Simplifying a cos x + b sin x
Learn to simplify a cos x + b sin x for Year 11 Specialist Mathematics in Queensland (QCAA). Any sum of a cosine and a sine wave can be rewritten as one single sinusoid, making its amplitude and phase shift plain to see.
You will find the amplitude and phase angle, convert to the R form, read off maximums and minimums, and use it to sketch graphs and solve equations like a cos x + b sin x = c — the key to modelling waves and oscillations later in the course.
Theory
Simplifying \(a\cos x+b\sin x\) means rewriting the sum of a cosine and a sine wave as one single sinusoid \(R\cos(x-\alpha)\) (or \(R\sin(x+\beta)\)) in Year 11 Specialist Mathematics (QCAA, Queensland). This one form reveals the amplitude, maximum, minimum and phase shift at a glance, and turns the equation \(a\cos x+b\sin x=c\) into a simple trig equation.
A sum such as \(a\cos x+b\sin x\) looks like two separate waves, but it is always a single sinusoid in disguise: one cosine (or sine) curve of a fixed amplitude, shifted sideways.
The amplitude \(R\) is the height of that combined wave, \(R=\sqrt{a^2+b^2}\). The phase angle \(\alpha\) is how far the wave is shifted, found from \(\tan\alpha=\dfrac{b}{a}\).
Writing \(a\cos x+b\sin x=R\cos(x-\alpha)\) is the R form (or auxiliary-angle form). Because a cosine sits between \(-1\) and \(1\), the sum runs between \(-R\) and \(R\): its maximum is \(R\) and its minimum is \(-R\).
The same expression can be written with sine, \(R\sin(x+\beta)\), where \(\tan\beta=\dfrac{a}{b}\). Cosine or sine, the amplitude \(R\) is identical — only the phase angle changes.
To write \(a\cos x+b\sin x\) as \(R\cos(x-\alpha)\) with \(R>0\), expand and match coefficients: \(a=R\cos\alpha\) and \(b=R\sin\alpha\). This gives
The combined wave then has a known maximum, minimum and turning points:
Convert \(a\cos x+b\sin x\) to the R form
- Read off \(a\) (the coefficient of \(\cos x\)) and \(b\) (the coefficient of \(\sin x\)).
- Amplitude: compute \(R=\sqrt{a^2+b^2}\) — keep it exact (a surd) unless a decimal is asked for.
- Phase angle: solve \(\tan\alpha=\dfrac{b}{a}\), then use the signs of \(a\) and \(b\) to fix the quadrant of \(\alpha\).
- Write and use it: state \(a\cos x+b\sin x=R\cos(x-\alpha)\); to solve \(=c\), divide by \(R\) and solve the resulting cosine equation over the interval.
Identify \(a=3\), \(b=4\); both positive, so \(\alpha\) is acute. Amplitude first:
| \(R\) | \(=\) | \(\sqrt{a^2+b^2}\) |
| \(=\) | \(\sqrt{3^2+4^2}\) | |
| \(=\) | \(\sqrt{9+16}\) | |
| \(=\) | \(\sqrt{25}=5\) |
Now the phase angle from \(\tan\alpha=\dfrac{b}{a}\):
| \(\tan\alpha\) | \(=\) | \(\dfrac{4}{3}\) |
| \(\alpha\) | \(=\) | \(\tan^{-1}\!\dfrac{4}{3}\) |
| \(=\) | \(53.13\ldots^\circ\approx 53^\circ\) |
\(3\cos x+4\sin x\approx 5\cos(x-53^\circ)\).
Here \(a=1\), \(b=\sqrt3\). Amplitude:
| \(R\) | \(=\) | \(\sqrt{1^2+(\sqrt3)^2}\) |
| \(=\) | \(\sqrt{1+3}\) | |
| \(=\) | \(\sqrt{4}=2\) |
Phase angle (both \(a,b>0\), so first quadrant):
| \(\tan\alpha\) | \(=\) | \(\dfrac{\sqrt3}{1}=\sqrt3\) |
| \(\alpha\) | \(=\) | \(\dfrac{\pi}{3}\) |
\(\cos x+\sqrt3\,\sin x=2\cos\!\left(x-\dfrac{\pi}{3}\right)\).
The combined wave runs between \(-R\) and \(R\), so find the amplitude:
| \(R\) | \(=\) | \(\sqrt{5^2+12^2}\) |
| \(=\) | \(\sqrt{25+144}\) | |
| \(=\) | \(\sqrt{169}=13\) |
The maximum is \(R\); the minimum is \(-R\):
| \(\max\) | \(=\) | \(R=13\) |
| \(\min\) | \(=\) | \(-R=-13\) |
Maximum \(13\), minimum \(-13\).
Write the left side as \(R\cos(x-\alpha)\): here \(a=b=1\), so
| \(R\) | \(=\) | \(\sqrt{1^2+1^2}=\sqrt2\) |
| \(\tan\alpha\) | \(=\) | \(\dfrac{1}{1}=1\) |
| \(\alpha\) | \(=\) | \(\dfrac{\pi}{4}\) |
Divide by \(\sqrt2\) and solve the cosine equation:
| \(\sqrt2\cos\!\left(x-\dfrac{\pi}{4}\right)\) | \(=\) | \(1\) |
| \(\cos\!\left(x-\dfrac{\pi}{4}\right)\) | \(=\) | \(\dfrac{1}{\sqrt2}\) |
| \(x-\dfrac{\pi}{4}\) | \(=\) | \(-\dfrac{\pi}{4},\ \dfrac{\pi}{4},\ \dfrac{7\pi}{4}\) |
| \(x\) | \(=\) | \(0,\ \dfrac{\pi}{2},\ 2\pi\) |
\(x=0,\ \dfrac{\pi}{2},\ 2\pi\) — three solutions.
Common pitfalls
Frequently asked questions
How do you write a cos x + b sin x as a single cosine?
Use \(a\cos x+b\sin x=R\cos(x-\alpha)\) with \(R=\sqrt{a^2+b^2}\) and \(\tan\alpha=\dfrac{b}{a}\), choosing \(\alpha\)'s quadrant from the signs of \(a\) and \(b\).
What is R in R cos(x minus alpha)?
\(R\) is the amplitude of the combined wave, \(R=\sqrt{a^2+b^2}\). It is always positive and is the same whether you use the cosine or the sine form.
How do you find the maximum of a cos x + b sin x?
The maximum equals \(R=\sqrt{a^2+b^2}\) and it occurs at \(x=\alpha\); the minimum is \(-R\) at \(x=\alpha+\pi\).
Can you use R sin(x + beta) instead of R cos(x minus alpha)?
Yes. Both give the same amplitude \(R=\sqrt{a^2+b^2}\); for the sine form \(\tan\beta=\dfrac{a}{b}\). Use whichever the question asks for.
How does the R form help solve a cos x + b sin x = c?
Rewrite the left side as \(R\cos(x-\alpha)\), divide both sides by \(R\), then solve the single cosine equation \(\cos(x-\alpha)=\dfrac{c}{R}\) over the given interval.
Why does my phase angle look wrong?
Almost always the quadrant. \(\tan\alpha=\dfrac{b}{a}\) has two possible angles; the signs of \(a=R\cos\alpha\) and \(b=R\sin\alpha\) tell you which one is correct.