Sqrt[2] * { Sin( x + π/4 ) * cos[ π/4 ] + Cos[ x + π/4] * sin[ π/4 ] = Sqrt[2] *sin (x + π/2)我这一步看不懂..

sin(sqrt(x))的不定积分 lim x->0(sqrt(sin(1/x^2)) 求极限 若π＜x＜3π/2,则sqrt(tanx+sinx)+sqrt(tanx-sinx)=?A.2·sqrt(tanx)·sin(x/2-π/4) B.2·sqrt(tanx)·sin(x/2+π/4)C.-2·sqrt(tanx)·sin(x/2-π/4) D.-2sqrt(tanx)`·sin(x/2+π/4)sqrt根号 Sqrt[2] * { Sin( x + π/4 ) * cos[ π/4 ] + Cos[ x + π/4] * sin[ π/4 ] = Sqrt[2] *sin (x + π/2)我这一步看不懂.. 高数极限题,x->0的极限?f=x^2/(sqrt(1+x*sin(x))-sqrt(cos(x))); 4/3 求工程! 求∫sqrt(1+sin(x)^2)dx 区间为 0 到 pie/2 ∫sqrt(4-4*sin^2(x)) dx 范围[2pi/3,0] sqrt(1-sin（4x）)求0到pi/2上定积分 z=sin(sqrt(x^2+y^2))/sqrt(x^2+y^2),-8≤x,y≤8； 用matlaB怎么编程! (sqrt(cos(x))*cos(200x)+sqrt(abs(x))-0.7)*(4-x*x)^0.01,sqrt(9-x^2),-sqrt(9 MATLAB z=100000*sin(y*pi/180)/(17*sqrt(3+sin(2*x*pi/180)));是不是缺几个点 lim(x->0)∫sin(sqrt(t))dt/x^a, y=sin x + sqrt(3)cos x的周期 Matlab出现is not a valid expression or equation.clear;syms theta x;solve('9.8*x*x/(tan(theta)*tan(theta))-(2*240000*(sqrt(7)-sqrt(4*cos(theta)+5))-196*(25+100)*(sin(theta)-sqrt(3)/2))*x/((1/3*5+100)*tan(theta))+9.8*x*x-2*10*sin(theta)*(240000*(sqrt 已知cos(x)=sqrt(10)/10,x属于(0,pi/2),求sin(pi/4+2x)的值 方程x=sqrt(x+1)+sqrt(x+2)+sqrt(x+3)+sqrt(x+4)的解是代数数吗?其中,sqrt是开平方根方的意思. 如何用matlab画二维图y=sqrt(25/sin(x)^2-(5cot(x)-1.8)^2)-5; matlab绘制函数曲线x的取值范围是0到90y=-sin(x/2)^4-4sin(x/2)^2+sin(x/2)^3sqrt(sin(x/2)^2+4)+sin(x/2)sqrt(sin(x/2)^2+4)是不是函数输入也有错误啊