Verify the identity 1 - cos( ) / = sin( ) / sin ( ) / 1 + cos ( ). Verify the identity. sin (pi / 6 + x) + sin (pi / 6 - x) = cos x Verify the identity: 4sin^2x -1/2 sin x + 1 = 2 sin x - 1 Verify the identity. sin 3x = 3 sin x cos^2 x - sin^...
√ (square) root 平方根 ∵ since; because 因为 ∴ hence 所以 ∷ equals,as (proportion) 等于,成比例 ∠ angle 角 ⌒ semicircle 半圆 ⊙ circle 圆 ○ circumference 圆周 π pi 圆周率 △ triangle 三角形 ∪ union of 并,合集 ∩ intersection of 交,通集 ∑ (sigma) summation of 总和 ° ...
[x, x', sin(theta),cos(theta), theta'] -> [x, x', theta, theta'] Args: state: Augmented state vector [state_size]. Returns: Reduced state size [reducted_state_size]. """ifstate.ndim ==1: x, x_dot, sin_theta, cos_theta, theta_dot = stateelse: x = state[...,0].resh...
Answer to: Write out the first four terms of the Taylor series for f(x) = cos x centered at pi/2. Then write this series using summation notation...
# 需要导入模块: from autograd import numpy [as 别名]# 或者: from autograd.numpy importcos[as 别名]defcalc_constraint(self, theta, a, b, c, d, e, f1, f2):return- (anp.cos(theta) * (f2 - e) - anp.sin(theta) * f1 -
square root 3 / 2 sin x - 1/2 cos x Simplify the expression. sin^2x + 12 sin x + 36/ sin x + 6 Let x = 9 sin \theta, - \frac{\pi}{2} less than \theta less than \frac{\pi}{2}. Simplify the expression. \frac{x}{\sq...
ctrl_cost = T.square(u).sum(axis=-1)# x: (batch_size, 8)# x[..., 0:4]: qpos# x[..., 4:8]: qvel, time derivatives of qpos, not used in the cost.theta = x[...,0]# qpos[0]: angle of joint 0phi = x[...,1]# qpos[1]: angle of joint 1target_xpos = x[....
# 需要导入模块: from autograd import numpy [as 别名]# 或者: from autograd.numpy importcos[as 别名]defcalc_constraint(self, theta, a, b, c, d, e, f1, f2):return- (anp.cos(theta) * (f2 - e) - anp.sin(theta) * f1 -