Two waves of the same frequency, however mixed, always collapse into a single wave — this is the algebra that proves it, and hands you the amplitude for free.
10 min read4 worked examples
Combining a cos θ + b sin θ into a single wave
An expression like $a\cos\theta + b\sin\theta$ mixes a cosine and a sine of the same angle — and it turns out this combination is always equivalent to a single, phase-shifted wave of the form $R\cos(\theta-\alpha)$ (or $R\sin(\theta+\alpha)$, or other sign variants), for suitable constants $R>0$ and $0<\alpha<90°$.
DerivationFinding R and α by expanding and comparing coefficients
1
Aim to write $a\cos\theta + b\sin\theta \equiv R\cos(\theta-\alpha)$, for constants $R>0$, $0<\alpha<90°$.
This is the target form; expand its right-hand side to see what R and α must satisfy.
2
Expand using the addition formula: $R\cos(\theta-\alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
cos(θ − α) = cos θ cos α + sin θ sin α.
3
Since this must equal $a\cos\theta+b\sin\theta$ for all $\theta$, match coefficients: $R\cos\alpha = a$ and $R\sin\alpha = b$.
Two expressions in cos θ and sin θ are identical for all θ only if their coefficients match exactly.
The same method (matching coefficients after expanding) works for any target form: $R\cos(\theta+\alpha)$, $R\sin(\theta\pm\alpha)$, and so on — only the addition formula used to expand, and hence the exact relationship between $R,\alpha$ and $a,b$, changes slightly. Always re-derive by expanding the specific target form given in a question, rather than memorising every variant.
Solving equations and finding maxima/minima
Once rewritten as $R\cos(\theta-\alpha)$, an equation like $a\cos\theta+b\sin\theta=c$ becomes a single trig equation, solvable by the usual methods. This form is equally powerful for optimisation: since $-1\le\cos(\theta-\alpha)\le1$ always, the expression $R\cos(\theta-\alpha)$ has:
Maximum and minimum
A maximum value of $R$, occurring when $\cos(\theta-\alpha)=1$, i.e. $\theta=\alpha$. A minimum value of $-R$, occurring when $\cos(\theta-\alpha)=-1$, i.e. $\theta=\alpha+180°$. This turns a calculus-style optimisation problem into simple reasoning about the range of cosine.
Worked example 1
Expressing in R cos(θ − α) form
Express $5\cos\theta + 12\sin\theta$ in the form $R\cos(\theta-\alpha)$, where $R>0$ and $0<\alpha<90°$. Give $\alpha$ to 1 decimal place.
State the maximum value of $5\cos\theta+12\sin\theta$, and the smallest positive value of $\theta$ at which it occurs.
MethodUse the R cos(θ − α) form found in Example 1
1
From Example 1, $5\cos\theta+12\sin\theta \equiv 13\cos(\theta-67.4°)$.
Reuse the R and α already found.
2
Since $-1\le\cos(\theta-67.4°)\le1$, the maximum value of the expression is $13$.
Maximum of R cos(θ − α) is R, since cosine's maximum value is 1.
3
This occurs when $\cos(\theta-67.4°)=1$, i.e. $\theta-67.4°=0° \Rightarrow \theta=67.4°$.
cos(x) = 1 exactly when x = 0° (the smallest non-negative solution).
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Worked example 3
Solving an equation using R cos/sin form
Solve $5\cos\theta+12\sin\theta = 6.5$ for $0 \le \theta < 360°$, giving answers to 1 d.p.
MethodSubstitute the R cos(θ − α) form, then solve as a standard equation
1
Using $13\cos(\theta-67.4°) = 6.5$ (from Example 1).
Rewrite the left-hand side using the derived R cos(θ − α) form.
2
$\cos(\theta-67.4°) = \dfrac{6.5}{13} = 0.5$.
Divide both sides by R = 13.
3
Let $u=\theta-67.4°$. Since $0\le\theta<360°$, then $-67.4°\le u <292.6°$. Solve $\cos u=0.5$ in this range: $u = \pm60°, 300°$, i.e. $u=-60°, 60°, 300°$.
Substitute to solve for u across the full transformed interval, checking all valid solutions including negative angles.
4
$\theta = u+67.4°$: $\theta = 7.4°, 127.4°, 367.4°$ (discard $367.4°$, out of range).
Reverse the substitution and discard any solution outside [0°, 360°).
5
$\theta = 7.4°$ or $\theta = 127.4°$ (1 d.p.).
Final answers within the given interval.
▪ solved
Examiner's eye
Widening the interval for the substituted variable $u$ (here to $-67.4° \le u < 292.6°$) is essential — solving $\cos u=0.5$ only in $[0°,360°)$ would miss the valid solution $u=-60°$, which corresponds to a genuine solution $\theta=7.4°$ in range.
Worked example 4
R sin(θ + α) form and a modelling context · Challenge
The height, in metres, of a tide above mean sea level is modelled by $h(t) = 3\sin(15t)° + 4\cos(15t)°$ for $t$ hours after midnight. Express $h(t)$ in the form $R\sin(15t°+\alpha)$, and hence find the maximum height and the first time after midnight it occurs.
MethodDerive R and α for the sin(θ+α) target form, then apply to the model
Maximum height is $R=5$ m, when $15t°+53.1°=90° \Rightarrow 15t°=36.9° \Rightarrow t\approx2.46$ hours ($\approx$ 02:28).
sin reaches its maximum of 1 when its argument is 90°; solve for the smallest positive t.
▪ solved
Look deeper — modelling power
This is exactly why R cos/sin form matters beyond pure algebra: physical systems that oscillate — tides, alternating current, sound waves — are frequently modelled as a sum of a sine and cosine term, and collapsing that sum into a single wave immediately reveals the true amplitude ($R$) and phase shift ($\alpha$) of the physical oscillation.
Common mistake
Mixing up which ratio gives $\tan\alpha$ depending on the target form — for $R\cos(\theta-\alpha)$ it's $\tan\alpha=\frac{b}{a}$, but for a different target form (e.g. $R\sin(\theta+\alpha)$) the roles of $a$ and $b$ can swap. Always re-derive by expanding the specific target form asked for in the question, rather than applying a memorised ratio blindly.
Check yourself
Express $8\cos\theta - 6\sin\theta$ in the form $R\cos(\theta+\alpha)$ where $R>0$, $0<\alpha<90°$, then state its minimum value and the smallest positive $\theta$ at which it occurs. (Hint: expand R cos(θ + α) using the addition formula and compare coefficients.)