Combining a sine and a cosine into a single wave
For real a and b, not both zero, write a sin x + b cos x as R sin(x + α) with R > 0. Choose α from both coefficient signs. Use the single-wave form to find its range, locate maxima and minima, and solve equations on a stated interval.
What a learner can do afterwards
- Combine same-angle sine and cosine terms by matching coefficients, including zero or negative coefficients, and check a proposed form
- Find the range and locations of maxima and minima of a combined wave, including a constant offset or a stated x interval
- Solve single-wave equations on a stated interval, checking the sine range and keeping all valid solution branches
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As with double-angle identities, start from the sine addition rule. Work in degrees here. Expand R sin(x + α) = R cos α sin x + R sin α cos x. To match a sin x + b cos x, choose R cos α = a and R sin α = b. Square and add: R²(cos²α + sin²α) = a² + b². Since the bracket is 1, choose R = √(a² + b²) > 0 when the coefficients are not both zero.
For a sin x + b cos x with R > 0, choose cos α = a/R and sin α = b/R. Their signs select the quadrant. If a ≠ 0, tan α = b/a, but inverse tangent alone may select the other quadrant. Choose 0° ≤ α < 360°, unless another interval is stated. Since sin 45° = cos 45° = 1/√2, equal positive values give α = 45°; a negative cosine value and a positive sine value of equal magnitude give α = 135°.
If a = 0 and b > 0, choose α = 90°; if a = 0 and b < 0, choose α = 270°. If a = b = 0, use R = 0 and do not divide by R. Both terms must use the same angle for this coefficient-matching rule. A shifted sine changes sign when x increases by 180°, since sin(y + 180°) = -sin y. An identity must hold at every x, so one unequal pair of values disproves it.
For 7 sin x + 24 cos x, R = √(49 + 576) = 25. Matching gives cos α = 7/25 and sin α = 24/25. Both are positive, so α is acute. Thus tan α = 24/7 and α ≈ 73.7°. The expression is 25 sin(x + α). Keep the unrounded angle for later calculations.
For R > 0 and unrestricted real x, the range is -R to R. Maxima occur at x + α = 90° + 360°n; minima occur at 270° + 360°n, for integer n. Adding c shifts the range to c - R through c + R. On a restricted interval, check allowed peaks, troughs and endpoints. Sine increases from 0° to 90°.
For y = x + α, these values give the lower endpoint, centre and upper endpoint.
To solve c + R sin(x + α) = d with R > 0, subtract c and divide by R: sin(x + α) = k, where k = (d - c)/R. If |k| > 1, no real solution exists. Otherwise set β = arcsin k, with -90° ≤ β ≤ 90°. For y = x + α, the solutions are y = β + 360°n or y = 180° - β + 360°n, for integer n. Subtract α, keep only x in the stated interval and remove repeated answers.
Solve 6 sin(x + 20°) = 3 for 0° ≤ x < 360°. Divide by 6 to get sin y = 0.5, where y = x + 20°. The shifted interval is 20° ≤ y < 380°. The sine solutions in it are 30° and 150°. Subtract 20° to get x = 10° and 130°. Both answers lie in the stated x interval.
Match coefficients to find R and the phase signs. Use the sine range for extrema. For equations, find both angle families, subtract the phase and check the interval.
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More to read
Matching the two coefficients
Expanding a shifted sine gives R sin(x + α) = R cos α sin x + R sin α cos x. To match a sin x + b cos x, the two coefficients must match separately: R cos α = a and R sin α = b.
Square both equations and add them. The result is R²(cos²α + sin²α) = a² + b². Since cos²α + sin²α = 1, the positive amplitude is R = √(a² + b²), provided at least one coefficient is nonzero.
For 10 sin x + 24 cos x, R = √(100 + 576) = 26. The phase must satisfy cos α = 10/26 and sin α = 24/26. Both values are positive, so an acute phase fits. The amplitude is not 34: sine and cosine cannot both equal 1 at the same angle.
This calculation assumes fixed coefficients and the same angle in both terms. It does not describe a fading amplitude by itself.
Check both phase signs
A phase angle must match both coefficients. For -12 sin x + 16 cos x, the positive amplitude is 20. Thus cos α = -12/20 and sin α = 16/20. Cosine is negative and sine is positive, placing α between 90° and 180°.
The tangent ratio is -4/3. A calculator's inverse tangent gives about -53.1°, whose sine and cosine signs do not match the required pair. Adding 180° gives about 126.9°, which does. Check both signs after using inverse tangent.
Zero coefficients also work. For -9 cos x, R = 9, cos α = 0 and sin α = -1. The phase 270° matches these values. If both coefficients vanish, the whole expression is zero and R = 0. Do not form ratios by dividing by that zero amplitude.
For any fixed phase, R sin(x + α) has opposite values at inputs 180° apart. This gives a useful identity check: if a proposed expression has the same nonzero value at both inputs, it cannot be that shifted sine.
Carry the interval through the solution
First isolate the sine factor. For 4 + 8 sin(x + 15°) = 8, subtract 4 and divide by 8. This gives sin(x + 15°) = 0.5.
Suppose 0° ≤ x < 180°. Set y = x + 15°. The allowed interval becomes 15° ≤ y < 195°. Sine equals 0.5 at y = 30° + 360°n or y = 150° + 360°n, where n is an integer. Only 30° and 150° lie in this shifted interval. Subtract 15° to obtain x = 15° and 135°. Substituting either answer gives a sine value of 0.5.
A normalized right side outside -1 to 1 gives no real solutions. For a value inside that range, check both angle families. At -1 or 1, the two families overlap, so do not list an answer twice.
For c + R sin(x + α) with R > 0, the unrestricted maximum is c + R. It is 12 for the expression above. The angles attaining that maximum may lie outside a stated interval. Check the allowed peaks and troughs and evaluate included endpoints before choosing the largest or smallest value.
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