Phasors: Rotating Arrows
At a right angle, 3 and 4 make 5
In the top plot, the light-blue arrow of length 3 (A1) is fixed at the 0-degree reference, and joined to its tip is the green arrow of length 4 (A2), turned by the phase φ. Combine the two sinusoids into one and its amplitude is the length R of the gold sum arrow from the start to the final tip; the thick curve in the lower graph is that sum sinusoid. Slide φ until the sum amplitude is exactly 5. That is precisely where the two arrows meet at a right angle (90 degrees).
One sinusoid, one arrow
A sinusoid A cos(ωt+φ) is the shadow of an arrow of length A, starting at angle φ, turning at a steady rate. When the arrow has turned by ωt, its shadow on the horizontal axis is the sinusoid's value at that instant. So a sinusoid that never stops swinging can be written, with the swinging set aside, by just two numbers, length and phase; this is the phasor A∠φ. At the same frequency every arrow turns at the same rate, so their angles relative to one another stay fixed.
Same frequency adds as arrows
Add two sinusoids of the same frequency and the result is also a sinusoid of that frequency; only its amplitude and phase are new. Both can be read straight off the vector sum of the two phasor arrows joined tip to tail. The sum arrow's horizontal component is A1 + A2 cosφ and its vertical component is A2 sinφ, so the sum amplitude is the hypotenuse R = √((A1+A2cosφ)² + (A2sinφ)²) = √(A1² + A2² + 2A1A2 cosφ). No trigonometric identities to memorize: one picture finishes the addition.
At a right angle, the root of the sum of squares
When the two arrows meet at 90 degrees, cosφ is 0, the middle term vanishes, and only R = √(A1² + A2²) remains. It is the hypotenuse of a right triangle, the familiar relation in which 3 and 4 make 5. This right-angle addition hides all over AC. In impedance, resistance R and reactance X add at 90 degrees to give |Z| = √(R²+X²); in power, real power P and reactive power Q add at 90 degrees to give apparent power S = √(P²+Q²). All of it comes from the same picture of arrows.
Back to the first screen
As you slid the phase φ, the sum arrow's length changed from 7 down to 1, and it was exactly 5 at φ = 90 degrees, when the two arrows met at a right angle. Lengths 3 and 4 at a right angle give a hypotenuse of 5, so that familiar right triangle was itself the addition of sinusoids. In the lower graph, too, the height of the thick sum curve is that very hypotenuse. One change of viewpoint, seeing a swinging sinusoid as a rotating arrow, turns a wrestling match with trigonometry into joining arrows, and impedance and the power triangle both come out of this one picture.