Tan <******@****.***> 1530081549 +0800

    public void test_reverse() throws Exception {
        Assert.assertEquals(Math.tan(1), SQLEvalVisitorUtils.evalExpr(JdbcConstants.MYSQL, "tan(1)"));
        Assert.assertEquals(Math.tan(1.001), SQLEvalVisitorUtils.evalExpr(JdbcConstants.MYSQL, "tan(1.001)"));
        Assert.assertEquals(Math.tan(0), SQLEvalVisitorUtils.evalExpr(JdbcConstants.MYSQL, "tan(0)"));
    }
    
    public void test_error() throws Exception {

b a b a b a sin cos cos sin ) sin( × - × = - b a b a b a sin sin cos cos ) cos( × - × = + b a b a b a sin sin cos cos ) cos( × + × = - b a b a b a tan tan 1 tan tan ) tan( × - + = + b a b a b a tan tan 1 tan tan ) tan( × + - = - 五、二倍角公式 a a a cos sin 2 2 sin = a a a a a 2 2 2 2 sin 2 1 1 cos 2 sin cos 2 cos - = - = - = … ) ( * a a a 2 tan 1 tan 2 2 tan - = 二倍角的余弦公式 ) ( * 有以下常用变形：（规律：降幂扩角，升幂缩角） a a 2 cos 2 2 cos 1 = + a a 2 sin 2 2 cos 1 = - 2 ) cos (sin 2 sin 1 a a a + = + 2 ) cos (sin 2 sin 1 a a...

                .append("  lin   rang  chan  xun   yan   lei   ba    none...

}

template <typename T>
C10_HOST_DEVICE inline T tan(T x) {
  static_assert(!std::is_same<T, double>::value, "this template must be used with float or less precise type");
#if defined(__CUDA_ARCH__) || defined(__HIP_ARCH__)
  // use __tanf fast approximation for peak bandwidth
  return __tanf(x);
#else
  return ::tan(x);
#endif
}

template <>
C10_HOST_DEVICE inline double tan<double>(double x) {
  return ::tan(x);
}

const logging = require('../../../../../shared/logging');

(async () => {
    let sum = 0;

    logging.info(`Starting CPU intensive task at ${new Date()}`);

    for (let i = 0; i < 100000000; i++) {
        sum += Math.tan(Math.tan(i));
    }

    logging.info(`Calculation result: ${sum}`);

    logging.info(`Finishing CPU intensive task at ${new Date()}`);

    RADIUS = 6378137
    # Equation parameters
    # Equation https://en.wikipedia.org/wiki/Haversine_formula#Formulation
    flattening = (AXIS_A - AXIS_B) / AXIS_A
    phi_1 = atan((1 - flattening) * tan(radians(lat1)))
    phi_2 = atan((1 - flattening) * tan(radians(lat2)))
    lambda_1 = radians(lon1)
    lambda_2 = radians(lon2)
    # Equation
    sin_sq_phi = sin((phi_2 - phi_1) / 2)
    sin_sq_lambda = sin((lambda_2 - lambda_1) / 2)

对: 对数函数
幂: 幂函数
指: 指数函数
三: 三角函数
46/58
例6. 求
Peking University Press
解: 令 ,cosln xu 
x
v
2cos
1
 , 则
,tan xu  xv tan
原式 = xx coslntan   xx dtan
2
xx coslntan    xx d)1(sec
2
xx coslntan  Cxx  tan
机动 目录 上页 下页 返回 结束
47/58
例7. 求
Peking University Press
解: 令 ,tx 则 ,2tx  ttx d2d 
原式 tet t d2
tet(2
Cxe x  )1(2

3.
2 2
1
1
dx
x x 
=
2 2
2 2 2
sec cos sin 1 1
tan sec sin sin sin
tdt t d t xdt C C
t t t t t x

         
4. 3 2sin
1 cos
x x dx
x


2
2
3 4sin cos 32 2 sec 2 tan
2 2 22cos
2
3 (tan ) 2 tan 3 tan 3 tan 2 tan
2 2 2 2 2
sin
23 tan 2 3 tan ln cos
2 2 2 2cos
2
x xx x xdx x dx dxx
x x x x xxd dx x dx dx
x

    #             3      2      2       2       2
    #        r = y *(T2+y *(T3+y *(...+y *(T12+y *T13))))
    #       then
    #                     3    2
    #        tan(y+z) = y + (T1*y + (y *(r+z)+z))
    #
    #   4. For y in [0.67434,pi/4],  let z = pi/4 - y, then
    #        tan(y) = tan(pi/4-z) = (1-tan(z))/(1+tan(z))
    #               = 1 - 2*(tan(z) - (tan(z)^2)/(1+tan(z)))