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 $\alpha$α, is $R\sin \left(x+\alpha \right)$R sin(x + α). Expand it with the compound angle formula: $R\sin \left(x+\alpha \right)=R\left(\sin x\cos \alpha +\cos x\sin \alpha \right)=\left(R\cos \alpha \right)\sin x+\left(R\sin \alpha \right)\cos x$R sin(x + α) = R(sin x cos α + cos x sin α) = (R cos α) sin x + (R sin α) 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 $\alpha$α 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 $R=\sqrt{16+2.25}=4.27$R = √(16 + 2.25) = 4.27, shifted left by $\alpha =20.6^{\circ }$α = 20.6°. On the right, the triangle with legs a and b has hypotenuse R and angle $\alpha$α. 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 $\left(R\cos \alpha \right)\sin x+\left(R\sin \alpha \right)\cos x$(R cos α) sin x + (R sin α) cos x at every x, the numbers in front of sin x must match, and the numbers in front of cos x must match: $a=R\cos \alpha$a = R cos α and $b=R\sin \alpha$b = R sin α.

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 $R\sin \alpha$R sin α. At x = 90°, sin x = 1 and cos x = 0, so the left side is a and the right side is $R\cos \alpha$R cos α.

Find R and $\alpha$α

Square both equations and add them: $a^2+b^2=R^2{\cos }^2\alpha +R^2{\sin }^2\alpha =R^2\left({\cos }^2\alpha +{\sin }^2\alpha \right)$a² + b² = R²cos²α + R²sin²α = R²(cos²α + sin²α). The bracket is 1, so $a^2+b^2=R^2$a² + b² = R², and $R=\sqrt{a^2+b^2}$R = √(a² + b²), taking R to be positive.

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

Both results come from one right triangle. Its legs are $a=R\cos \alpha$a = R cos α and $b=R\sin \alpha$b = R sin α, its hypotenuse is R, and $\alpha$α is the angle between the hypotenuse and the leg a. R is Pythagoras on that triangle, and $\tan \alpha$tan α 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 $R=\sqrt{9+16}=5$R = √(9 + 16) = 5, and the angle $\alpha$α has $\tan \alpha =\frac{4}{3}$tan α = 4/3.

3 sin x + 4 cos x as one wave

Here a = 3 and b = 4. So $R=\sqrt{3^2+4^2}=\sqrt{25}=5$R = √(3² + 4²) = √25 = 5, and $\tan \alpha =\frac{4}{3}$tan α = 4/3. Both $R\cos \alpha =3$R cos α = 3 and $R\sin \alpha =4$R sin α = 4 are positive, so $\alpha$α is acute, and $\alpha ={\tan }^{-1}\left(\frac{4}{3}\right)=53.13^{\circ }$α = tan⁻¹(4/3) = 53.13° 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 $\sqrt{3}\sin x+\cos x$√3 sin x + cos x has $R=\sqrt{3+1}=2$R = √(3 + 1) = 2 and $\tan \alpha =\frac{1}{\sqrt{3}}$tan α = 1/√3, so $\alpha =30^{\circ }$α = 30°, and $\sqrt{3}\sin x+\cos x=2\sin \left(x+30^{\circ }\right)$√3 sin x + cos x = 2 sin(x + 30°). 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 $R\sin \left(x+\alpha \right)$R sin(x + α) 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 $\alpha$α from the signs of $\cos \alpha$cos α and $\sin \alpha$sin α, not from the inverse tangent alone. For 3 sin x − 4 cos x, $R\cos \alpha =3$R cos α = 3 and $R\sin \alpha =-4$R sin α = −4. R is still 5. The cosine of $\alpha$α is positive and its sine is negative, so $\alpha$α is a negative acute angle: $\alpha =-53.13^{\circ }$α = −53.13°, 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, $R\cos \left(x-\beta \right)$R cos(x − β). Expand it: $R\cos \left(x-\beta \right)=\left(R\cos \beta \right)\cos x+\left(R\sin \beta \right)\sin x$R cos(x − β) = (R cos β) cos x + (R sin β) sin x. Matching with 3 sin x + 4 cos x gives $R\cos \beta =4$R cos β = 4 and $R\sin \beta =3$R sin β = 3, so R = 5 again and $\tan \beta =\frac{3}{4}$tan β = 3/4, which gives $\beta =36.87^{\circ }$β = 36.87°.

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 $\sqrt{3^2+4^2}=5$√(3² + 4²) = 5, not 3 + 4 = 7, and not $\sqrt{3+4}=\sqrt{7}$√(3 + 4) = √7. Square, add, then take the root.

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

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

Taking the greatest value where $x=\alpha$x = α. The peak is where $x+\alpha =90^{\circ }$x + α = 90°, 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)$5 sin(x + α) shows the peak current is 5 amps, not 7, and $\alpha =0.927$α = 0.927 radians is how far the combined wave is shifted.