We have, f(x)=sinx.cosx=1/2sin2x therefore f^(')(x)=1/2.cos2x.2=cos2x Now, f^(')(x)=0 rArr cos2x=0 rArr cos2x=cospi/2 rArr x=pi/4 Also, f^('')(x)("at x"=pi//4) =-2 sin 2.pi/4=-2sinpi/2=-2lt0 At pi/4, f(x) is maximum and pi/4 is point ...
Maximum value of cosx (sinx +cos x) is equal to : 07:10 Find the maximum value of sinx +cosx and the value of x for which it i... 05:48 The maximum value of sinx+cosx is : 01:34 Maximum value of f(x) =sinx+cosx is : 01:23 What is the maximum value of the function si...
百度试题 结果1 题目Find the exact value of cosx if sinx=1/5andπ/(2)xπ 相关知识点: 试题来源: 解析 cosx=-(2√6)/5 反馈 收藏
In mathematics, the maximum value of a function is the largest value that the function takes on. We can determine the maximum value of a function by examining its range, where the range of a function, f(x), is all of the values of the outputs of the function, or all of the values ...
Let f(x)=sinx. The maximum value attained by f is. Select one: a. 1 b. 3 c. 5 d. 2 Range of Trigonometric Functions: The six trigonometric functions are defined based on the unit circle. Because of this, their domains and ranges can be quickly determined, and...
得cosx=x/r sinx=y/r 带入极坐标方程 r=1/(x/r+y/r) 化简得到直角坐标方程为 x+y=1 也就是直线 y=1-x与两个坐标轴围成的面积 很简单了 答案是1/2结果一 题目 SAT2数学题The area of the region enclosed by the graph of the polar curve r=1/(sinx+cosx)and the x- and y-axes is(...
Find the second derivative of y=(sinx-cosx)2.There are 2 steps to solve this one. Solution Share Step 1 Given that; y=(sin(x)−cos(x))2 Differentiate using the chain rule, where f(x)=x2 and g(x)=sin(x)−cos(x). 2(sin(x)−cos(x))d...
2 The equation of a curvcisy=(sinx)/(1+cosx) , for -x x元. Show that the gradient of the curve is positive片for all x in the given interval.[4] 相关知识点: 试题来源: 解析 Use correct quotient or product rule Obtain correct derivative in any form Use Pythagoras to simplify the ...
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Error Analysis of Numerical Method for sinx and cosx Using the Approximation to Bessel Functions Obtained by the Recurrence TechniqueThe functions sin x and cos x can be expressed as sin x = Σ~∞_K= σKJ_v+2K+1(x), cos x = Σ~∞_K=o γK Jv+2k(x) (σk, γk: coefficients ...