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Results 1 - 10 of 448 for Tan (0.25 sec)

  1. 深度学习500问-Tan-00目录.docx

    Tan <******@****.***> 1530081549 +0800
    MS Word
    - Registered: 2021-01-17 15:33
    - Last Modified: 2018-06-27 07:00
    - 32.3K bytes
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  2. src/test/java/com/alibaba/druid/bvt/sql/eval/EvalMethodTanTest.java

        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 {
    Java
    - Registered: 2020-10-26 07:39
    - Last Modified: 2013-06-16 16:34
    - 1.2K bytes
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  3. 专业课/高等数学D/三角函数公式汇总.doc

    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...
    MS Word
    - Registered: 2021-01-17 02:06
    - Last Modified: 2018-11-22 14:19
    - 200.5K bytes
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  4. lib/subutil/src/main/java/com/blankj/subutil/util/PinyinUtils.java

                    .append("  lin   rang  chan  xun   yan   lei   ba    none...
    Java
    - Registered: 2021-01-15 00:33
    - Last Modified: 2019-07-10 11:46
    - 129.1K bytes
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  5. aten/src/ATen/NumericUtils.h

    }
    
    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);
    }
    
    C
    - Registered: 2021-01-17 05:39
    - Last Modified: 2020-11-02 19:49
    - 2.7K bytes
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  6. core/server/web/api/testmode/jobs/cpu-hog.js

    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()}`);
    JavaScript
    - Registered: 2021-01-12 17:59
    - Last Modified: 2020-11-23 04:17
    - 368 bytes
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  7. geodesy/haversine_distance.py

        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)
    Python
    - Registered: 2021-01-14 19:18
    - Last Modified: 2020-02-16 09:55
    - 2.3K bytes
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  8. 专业课/高等数学D/高数课件/Chap_4_2_不定积分的计算.pdf

    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
    PDF
    - Registered: 2021-01-17 02:06
    - Last Modified: 2018-11-22 14:19
    - 879.4K bytes
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  9. 专业课/高等数学D/历年考题/2010年高数期末试题(含答案).pdf

    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
    PDF
    - Registered: 2021-01-17 02:06
    - Last Modified: 2018-11-22 14:19
    - 380.7K bytes
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  10. base/special/trig.jl

        #             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)))
    
    Others
    - Registered: 2021-01-18 02:21
    - Last Modified: 2020-12-07 13:15
    - 40.3K bytes
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