First-principles sine and cosine derivatives
Введено| English | Значение |
|---|---|
| radian limit/ˈreɪdɪən ˈlɪmɪt/ | предельный радиан |
How do its tiny coordinate changes lead to exact derivative formulas?
- A rotating point has sine and cosine coordinates.
- This lesson studies radian limit 弧度极限: A small-angle limit used with angles measured in radians.
The method
- In radians, sin h/h tends to 1 and (cos h−1)/h tends to 0 as h tends to zero.
- Using angle-addition formulas in the difference quotient gives d(sin x)/dx=cos x and d(cos x)/dx=−sin x.
- These limits determine the tangent gradients; they require radian input.
Why do the derivative formulas for sin and cos require radians?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The geometric limit sin h<h<tan h is a statement about arc lengths in radians. In degrees the limit is π/180, and every derivative picks up that factor.
A worked case
For sin x, [sin(x+h)−sin x]/h=sin x[(cos h−1)/h]+cos x[sin h/h], tending to cos x. For cos x the quotient is cos x[(cos h−1)/h]−sin x[sin h/h], tending to −sin x. Near zero, the unit-circle inequality sin h<h<tan h for 0<h<π/2 bounds sin h/h between cos h and 1; negative h has the same ratio. Also (cos h−1)/h=−2sin²(h/2)/h tends to zero, using sin(h/2)/(h/2)→1. At x=π/6, the sine gradient is √3/2 and the cosine gradient is −1/2. For sine at x=0, h=0.1 gives sin(0.1)/0.1≈0.998334; at h=−0.1 the ratio is the same.
First-principles sine and cosine derivatives
In radians, sin h/h tends to 1 and (cos h−1)/h tends to 0 as h tends to zero
Connect a derivative calculation to its limit or gradient sign interpretation.
Find the derivative of sin x at x=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The sine proof gives cos0=1.
Where it goes wrong
- A small-angle approximation is not an exact finite-h equality.
- The cosine derivative has a minus sign.
- If a variable θ is measured in degrees, d sin(πθ/180)/dθ=(π/180)cos(πθ/180); the radian formula cannot be copied without that factor.
- Do not use the derivative formula itself to prove the starting limit.
Not true: The derivative of cos x is positive sin x for radian x. The derivative of cos x is −sin x.
Find the derivative of cos x at x=π/2.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The cosine proof gives −sin(π/2)=−1.
The derivative of cos x is −sin x.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The derivative of cos x is −sin x.
Applying it
- Start with the original difference quotient and use angle addition before applying the two radian limits.
- Retain the fixed sin x/cos x factors and their signs.
- Check exact-point gradients against whether the graph is rising or falling; the proof applies at any real radian x, not only near x=0.
Find the derivative of sin x at x=π.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The sine proof gives cosπ=−1.
Match each radian limit to its value as h→0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The difference quotient for sine splits into cos x·(sin h/h)+sin x·((cos h-1)/h), and the two limits deliver cos x.
In your course
- AQA A-level Mathematics (7357), specification section G.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Radian limit: A small-angle limit used with angles measured in radians.
- In radians, sin h/h tends to 1 and (cos h−1)/h tends to 0 as h tends to zero.
- Watch for: A small-angle approximation is not an exact finite-h equality.