Harmonic form: writing a cos x + b sin x as R cos(x minus alpha)

Harmonic form combines a sine term and a cosine term of the same angle into one shifted trigonometric function. For real constants $a$ and $b$, not both zero,

$$a\cos x+b\sin x=R\cos(x-\alpha),$$

where

$$\boxed{R=\sqrt{a^2+b^2}}, \qquad \boxed{R\cos\alpha=a}, \qquad \boxed{R\sin\alpha=b}.$$

The number $R>0$ is the amplitude and $\alpha$ is the phase shift. This form makes ranges, maximum and minimum values, equations and periodic models much easier to analyse.

Prerequisites

You should be able to:

• expand $\cos(A-B)$ and $\sin(A+B)$ using the compound-angle formulae;
• use exact trigonometric values and inverse trigonometric functions;
• choose an angle from the correct quadrant;
• solve basic trigonometric equations;
• interpret transformations of trigonometric graphs.

Why the method works

Expand the proposed form:

$$\begin{aligned} R\cos(x-\alpha) &=R(\cos x\cos\alpha+\sin x\sin\alpha)\\ &=(R\cos\alpha)\cos x+(R\sin\alpha)\sin x. \end{aligned}$$

For this to equal $a\cos x+b\sin x$ for every $x$, corresponding coefficients must match:

$$R\cos\alpha=a, \qquad R\sin\alpha=b.$$

Squaring and adding gives

$$R^2(\cos^2\alpha+\sin^2\alpha)=a^2+b^2,$$

so $R^2=a^2+b^2$. By convention $R$ is positive, hence $R=\sqrt{a^2+b^2}$.

The coefficient pair $(a,b)$ can be regarded as a vector of length $R$ making angle $\alpha$ with the positive horizontal axis. This explains both the Pythagorean formula for $R$ and why the signs of $a$ and $b$ determine the quadrant of $\alpha$.

Do not rely on tangent alone

Dividing the matching equations gives

$$\tan\alpha=\frac ba,$$

but tangent cannot distinguish opposite quadrants. Check that $\cos\alpha$ has the sign of $a$ and $\sin\alpha$ has the sign of $b$. A calculator’s two-input angle function, often written $\operatorname{atan2}(b,a)$, handles this directly.

Converting to cosine harmonic form

Use this reliable procedure:

1. Write $R\cos(x-\alpha)$ and expand it.
2. Match $R\cos\alpha=a$ and $R\sin\alpha=b$.
3. Calculate $R=\sqrt{a^2+b^2}$.
4. Find $\alpha$ in the quadrant fixed by the signs of $a$ and $b$.
5. Check by expanding your answer.

Worked example 1: exact phase angle

Write $3\cos x+3\sqrt3\sin x$ in the form $R\cos(x-\alpha)$, where $R>0$ and $0\leq\alpha<2\pi$.

Coefficient matching gives

$$R\cos\alpha=3, \qquad R\sin\alpha=3\sqrt3.$$

Therefore

$$R=\sqrt{3^2+(3\sqrt3)^2}=\sqrt{36}=6.$$

Both coefficients are positive, so $\alpha$ is in the first quadrant. Also

$$\cos\alpha=\frac36=\frac12, \qquad \sin\alpha=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2},$$

giving $\alpha=\pi/3$. Hence

$$\boxed{3\cos x+3\sqrt3\sin x=6\cos\left(x-\frac\pi3\right).}$$

Worked example 2: a phase angle in another quadrant

Write $-5\cos x+12\sin x$ as $R\cos(x-\alpha)$, with $0\leq\alpha<2\pi$. Give $\alpha$ to three decimal places.

First,

$$R=\sqrt{(-5)^2+12^2}=13.$$

Since $R\cos\alpha=-5$ and $R\sin\alpha=12$, cosine is negative while sine is positive. Thus $\alpha$ lies in the second quadrant. The reference angle is

$$\tan^{-1}\left(\frac{12}{5}\right)\approx1.176.$$

Therefore

$$\alpha=\pi-1.176\ldots=1.966\ldots,$$

and

$$\boxed{-5\cos x+12\sin x=13\cos(x-1.966)}$$

to three decimal places. The common incorrect answer $\alpha=-1.176$ has both sine and cosine negative after coefficient matching, so it cannot work.

Self-check 1

Write $4\cos x-4\sin x$ in the form $R\cos(x-\alpha)$, where $R>0$ and $0\leq\alpha<2\pi$.

Answer

Here $R=\sqrt{4^2+(-4)^2}=4\sqrt2$. Cosine is positive and sine is negative, so $\alpha$ is in the fourth quadrant. Hence $\alpha=7\pi/4$ and

$$\boxed{4\cos x-4\sin x=4\sqrt2\cos\left(x-\frac{7\pi}{4}\right).}$$

Equivalently, allowing a negative phase angle,

$$4\sqrt2\cos\left(x+\frac\pi4\right)$$

is the same expression.

Sine harmonic form

The same expression may instead be written as

$$a\cos x+b\sin x=R\sin(x+\beta).$$

Expanding gives

$$R\sin(x+\beta)=(R\sin\beta)\cos x+(R\cos\beta)\sin x,$$

so

$$\boxed{R\sin\beta=a}, \qquad \boxed{R\cos\beta=b}.$$

Notice that these coefficients occur in the opposite order from cosine harmonic form.

Worked example 3: choosing sine form

Write $5\cos x-5\sqrt3\sin x$ as $R\sin(x+\beta)$, where $R>0$ and $-\pi<\beta\leq\pi$.

We need

$$R\sin\beta=5, \qquad R\cos\beta=-5\sqrt3.$$

Thus $R=10$. Since sine is positive and cosine is negative, $\beta$ is in the second quadrant. The exact values

$$\sin\beta=\frac12, \qquad \cos\beta=-\frac{\sqrt3}{2}$$

give $\beta=5\pi/6$. Therefore

$$\boxed{5\cos x-5\sqrt3\sin x=10\sin\left(x+\frac{5\pi}{6}\right).}$$

Range and extreme values

Since $-1\leq\cos(x-\alpha)\leq1$,

$$\boxed{-R\leq a\cos x+b\sin x\leq R.}$$

Both endpoints are attained when $x$ is unrestricted. If a constant $c$ is added, then

$$\boxed{c-R\leq c+a\cos x+b\sin x\leq c+R.}$$

Worked example 4: maximum, minimum and where they occur

Find the maximum and minimum values of

$$f(x)=7+8\cos x+6\sin x,$$

and find the smallest non-negative $x$ at which each occurs.

Here

$$R=\sqrt{8^2+6^2}=10, \qquad \tan\alpha=\frac68=\frac34.$$

Both coefficients are positive, so $\alpha=\tan^{-1}(3/4)$. Thus

$$f(x)=7+10\cos(x-\alpha).$$

The maximum occurs when $\cos(x-\alpha)=1$:

$$x-\alpha=2k\pi.$$

Its value is $7+10=17$, first attained at

$$x=\alpha=\tan^{-1}\left(\frac34\right).$$

The minimum occurs when $\cos(x-\alpha)=-1$:

$$x-\alpha=(2k+1)\pi.$$

Its value is $7-10=-3$, first attained at

$$x=\pi+\tan^{-1}\left(\frac34\right).$$

Therefore

$$\boxed{\max f=17,\qquad \min f=-3.}$$

Restricted domains

On a restricted interval, $c+R$ and $c-R$ are only possible extreme values, not guaranteed ones. Check whether the required phase positions lie inside the interval. If they do not, compare the values at relevant endpoints and stationary phase positions.

Self-check 2

Find the range of $2-3\cos x+4\sin x$ for unrestricted real $x$.

Answer

The amplitude is

$$R=\sqrt{(-3)^2+4^2}=5.$$

Therefore

$$-5\leq-3\cos x+4\sin x\leq5,$$

and adding $2$ gives

$$\boxed{-3\leq2-3\cos x+4\sin x\leq7.}$$

Solving equations using harmonic form

Once two terms have been combined, solve the resulting single trigonometric equation. Keep the phase angle unrounded until the final answer.

Worked example 5: all solutions in an interval

Solve

$$3\cos x+4\sin x=2, \qquad 0\leq x<2\pi.$$

Write the left side as $5\cos(x-\alpha)$, where

$$\cos\alpha=\frac35, \qquad \sin\alpha=\frac45, \qquad \alpha=\tan^{-1}\left(\frac43\right).$$

The equation becomes

$$\cos(x-\alpha)=\frac25.$$

Let $y=x-\alpha$. The general cosine solutions are

$$y=\pm\cos^{-1}\left(\frac25\right)+2k\pi.$$

Therefore

$$x=\alpha\pm\cos^{-1}\left(\frac25\right)+2k\pi.$$

Numerically, $\alpha\approx0.9273$ and $\cos^{-1}(2/5)\approx1.1593$. Selecting values in $0\leq x<2\pi$ gives

$$x\approx0.9273+1.1593=2.0866$$

and, using $k=1$ for the negative branch,

$$x\approx0.9273-1.1593+2\pi=6.0512.$$

Hence

$$\boxed{x=2.087\text{ or }x=6.051}$$

to three decimal places.

Transform the interval too

If you solve in the substituted variable $y=x-\alpha$, then $0\leq x<2\pi$ becomes

$$-\alpha\leq y<2\pi-\alpha.$$

Using $0\leq y<2\pi$ without adjustment can lose one solution and introduce another outside the original interval.

Worked example 6: proving an equation has no solution

Determine whether

$$5\cos x-12\sin x=14$$

has any real solutions.

The amplitude of the left side is

$$R=\sqrt{5^2+(-12)^2}=13.$$

Therefore its range is $[-13,13]$. Since $14$ lies outside this range, the equation has

$$\boxed{\text{no real solutions}.}$$

This range check should come before any inverse trigonometric calculation.

Self-check 3

Solve

$$\cos x+\sin x=1, \qquad 0\leq x<2\pi.$$

Answer

Since

$$\cos x+\sin x=\sqrt2\cos\left(x-\frac\pi4\right),$$

we solve

$$\cos\left(x-\frac\pi4\right)=\frac1{\sqrt2}.$$

Thus

$$x-\frac\pi4=\pm\frac\pi4+2k\pi,$$

which gives

$$\boxed{x=0\text{ or }x=\frac\pi2.}$$

Modelling periodic quantities

An expression such as

$$M+A\cos(\omega t)+B\sin(\omega t)$$

has midline $M$, amplitude $\sqrt{A^2+B^2}$ and period $2\pi/|\omega|$. Harmonic form also identifies the time shift. The two trigonometric terms must have the same angle before they can be combined directly.

Worked example 7: interpreting a model

A model for the temperature, in degrees Celsius, $t$ hours after midnight is

$$T(t)=15-4\cos\left(\frac{\pi t}{12}\right)+3\sin\left(\frac{\pi t}{12}\right).$$

Let $x=\pi t/12$. The oscillating part has amplitude

$$R=\sqrt{(-4)^2+3^2}=5.$$

Hence the model predicts a minimum of $15-5=10^\circ\text{C}$ and a maximum of $15+5=20^\circ\text{C}$.

For cosine form, $\cos\alpha=-4/5$ and $\sin\alpha=3/5$, so $\alpha$ is in the second quadrant:

$$\alpha=\pi-\tan^{-1}\left(\frac34\right).$$

Thus

$$T(t)=15+5\cos\left(\frac{\pi t}{12}-\alpha\right).$$

The first maximum after midnight occurs when the cosine argument is zero:

$$\frac{\pi t}{12}=\alpha.$$

Therefore

$$t=\frac{12}{\pi}\left(\pi-\tan^{-1}\left(\frac34\right)\right)\approx9.54.$$

The predicted maximum occurs about $9.54$ hours after midnight, approximately 09:32.

Common misconceptions

• Using $R=a+b$. The amplitude is $\sqrt{a^2+b^2}$ because the coefficients are perpendicular components.
• Writing $R^2=a^2-b^2$. Squaring and adding the matched equations always produces a sum.
• Forgetting the sign in the compound angle. $\cos(x-\alpha)$ produces a positive sine coefficient, while $\cos(x+\alpha)$ produces a negative one.
• Accepting the calculator’s first arctangent value. Use the signs of both matched coefficients to select the quadrant.
• Rounding $\alpha$ too early. Store the full calculator value and round only final numerical answers.
• Combining different frequencies. In general, $a\cos x+b\sin 2x$ cannot be written as one sinusoid because the angles are different.
• Assuming $R$ is the maximum after a vertical shift. The range of $c+R\cos(x-\alpha)$ is $[c-R,c+R]$.

Mixed self-check

1. Write $-8\cos x-6\sin x$ as $R\cos(x-\alpha)$, where $0\leq\alpha<2\pi$.
2. Find the exact maximum and minimum values of $4+5\cos x+12\sin x$.
3. Solve $2\cos x-2\sin x=1$ for $0\leq x<2\pi$, giving answers to three decimal places.
4. Explain why $3\cos x+4\sin x=6$ has no real solution.

Answers

1. $R=10$. Both matched coefficients are negative, so $\alpha$ is in the third quadrant:

   $$\alpha=\pi+\tan^{-1}\left(\frac34\right).$$

   Therefore

   $$\boxed{-8\cos x-6\sin x=10\cos\left(x-\pi-\tan^{-1}\left(\frac34\right)\right).}$$

2. The amplitude is $\sqrt{5^2+12^2}=13$, so

   $$\boxed{\max=4+13=17,\qquad \min=4-13=-9.}$$

3. Use

   $$2\cos x-2\sin x=2\sqrt2\cos\left(x+\frac\pi4\right).$$

   Then

   $$\cos\left(x+\frac\pi4\right)=\frac1{2\sqrt2}.$$

   Selecting the solutions in the stated interval gives

   $$\boxed{x=0.424\text{ or }x=4.288.}$$

4. The amplitude is $\sqrt{3^2+4^2}=5$, so the left side can only lie between $-5$ and $5$. Since $6>5$, there is no real solution.

Next steps

• Practise using harmonic form inside trigonometric equations.
• Connect phase shifts and amplitudes with trigonometric graphs.
• Apply the method to periodic contexts in trigonometric proof and modelling.
• Review the identities behind the expansion in compound-angle and double-angle formulae.