An expression such as asintheta+bcostheta mixes a sine and a cosine of the same angle. The R-formula rewrites it as a single trigonometric function, so its greatest and least values can be read off directly. The graph of y=sintheta+costheta is itself a single sine wave whose amplitude R is bigger than 1 — the R-formula finds that single-wave form exactly. **The R-formula (a>0, b>0, R>0, alpha acute)** **Expression** | **Single-function form** asintheta+- bcostheta | Rsin(theta+-alpha) acostheta+- bsintheta | Rcos(theta) In both cases R=sqrt(a2+b2) and tanalpha= ba, where a is the coefficient of the ratio named in the answer. - Write the target form and expand it: Rsin(theta+alpha)=Rsin +Rcos. - Compare the coefficients of sintheta and costheta to get two equations. - Square and add the two equations — cos2alpha+sin2alpha=1 leaves R2=a2+b2. - Divide the equations to get tanalpha= ba, then take the acute value of alpha. **Comparison in full for asintheta+bcostheta=Rsin(theta+alpha)** **Step** | **Working** Match sintheta and costheta | Rcosalpha=a —— \(①\) Rsinalpha=b —— \(②\) \(①\)² + \(②\)² | R2(cos2alpha+sin2alpha)=a2+b2 R=sqrt(a2+b2) \(②\) ÷ \(①\) | tanalpha= ba alpha=tan-1 ba **Maximum and minimum values** Since -1 0 and b>0: match 4sintheta-3costheta to asintheta-bcostheta=Rsin(theta-alpha) with a=4, b=3; never carry negative values into R or alpha.

The R-formula

Secondary 4

TGM Original Questions

An expression such as \(a\sin\theta+b\cos\theta\) mixes a sine and a cosine of the same angle. The R-formula rewrites it as a single trigonometric function, so its greatest and least values can be read off directly.

The graph of \(y=\sin\theta+\cos\theta\) is itself a single sine wave whose amplitude \(R\) is bigger than \(1\) — the R-formula finds that single-wave form exactly.

The R-formula (\(a>0\), \(b>0\), \(R>0\), \(\alpha\) acute)

Expression	Single-function form
\(a\sin\theta\pm b\cos\theta\)	\(R\sin(\theta\pm\alpha)\)
\(a\cos\theta\pm b\sin\theta\)	\(R\cos(\theta\mp\alpha)\)

In both cases \(R=\sqrt{a^2+b^2}\) and \(\tan\alpha=\dfrac{b}{a}\), where \(a\) is the coefficient of the ratio named in the answer.

Method: Expand and compare coefficients

Write the target form and expand it: \(R\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha\).
Compare the coefficients of \(\sin\theta\) and \(\cos\theta\) to get two equations.
Square and add the two equations — \(\cos^2\alpha+\sin^2\alpha=1\) leaves \(R^2=a^2+b^2\).
Divide the equations to get \(\tan\alpha=\dfrac{b}{a}\), then take the acute value of \(\alpha\).

Comparison in full for \(a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)\)

Step	Working
Match \(\sin\theta\) and \(\cos\theta\)	\(R\cos\alpha=a\) —— \(①\) \(R\sin\alpha=b\) —— \(②\)
\(①\)² + \(②\)²	\(R^2(\cos^2\alpha+\sin^2\alpha)=a^2+b^2\) \(R=\sqrt{a^2+b^2}\)
\(②\) ÷ \(①\)	\(\tan\alpha=\dfrac{b}{a}\) \(\alpha=\tan^{-1}\dfrac{b}{a}\)

Maximum and minimum values

Since \(-1\le\sin(\theta+\alpha)\le 1\), the whole expression lies between \(-R\) and \(R\).

Single-function form	Maximum	Minimum
\(R\sin(\theta+\alpha)\)	\(R\), when \(\theta+\alpha=90^\circ\)	\(-R\), when \(\theta+\alpha=270^\circ\)
\(R\cos(\theta-\alpha)\)	\(R\), when \(\theta-\alpha=0^\circ\)	\(-R\), when \(\theta-\alpha=180^\circ\)

Exam check

The R-formula is not in the exam formula list — derive it each time by expanding and comparing coefficients. The template needs \(a>0\) and \(b>0\): match \(4\sin\theta-3\cos\theta\) to \(a\sin\theta-b\cos\theta=R\sin(\theta-\alpha)\) with \(a=4\), \(b=3\); never carry negative values into \(R\) or \(\alpha\).

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