The Harmonic Form R sin(x + α)

Two waves that add to one

Take the expression 3 sin x + 4 cos x. It is the sum of two waves: a sine wave of height 3 and a cosine wave of height 4. They do not peak at the same x, since 3 sin x peaks at 90° and 4 cos x peaks at 0°. Added together, they make a single wave of a new height, shifted along the x-axis.

A single sine wave of height R, shifted to the left by $\alpha$, is $R \sin\left(x + \alpha\right)$. Expand it with the compound angle formula: $R \sin\left(x + \alpha\right) = R\left(\sin x \cos \alpha + \cos x \sin \alpha\right) = \left(R \cos \alpha\right) \sin x + \left(R \sin \alpha\right) \cos x$.

That has the same shape as a sin x + b cos x: a number times sin x plus a number times cos x. So the question is which R and which $\alpha$ make the two numbers equal to a and b.

With a = 4 and b = 1.5, the two plain dashed waves are 4 sin x and 1.5 cos x, and the gold wave is their sum. It is one wave of height $R = \sqrt{16 + 2.25} = 4.27$, shifted left by $\alpha = 20.6^\circ$. On the right, the triangle with legs a and b has hypotenuse R and angle $\alpha$. Drag its corner to a = 3 and b = 4: the sum becomes 5 sin(x + 53.1°).

Match the coefficients

For a sin x + b cos x to equal $\left(R \cos \alpha\right) \sin x + \left(R \sin \alpha\right) \cos x$ at every x, the numbers in front of sin x must match, and the numbers in front of cos x must match: $a = R \cos \alpha$ and $b = R \sin \alpha$.

To see why they must match, put in two values of x. At x = 0, sin x = 0 and cos x = 1, so the left side is b and the right side is $R \sin \alpha$. At x = 90°, sin x = 1 and cos x = 0, so the left side is a and the right side is $R \cos \alpha$.

Find R and $\alpha$

Square both equations and add them: $a^{2} + b^{2} = R^{2}\cos^{2}\alpha + R^{2}\sin^{2}\alpha = R^{2}\left(\cos^{2}\alpha + \sin^{2}\alpha\right)$. The bracket is 1, so $a^{2} + b^{2} = R^{2}$, and $R = \sqrt{a^{2} + b^{2}}$, taking R to be positive.

Divide the second equation by the first: $R \sin \alpha / R \cos \alpha = b / a$. The R cancels, leaving $\tan \alpha = \frac{b}{a}$.

Both results come from one right triangle. Its legs are $a = R \cos \alpha$ and $b = R \sin \alpha$, its hypotenuse is R, and $\alpha$ is the angle between the hypotenuse and the leg a. R is Pythagoras on that triangle, and $\tan \alpha$ is the opposite leg over the adjacent one.

For 3 sin x + 4 cos x, the triangle has a horizontal leg of a = 3 and a vertical leg of b = 4. Its hypotenuse is $R = \sqrt{9 + 16} = 5$, and the angle $\alpha$ has $\tan \alpha = \frac{4}{3}$.

3 sin x + 4 cos x as one wave

Here a = 3 and b = 4. So $R = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$, and $\tan \alpha = \frac{4}{3}$. Both $R \cos \alpha = 3$ and $R \sin \alpha = 4$ are positive, so $\alpha$ is acute, and $\alpha = \tan^{-1}\left(\frac{4}{3}\right) = 53.13^\circ$ to 2 decimal places, which is 0.927 radians. So 3 sin x + 4 cos x = 5 sin(x + 53.13°).

Check by substitution. At x = 0 the left side is 3 × 0 + 4 × 1 = 4, and 5 sin 53.13° = 5 × 0.8 = 4. At x = 0.3 radians the left side is 3 sin 0.3 + 4 cos 0.3 = 4.7079, and 5 sin(0.3 + 0.927) = 4.7079 as well. At x = 1.1 radians both sides are 4.4880.

In the same way $\sqrt{3} \sin x + \cos x$ has $R = \sqrt{3 + 1} = 2$ and $\tan \alpha = \frac{1}{\sqrt{3}}$, so $\alpha = 30^\circ$, and $\sqrt{3} \sin x + \cos x = 2 \sin\left(x + 30^\circ\right)$. At x = 0 the left side is 1, and 2 sin 30° = 1.

The greatest and least values

A sine is never more than 1 or less than −1, so $R \sin\left(x + \alpha\right)$ lies between −R and R. The greatest value of 3 sin x + 4 cos x is 5, not 3 + 4 = 7: the two waves never reach their peaks at the same x.

The greatest value comes where the bracket is 90°: x + 53.13° = 90°, so x = 36.87°. Check: 3 sin 36.87° + 4 cos 36.87° = 3 × 0.6 + 4 × 0.8 = 5. The least value, −5, comes where the bracket is 270°, at x = 216.87°.

Over one turn, with one square across for every 90°: the plain curve that starts at 0 is y = 3 sin x, the plain curve that starts at 4 is y = 4 cos x, and the gold curve is their sum, y = 3 sin x + 4 cos x. The gold wave has height 5, with its highest point, the first dot, at 36.87° and its lowest, the second dot, at 216.87°.

A negative coefficient

When a or b is negative, find $\alpha$ from the signs of $\cos \alpha$ and $\sin \alpha$, not from the inverse tangent alone. For 3 sin x − 4 cos x, $R \cos \alpha = 3$ and $R \sin \alpha = -4$. R is still 5. The cosine of $\alpha$ is positive and its sine is negative, so $\alpha$ is a negative acute angle: $\alpha = -53.13^\circ$, and 3 sin x − 4 cos x = 5 sin(x − 53.13°). At x = 0.3 radians both sides are −2.9348.

The same wave as a cosine

The same expression can be written as a shifted cosine wave, $R \cos\left(x - \beta\right)$. Expand it: $R \cos\left(x - \beta\right) = \left(R \cos \beta\right) \cos x + \left(R \sin \beta\right) \sin x$. Matching with 3 sin x + 4 cos x gives $R \cos \beta = 4$ and $R \sin \beta = 3$, so R = 5 again and $\tan \beta = \frac{3}{4}$, which gives $\beta = 36.87^\circ$.

So 3 sin x + 4 cos x = 5 cos(x − 36.87°). A cosine is greatest where its angle is 0, so this form shows the peak directly: at x = 36.87°, the same place as before. At x = 0.3 radians it gives 4.7079 too. A question names the form it wants, and the method is the same for each: expand, match coefficients, then find R and the angle.

The usual mistakes

Adding the coefficients. R is $\sqrt{3^{2} + 4^{2}} = 5$, not 3 + 4 = 7, and not $\sqrt{3 + 4} = \sqrt{7}$. Square, add, then take the root.

Putting $\tan \alpha$ the wrong way up. b goes with $\sin \alpha$, so $\tan \alpha = \frac{b}{a} = \frac{4}{3}$. With $\frac{3}{4}$ the shift would be 36.87°, which belongs to the cosine form.

Mixing degrees and radians. $\alpha = 53.13^\circ$ is 0.927 radians; with x in radians, write 5 sin(x + 0.927).

Taking the greatest value where $x = \alpha$. The peak is where $x + \alpha = 90^\circ$, at x = 36.87°, not at 53.13°.

Two currents in one wire

In the application below, currents of 3 sin x and 4 cos x amps join in one wire. Writing the total as $5 \sin\left(x + \alpha\right)$ shows the peak current is 5 amps, not 7, and $\alpha = 0.927$ radians is how far the combined wave is shifted.