2.3, 3.3 - Combining sine and cosine

2.3, 3.3 - Combining sine and cosine

A sum of sine and cosine with the same angle is another shifted sine or cosine. In the shared trigonometry of 2.3 and 3.3, you will find its amplitude and angle shift, use them to determine attainable extreme values, and solve equations in a specified interval.

Matching coefficients

Can 3sin⁡θ+4cos⁡θ3\sin\theta+4\cos\theta ever be larger than 44? Taking θ=0∘\theta=0^\circ gives 44, but θ=30∘\theta=30^\circ gives 3/2+23=4.964…3/2+2\sqrt3=4.964\ldots. The two terms can combine to give more than either coefficient. Their separate maxima do not occur at the same angle, so adding 3+43+4 does not give the true maximum either.

Expand a shifted sine before choosing any constants:

Rsin⁡(θ+α)=Rcos⁡αsin⁡θ+Rsin⁡αcos⁡θ.R\sin(\theta+\alpha) =R\cos\alpha\sin\theta+R\sin\alpha\cos\theta.

To make this equal asin⁡θ+bcos⁡θa\sin\theta+b\cos\theta for every θ\theta, match the sine coefficients and the cosine coefficients separately. These two functions cannot substitute for each other: setting θ=0\theta=0 isolates cosine, and setting θ=90∘\theta=90^\circ isolates sine.

Amplitude and auxiliary angle

For real a,ba,b, not both zero, write

asin⁡θ+bcos⁡θ=Rsin⁡(θ+α),R>0.a\sin\theta+b\cos\theta=R\sin(\theta+\alpha),\qquad R>0.

Matching coefficients gives

Rcos⁡α=a,Rsin⁡α=b,R=a2+b2.R\cos\alpha=a,\qquad R\sin\alpha=b,\qquad R=\sqrt{a^2+b^2}.

Choose α\alpha with both required signs, in any range specified by the question.

Not in the formula booklet: derive it by expanding the supplied compound-angle formula.

Squaring and adding the coefficient equations gives R2(cos⁡2α+sin⁡2α)=a2+b2R^2(\cos^2\alpha+\sin^2\alpha)=a^2+b^2, which proves the amplitude formula. The zero expression a=b=0a=b=0 has amplitude zero and no unique shift; it is excluded from the convention R>0R>0.

Worked example 1: a positive sine shift

Question. Express 3sin⁡θ+4cos⁡θ3\sin\theta+4\cos\theta as Rsin⁡(θ+α)R\sin(\theta+\alpha), where R>0R>0 and 0∘<α<90∘0^\circ<\alpha<90^\circ. Give α\alpha to 22 decimal places.

Plan. Expand the requested form, match coefficients, and use both signs to choose the angle.

Rcos⁡α=3,Rsin⁡α=4R2=32+42=25,R=5tan⁡α=43,α=53.1301…∘.\begin{aligned} R\cos\alpha&=3,&R\sin\alpha&=4\\ R^2&=3^2+4^2=25,&R&=5\\ \tan\alpha&=\frac43,&\alpha&=53.1301\ldots^\circ. \end{aligned}

Both sine and cosine of α\alpha are positive, agreeing with the required acute angle.

Answer. 5sin⁡(θ+53.13∘)5\sin(\theta+53.13^\circ), with the exact shift specified by cos⁡α=3/5\cos\alpha=3/5, sin⁡α=4/5\sin\alpha=4/5.

Check. Expanding with the unrounded shift gives 5(3/5)sin⁡θ+5(4/5)cos⁡θ5(3/5)\sin\theta+5(4/5)\cos\theta, exactly the original expression.

Change only the sign of the cosine term. Use a minus shift to keep the auxiliary angle acute.

Worked example 2: a negative sine shift

Question. Express 3sin⁡θ−4cos⁡θ3\sin\theta-4\cos\theta as Rsin⁡(θ−α)R\sin(\theta-\alpha), where R>0R>0 and 0∘<α<90∘0^\circ<\alpha<90^\circ. Give α\alpha to 22 decimal places.

Plan. Expand the minus-shift form before matching its coefficients.

Rsin⁡(θ−α)=Rcos⁡αsin⁡θ−Rsin⁡αcos⁡θRcos⁡α=3,Rsin⁡α=4R=5,α=53.1301…∘.\begin{aligned} R\sin(\theta-\alpha)&=R\cos\alpha\sin\theta-R\sin\alpha\cos\theta\\ R\cos\alpha&=3,\qquad R\sin\alpha=4\\ R&=5,\qquad\alpha=53.1301\ldots^\circ. \end{aligned}

Answer. 5sin⁡(θ−53.13∘)5\sin(\theta-53.13^\circ), retaining the unrounded shift in later calculations.

Check. At θ=0∘\theta=0^\circ, the original is −4-4, and 5sin⁡(−α)=−45\sin(-\alpha)=-4. The expansion checks equality for every angle.

A cosine form works in exactly the same way, but its sine and cosine coefficients swap places. This table comes from expanding each requested form; it is safer than memorising a single inverse-tangent ratio.

Requested formCoefficient of sin⁡θ\sin\thetaCoefficient of cos⁡θ\cos\theta
Rsin⁡(θ+α)R\sin(\theta+\alpha)Rcos⁡αR\cos\alphaRsin⁡αR\sin\alpha
Rsin⁡(θ−α)R\sin(\theta-\alpha)Rcos⁡αR\cos\alpha−Rsin⁡α-R\sin\alpha
Rcos⁡(θ−α)R\cos(\theta-\alpha)Rsin⁡αR\sin\alphaRcos⁡αR\cos\alpha
Rcos⁡(θ+α)R\cos(\theta+\alpha)−Rsin⁡α-R\sin\alphaRcos⁡αR\cos\alpha

For any of these forms, R=a2+b2R=\sqrt{a^2+b^2}. The equations determine the quadrant of the shift, including when a coefficient is negative or zero. For example, −3sin⁡θ−4cos⁡θ=5sin⁡(θ+α)-3\sin\theta-4\cos\theta=5\sin(\theta+\alpha) needs both cos⁡α<0\cos\alpha<0 and sin⁡α<0\sin\alpha<0, so α=233.1301…∘\alpha=233.1301\ldots^\circ in [0∘,360∘)[0^\circ,360^\circ). Taking only the principal value of tan⁡−1(4/3)\tan^{-1}(4/3) would give the wrong signs.

Worked example 3: a cosine with a minus shift

Question. Express 4cos⁡θ+3sin⁡θ4\cos\theta+3\sin\theta as Rcos⁡(θ−β)R\cos(\theta-\beta), with R>0R>0 and acute β\beta.

Plan. Expand the requested cosine form and match its two coefficients.

Rcos⁡(θ−β)=Rcos⁡βcos⁡θ+Rsin⁡βsin⁡θRcos⁡β=4,Rsin⁡β=3R=5,β=tan⁡−1(3/4)=36.8698…∘.\begin{aligned} R\cos(\theta-\beta)&=R\cos\beta\cos\theta+R\sin\beta\sin\theta\\ R\cos\beta&=4,\qquad R\sin\beta=3\\ R&=5,\qquad\beta=\tan^{-1}(3/4)=36.8698\ldots^\circ. \end{aligned}

Answer. 5cos⁡(θ−36.87∘)5\cos(\theta-36.87^\circ), where the shift is quoted to 22 decimal places.

Check. At θ=90∘\theta=90^\circ, the original is 33, and 5cos⁡(90∘−β)=5sin⁡β=35\cos(90^\circ-\beta)=5\sin\beta=3. Expansion checks both coefficients exactly.

Change only the sine coefficient's sign. The cosine form then needs a plus shift.

Worked example 4: a cosine with a plus shift

Question. Express 4cos⁡θ−3sin⁡θ4\cos\theta-3\sin\theta as Rcos⁡(θ+β)R\cos(\theta+\beta), with R>0R>0 and acute β\beta.

Plan. Match the minus sign in the cosine expansion to the negative sine coefficient.

Rcos⁡(θ+β)=Rcos⁡βcos⁡θ−Rsin⁡βsin⁡θRcos⁡β=4,Rsin⁡β=3R=5,β=36.8698…∘.\begin{aligned} R\cos(\theta+\beta)&=R\cos\beta\cos\theta-R\sin\beta\sin\theta\\ R\cos\beta&=4,\qquad R\sin\beta=3\\ R&=5,\qquad\beta=36.8698\ldots^\circ. \end{aligned}

Answer. 5cos⁡(θ+36.87∘)5\cos(\theta+36.87^\circ).

Check. At θ=90∘\theta=90^\circ, the original is −3-3, and the new form gives −5sin⁡β=−3-5\sin\beta=-3. At θ=0∘\theta=0^\circ, both give 44.

The requested form determines which coefficient belongs to which ratio.

Watch out

For Rsin⁡(θ+α)R\sin(\theta+\alpha), tan⁡α=b/a\tan\alpha=b/a; for Rcos⁡(θ−α)R\cos(\theta-\alpha), it is a/ba/b. These quotients apply only when their denominators are nonzero, and neither quotient alone determines the quadrant. Expand and match first.

Try the cosine plus-shift form with fresh coefficients.

Ranges and extreme values

The amplitude gives possible highest and lowest values, but the allowed angles decide whether they occur. Since −1≤sin⁡u≤1-1\le\sin u\le1, the expression Rsin⁡uR\sin u, with R>0R>0, lies between −R-R and RR. A full cycle reaches both extremes; a restricted interval may reach neither.

Amplitude

The amplitude of Rsin⁡(θ±α)R\sin(\theta\pm\alpha) or Rcos⁡(θ±α)R\cos(\theta\pm\alpha), where R>0R>0, is RR: its maximum distance from the horizontal midline. Adding a constant moves that midline without changing the amplitude.

For c+Rsin⁡uc+R\sin u, the full-cycle range is [c−R,c+R][c-R,c+R]. To check attainment, solve sin⁡u=1\sin u=1 for the maximum and sin⁡u=−1\sin u=-1 for the minimum, then see whether the corresponding original angles are permitted. In degrees, these occur at u=90∘+360∘ku=90^\circ+360^\circ k and 270∘+360∘k270^\circ+360^\circ k. For cosine, the corresponding angles are 360∘k360^\circ k and 180∘+360∘k180^\circ+360^\circ k.

[DIAGRAM: asset_name: Full and restricted ranges; asset_slug: auxiliary-angle-range; caption: The same amplitude can give different attainable extreme values on different intervals; description: Two exact plots of 3 sine theta plus 4 cosine theta. The full-cycle panel from 0 to 360 degrees reaches 5 at 36.9 degrees and minus 5 at 216.9 degrees. The panel restricted to 0 through 30 degrees rises from 4 to 4.964 and does not reach the amplitude 5. Dashed horizontal lines mark plus and minus 5.]
The same amplitude can give different attainable extreme values on different intervals

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