Simplifying a cos x + b sin x
Master how to simplify \(a\cos x + b\sin x\) in Year 11 VCE Specialist Mathematics. Any sum of a cosine and a sine wave of the same period can be rewritten as one single sinusoid, making its amplitude and phase shift plain to see. It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Trigonometry topic of Unit 2.
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. 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.