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For example, the derivative of the sine function is written sin ′ (a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle. All derivatives of circular trigonometric functions can be found from those of sin( x ) and cos( x ) by means of the quotient rule applied to functions such ...
For the sine function, we can handle other values. If θ > π /2, then θ > 1. But sin θ ≤ 1 (because of the Pythagorean identity), so sin θ < θ. So we have < <. For negative values of θ we have, by the symmetry of the sine function
A calculation confirms that z(0) = 1, and z is a constant so z = 1 for all x, so the Pythagorean identity is established. A similar proof can be completed using power series as above to establish that the sine has as its derivative the cosine, and the cosine has as its derivative the negative sine.
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle.
The sine and tangent small-angle approximations are used in relation to the double-slit experiment or a diffraction grating to develop simplified equations like the following, where y is the distance of a fringe from the center of maximum light intensity, m is the order of the fringe, D is the distance between the slits and projection screen ...
The derivative of x is the constant function with ... One proof of the chain rule begins by defining the derivative of the composite ... = x 2 sin(1/x) otherwise.
The original proof is based on the Taylor series expansions of the exponential function e z (where z is a complex number) and of sin x and cos x for real numbers x . In fact, the same proof shows that Euler's formula is even valid for all complex numbers x.
By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function x ↦ x 2 sin ( 1 / x ) {\displaystyle x\mapsto x^{2}\sin(1/x)} is a Darboux function even though it is not continuous at one point.