How this instrument works
sin(2θ) = 2sinθcosθ falls out of the ordinary sine-addition rule, sin(A + B) = sinA cosB + cosA sinB, the moment you stop treating A and B as different angles and let both equal θ. The two product terms on the right, sinθcosθ and cosθsinθ, are then the same expression written twice, so the addition rule doesn't hand back a sum of two different things — it hands back one thing, counted twice, which is what doubling means here in a very literal sense.
The identity shows up outside pure trigonometry in a way its cosine counterpart never quite matches: a projectile launched at speed v and angle θ above level ground lands at range R = v²sin(2θ) ⁄ g, ignoring air resistance, so that doubled-angle value sets the distance directly. A cannonball fired at 30° travels exactly as far as one fired at 60° — a mirror-image pairing that any two supplementary angles share — and the true maximum arrives at 45°, the one angle where doubling lands squarely on 90° and sine peaks at 1.
At the boundaries the identity behaves exactly as geometry demands: θ = 0° gives sin(2θ) = 0, since a zero angle has no sine to double, and θ = 90° returns to 0 again because 2θ has already swept a full 180°, where sine closes back to zero. Between those flat endpoints the curve rises to its single peak of 1 at 45°, then eases back down by 90° — one full arch traced in the same span where plain sinθ has only just reached its own peak and is still climbing toward it.
- Angle, θ takes your input — leave the dropdown on degrees for a plain 30 to mean 30°, or pick radians or a turn count instead.
- sin(2θ) updates the moment you finish typing; there is no separate button to press.
- Read the doubled angle's sine straight off the sin(2θ) field — it always lands somewhere between −1 and 1.
- Punch in θ = 45° as a checkpoint: sin(2θ) should come back as exactly 1, sine's own ceiling, confirming the unit you picked is being read correctly.
- Follow that with θ = 90° and watch sin(2θ) drop straight back to 0 — a full swing ahead of where θ itself has only just gotten to.
Worked example — θ = 30° gives sin(2θ) = 0.8660254
Picture the Angle, θ field holding 30°, a natural landmark for sine. The sheet converts that to radians — 0.5235987755982988 — then works out two floating-point figures: a sine of 0.49999999999999994 and a cosine of 0.8660254037844387, each one bit shy of the clean textbook values of 0.5 and √3⁄2. Multiplying those two together and doubling the result gives 0.8660254037844386, matching sin(60°) precisely, with no 60° angle ever entering the calculation.
The same figure sets how far a projectile travels: range R = v²sin(2θ) ⁄ g, ignoring air resistance. A ball leaving a launcher at 20 m/s and this same 30° angle travels R = 20² × 0.8660254037844386 ⁄ 9.8 ≈ 35.35 m — while a launch at 60° instead sends it exactly as far, since 60° and 120° are supplementary angles and their sines coincide. Only 45° pushes the doubled angle to exactly 90°, where sin(2θ) peaks at 1 and the range is greatest for that speed.
Questions
What does sin(2θ) actually mean?
It's the sine taken after the angle θ has already been doubled — very different from finding sinθ first and doubling that number afterward. Set θ = 30° and 2θ becomes 60°, whose sine is about 0.8660254; the identity 2sinθcosθ arrives at that same figure straight from what θ's sine and cosine already are, with no 60° lookup involved anywhere in the process.
Why isn't sin(2θ) just 2 times sin(θ)?
Sine curves rather than runs in a straight line, so scaling its input by 2 doesn't scale its output by 2 as well. Take θ = 30°: sinθ = 0.5, and doubling that naively suggests 1, yet sin(2θ) = sin(60°) is actually about 0.866, well below the guess. Push to θ = 60° and doubling sinθ ≈ 0.866 predicts roughly 1.732 — past sine's own ceiling of 1 — while the real sin(120°) settles back near 0.866.
Where does the 2sinθcosθ identity come from?
It's a direct consequence of expanding sin(A+B), not a rule needing its own separate derivation. Write out sinAcosB + cosAsinB, then let B stand for the same value as A: both cross terms become sinθcosθ, and adding a quantity to itself is exactly what doubling means — which is how the plain sum turns into 2sinθcosθ.
How does sin(2θ) connect to cos(2θ)?
They come from the same angle-addition step, one for sine and one for cosine, so sin²(2θ) + cos²(2θ) = 1 still holds, the same as for any angle. Where sin(2θ) peaks at 1 when θ = 45°, cos(2θ) hits exactly 0 at that same angle — the two doubled-angle curves reach their extreme and their zero at precisely the same spot.
Why does sin(2θ) reach its maximum at θ = 45°?
Because 2θ becomes exactly 90° there, the one angle where sine itself tops out at 1. Nudge θ even slightly past 45° and 2θ moves past 90°, so sin(2θ) starts falling again immediately — the peak is a single point, not a plateau, which is also why a projectile's range falls off quickly on either side of a 45° launch angle.
What does sin(2θ) have to do with projectile range?
The range of an object launched at speed v and angle θ, ignoring air resistance, is R = v²sin(2θ) ⁄ g — so the range depends on the doubled angle's sine, not θ's own sine. That is why 30° and 60° launches send a projectile the same distance, since sin(60°) equals sin(120°), while 45° alone reaches the maximum, because only then does 2θ land on 90° where sine peaks.