first expression is also 5sin⁡(x+γ)5\sin(x+\gamma), where sin⁡γ=3/5\sin\gamma=3/5, cos⁡γ=4/5\cos\gamma=4/5 and γ=36.8698…∘\gamma=36.8698\ldots^\circ. The second is 5sin⁡(x−γ)5\sin(x-\gamma). The same amplitude can therefore accompany different correctly matched phase angles.

Solving in a given interval

Once the terms are combined, solve a single shifted sine or cosine equation. Transform the whole interval for the shifted angle before choosing branches, then return to the original variable.

For 5cos⁡(x−α)=c5\cos(x-\alpha)=c, write u=x−αu=x-\alpha. The equation becomes cos⁡u=c/5\cos u=c/5. If ∣c/5∣>1|c/5|>1, there are no real solutions. At c/5=1c/5=1 or −1-1, the two usual cosine branches coincide modulo a full turn; do not count them twice.

In the following pair, retain α=tan⁡−1(4/3)=53.1301…∘\alpha=\tan^{-1}(4/3)=53.1301\ldots^\circ without rounding during the solution.

Worked example 3

Solve 3cos⁡x+4sin⁡x=5/23\cos x+4\sin x=5/2 for 0∘≤x<360∘0^\circ\le x<360^\circ. Give answers to one decimal place.

Plan. Use the combined form, find all shifted-angle cosine solutions, then add back α\alpha.

5cos⁡(x−α)=52⟹cos⁡u=12,−α≤u<360∘−α.5\cos(x-\alpha)=\frac52 \quad\Longrightarrow\quad\cos u=\frac12, \qquad-\alpha\le u<360^\circ-\alpha.

The interval is approximately [−53.13∘,306.87∘)[-53.13^\circ,306.87^\circ), so u=60∘,300∘u=60^\circ,300^\circ. Therefore

x=60∘+α=113.1301…∘,x=300∘+α=353.1301…∘.x=60^\circ+\alpha=113.1301\ldots^\circ, \quad x=300^\circ+\alpha=353.1301\ldots^\circ.

Answer. x=113.1∘,353.1∘x=113.1^\circ,353.1^\circ.

Check. Substitution of the unrounded angles into 3cos⁡x+4sin⁡x3\cos x+4\sin x gives 2.52.5 each. Both are inside the original interval. The value −6.8698…∘-6.8698\ldots^\circ, obtained from u=−60∘u=-60^\circ, lies outside it.

Change only the right-hand side's sign. The amplitude and shift are unchanged, but the shifted-angle quadrants move.

Worked example 4

Solve 3cos⁡x+4sin⁡x=−5/23\cos x+4\sin x=-5/2 for 0∘≤x<360∘0^\circ\le x<360^\circ. Give answers to one decimal place.

Plan. Solve cos⁡u=−1/2\cos u=-1/2 in the same shifted interval.

cos⁡u=−12,−53.1301…∘≤u<306.8698…∘.\cos u=-\frac12,\qquad-53.1301\ldots^\circ\le u<306.8698\ldots^\circ.

Here u=120∘,240∘u=120^\circ,240^\circ, so

x=173.1301…∘,293.1301…∘.x=173.1301\ldots^\circ,293.1301\ldots^\circ.

Answer. x=173.1∘,293.1∘x=173.1^\circ,293.1^\circ.

Check. The shifted angles have cosine −1/2-1/2, giving the required −2.5-2.5. Both original angles satisfy the stated interval, and a full period gives no further distinct solutions there.

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