Practice: Trig derivatives from geometry
Self-check
Section titled “Self-check”Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.
1. What are sin(x) and cos(x), geometrically, and what does differentiating them mean?
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A point on the unit circle at angle x (radians, counterclockwise from the positive horizontal axis) sits at (cos(x), sin(x)): cosine is its horizontal coordinate, sine its vertical coordinate. Differentiating them means asking how fast those two coordinates change as the point moves around the circle, which is the point’s velocity.
2. Why does the point move at “unit speed,” and why does that need radians?
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A radian is defined so that arc length along the unit circle equals the angle. So nudging the angle by dx moves the point an arc length of dx: it covers dx of distance per dx of angle, which is unit speed. In degrees, a full turn is 360 units of angle for 2π of arc, so the point would move at speed π/180, and every trig derivative would carry that ugly factor.
3. How do you get the velocity vector from the position vector?
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A point circling counterclockwise moves tangent to the circle, perpendicular to the radius, at length 1 (unit speed), pointing 90° counterclockwise from the position. Rotating any vector (a, b) by 90° counterclockwise gives (-b, a). Applied to the position (cos x, sin x), the velocity is (-sin x, cos x).
4. Read off both derivatives, and explain why the minus sign lands on cosine.
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The velocity components are the coordinates’ rates of change. Horizontal coordinate cos x changes at -sin x, so d/dx(cos x) = -sin x. Vertical coordinate sin x changes at cos x, so d/dx(sin x) = cos x. The minus is on cosine because as the point climbs counterclockwise through the right side of the circle, its horizontal coordinate is shrinking (moving leftward), so cosine’s rate is negative there.
5. Give one sanity check of d/dx(sin x) = cos x against the shape of the sine curve.
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At x = 0, the sine graph crosses the origin climbing; the formula says its slope is cos(0) = 1, a slope-1 climb, which matches. At x = π/2, sine is at its peak (slope should be 0); the formula gives cos(π/2) = 0, flat at the top, which matches. The derivative formula describes the slopes you can already see in the curve.
6. What is the small-angle approximation sin(x) ≈ x, and where does it come from?
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For small x, sin(x) is approximately x. It comes from the derivative at zero: sin passes through sin(0) = 0 with slope cos(0) = 1, so near zero the curve is nearly the straight line 0 + 1·x = x. It is the derivative telling you the curve’s initial direction (the first sliver of the Taylor-series idea).
Try it yourself, part 1: read the slopes at a new angle
Section titled “Try it yourself, part 1: read the slopes at a new angle”Pen and paper, about 6 minutes. Do the whole derivation at the specific angle x = π (pointing left along the negative horizontal axis), then check against the graphs.
Steps. (1) Write the position (cos π, sin π). (2) Rotate it 90° counterclockwise ((a, b) -> (-b, a)) to get the velocity. (3) Read off d/dx(sin x) and d/dx(cos x) at x = π. (4) Sanity-check each against the shape of its curve at x = π.
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position = (cos π, sin π) = (-1, 0)velocity = rotate (-1, 0) by 90° CCW = (-sin π, cos π) = (0, -1)Reading the components: d/dx(sin x) at π is cos π = -1; d/dx(cos x) at π is -sin π = 0.
Sanity check. At x = π, sine is crossing zero on its way down (from positive to negative), so its slope should be negative, and indeed cos π = -1 (descending at slope 1). Cosine at x = π is at its trough (-1, its lowest point), so its slope should be 0, flat, and indeed -sin π = 0. Both match the curves. The velocity (0, -1) says the circling point is moving straight down at that moment, which is exactly right when it is at the far-left of the circle.
Try it yourself, part 2: the small-angle approximation
Section titled “Try it yourself, part 2: the small-angle approximation”About 3 minutes (a calculator helps). The slope of sin at x = 0 is cos(0) = 1, so near zero sin(x) ≈ x. Use that to estimate sin(0.1) (radians), then compare to the actual value.
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The approximation gives sin(0.1) ≈ 0.1. The actual value is sin(0.1) ≈ 0.09983 (radians). The estimate is off by only about 0.0002, under a quarter of a percent, because near zero the sine curve hugs the line y = x whose slope is the derivative cos(0) = 1. The smaller the angle, the better the approximation, which is why sin(x) ≈ x is used constantly in physics and signal processing for small angles.
Flashcards
Section titled “Flashcards”Nine cards. Click any card to reveal the answer. Use the Print flashcards button to lay out the full set as one card per page, ready to print or save as a PDF for offline review.
Q. What are sin(x) and cos(x) geometrically?
The coordinates of a point on the unit circle at angle x: the point sits at (cos x, sin x). Cosine is the horizontal coordinate, sine the vertical. Their derivatives are the components of the point’s velocity as it circles.
Q. Why does the circling point move at unit speed, and why radians?
A radian makes arc length equal the angle, so nudging the angle by dx moves the point arc length dx: unit speed. In degrees the point would move at speed π/180, putting that ugly factor on every trig derivative.
Q. How do you get the velocity from the position on a circle?
Velocity is tangent to the circle (perpendicular to the radius), length 1, pointing 90° counterclockwise from the position. Rotating (a, b) by 90° CCW gives (-b, a), so the position (cos x, sin x) gives velocity (-sin x, cos x).
Q. What are the derivatives of sin and cos, and why the minus on cosine?
d/dx(sin x) = cos x and d/dx(cos x) = -sin x. The minus is on cosine because, as the point climbs counterclockwise through the right of the circle, its horizontal coordinate shrinks (moves left), so cosine’s rate is negative.
Q. Sanity-check d/dx(sin x) = cos x at x = 0 and x = π/2.
At x = 0, sine crosses the origin climbing with slope cos 0 = 1. At x = π/2, sine is at its peak with slope cos(π/2) = 0 (flat). Both match the curve.
Q. What is the small-angle approximation, and where does it come from?
sin(x) ≈ x for small x. It comes from the slope of sine at zero being cos(0) = 1, so near zero the curve hugs the line y = x. (The first sliver of Taylor series.)
Q. Why is sine the universal shape of oscillation in physics?
Differentiating sine twice gives -sin, so f'' = -f: a function whose acceleration pulls it back toward center in proportion to displacement. That equation governs springs, pendulums, sound, AC current, and light, all sine-shaped.
Q. What breaks if you use degrees instead of radians?
The point no longer moves at unit speed (it moves at π/180), so every trig derivative picks up a factor of π/180: d/dx(sin x) = (π/180)·cos x. Radians exist to make that factor 1 and keep trig calculus clean.
Q. Where do trig derivatives show up in machine learning?
Transformer positional encodings (Vaswani et al., 2017) use sin/cos waves to encode token position; 3D rotation matrices (differentiable rendering, pose estimation) use sin/cos; and Fourier analysis decomposes signals into sin/cos. Differentiating through any of these uses these derivatives.