Auxiliary angle: writing $a \sin x + b \cos x$ as $R \sin(x + \alpha)$

What this dot point is asking

NESA wants you to combine $a \sin x + b \cos x$ into a single sinusoidal expression of the form $R \sin(x + \alpha)$ or $R \cos(x \pm \alpha)$, find the amplitude $R$ and the phase $\alpha$, and use this to solve equations or find extreme values. The whole method rests on one fact: the sum of two sinusoids of the same frequency is itself a single sinusoid of that frequency. Adding a sine wave and a cosine wave does not give a lumpy curve; it gives one clean sine wave, just shifted sideways and scaled up. Finding $R$ and $\alpha$ is finding the amplitude and the sideways shift of that combined wave.

The answer

The two waves combine into one

It helps to see the result before the algebra. Take a concrete case, $3 \sin x + 4 \cos x$. Plot the two pieces separately.

Now add them at every value of $x$ (add the two heights). The result is a new wave.

That summed wave is not a new kind of curve. It is exactly a single sine wave of amplitude $5$, shifted to the left. In symbols, $3\sin x + 4\cos x = 5\sin(x + \alpha)$.

So the job is just to find the amplitude $R$ (here $5$) and the phase shift $\alpha$. Both come straight out of a right triangle, as we will see.

Deriving the form $R \sin(x + \alpha)$

Expand the target using the sine sum identity:

For this to equal $a \sin x + b \cos x$ for all $x$, the coefficient of $\sin x$ and the coefficient of $\cos x$ must match separately:

Square and add the two equations. Since $\cos^2\alpha + \sin^2\alpha = 1$,

Divide the two equations:

Geometrically, $a$, $b$ and $R$ are the two legs and the hypotenuse of a right triangle, and $\alpha$ is the angle at the origin. That is the picture to draw.

Crucially, $\tan\alpha = \frac{b}{a}$ alone has two candidate angles per turn, so choose $\alpha$ in the quadrant where $R\cos\alpha$ has the sign of $a$ and $R\sin\alpha$ has the sign of $b$. When both $a$ and $b$ are positive, $\alpha$ is in the first quadrant, which is the most common exam case.

The other equivalent forms

Depending on what the question asks, the same expression can be written with cosine or with a minus sign. Each comes from matching coefficients against the corresponding expansion:

The amplitude $R = \sqrt{a^2+b^2}$ is the same in every form; only the placement and sign of $\alpha$ change. If the exam does not specify a form, $R\sin(x+\alpha)$ with $R > 0$ and $\alpha$ in $\left(-\pi, \pi\right]$ is the safe default.

Why this matters: maxima, minima and equations

Once $a \sin x + b \cos x$ is a single sinusoid $R\sin(x + \alpha)$, three things become easy because $\sin$ of anything lives between $-1$ and $1$:

• The maximum value is $R$, reached when $x + \alpha = \frac{\pi}{2} + 2n\pi$ (i.e. $\sin = 1$).
• The minimum value is $-R$, reached when $\sin = -1$.
• An equation $a \sin x + b \cos x = c$ collapses to $\sin(x + \alpha) = \frac{c}{R}$, which you solve by the standard general solution and then restrict.
• Sketching is reduced to drawing one amplitude-$R$ sine curve shifted by $\alpha$.

A neat consequence: $a\sin x + b\cos x = c$ has no solution at all when $|c| > R = \sqrt{a^2+b^2}$, because $\sin(x+\alpha)$ cannot exceed $1$. Spotting that saves you from chasing roots that do not exist.

How exam questions ask about the auxiliary angle

• "Express $a\sin x + b\cos x$ in the form $R\sin(x+\alpha)$": match coefficients, give $R = \sqrt{a^2+b^2}$ and $\alpha$ in the stated quadrant. Watch the requested form and the stated range for $\alpha$.
• "Hence find the maximum/minimum value (and where it occurs)": the max is $R$, the min is $-R$; solve $x + \alpha = \frac{\pi}{2}$ (or $-\frac{\pi}{2}$) for the location.
• "Hence solve $a\sin x + b\cos x = c$": divide by $R$, take both general-solution branches of $\sin$, then restrict to the interval.
• "Sketch $y = a\sin x + b\cos x$": read off amplitude $R$, period $2\pi$, and a phase shift of $\alpha$ to the left.
• "Find the values of $c$ for which $a\sin x + b\cos x = c$ has solutions": the condition is $|c| \le R$.

Edge cases worth knowing

• A pure $R\cos$ answer is sometimes neater. If the question gives $a\sin x + b\cos x$ with $b$ much larger, the $R\cos(x - \alpha)$ form (with $\tan\alpha = \frac{a}{b}$) may put $\alpha$ in a tidier place. The amplitude is identical.
• A negative leading coefficient. For $-3\sin x + 4\cos x$ the matched equations give $R\cos\alpha = -3 < 0$ and $R\sin\alpha = 4 > 0$, so $\alpha$ is in the second quadrant. Trust the signs, not just $\tan\alpha$.
• Adding a constant shifts the extremes, not the amplitude. The maximum of $R\sin(x+\alpha) + k$ is $R + k$ and the minimum is $k - R$; the amplitude is still $R$.