We know that [(cos theta,- sin theta),(sin theta,cos theta)]^(n)=[(cos n theta,-sin n theta),(sin n theta,cos n theta)] :. [("cos" (2pi)/7,-"sin" (2pi)/7),("sin"(2pi)/7,"cos"(2pi)/7)]^(k)=[("cos"(2kpi)/7,-"sin"(2kpi)/7),("sin" (2kpi)/7,"...
सिद्ध कीजिए कि : cos.(2pi)/(3)cos.(pi)/(4)-sin.(2pi)/(3)sin.(pi)/(4)=(-(sqrt3+1))/(2sqrt2)
Answer to: Verify the identity: sin(pi/2 + x) = cos x By signing up, you'll get thousands of step-by-step solutions to your homework questions. You...
sin-pi/5=?cos-pi/5=?总共15个问题.问题可能比较多,但有30分奖励,加油吧大侠们.注意:这里的pi是指圆周率3.14.,2pi就是2乘pi的意思,pi/2就是pi除以2的意思,“-”表示负号的意思,例如-pi/3表示负的pi除以3.我记得初中有份表的,但不记得了,请高手门帮帮手.答案形式就是例如:cospi= -1,sinpi= 0...
Answer to: Find all solutions in the interval [0,2\pi] \cos 2x + \sin x = 1 x = By signing up, you'll get thousands of step-by-step solutions...
函数f(x)=sin(x)cos^2(x)在区间[-π,π]上的傅里叶级数展开为一个简洁的形式。首先，我们通过三角恒等变换将原函数简化为一个可直接识别的形式：sin(x)cos^2(x)=1/2sin(2x)cos(x)。进一步应用三角恒等式，可以将其转换为1/4(sin(3x)+sin(x))。由此，我们可以看出函数f(x)在特定区间...
百度试题 结果1 题目若\sin (\frac{\pi}{2} \alpha )=-\frac{3}{5},则\cos (2\pi-\alpha )=(\quad)A. -\frac{3}{5} B. \frac{3}{5} C. -\frac{4}{5} D. \frac{4}{5} 相关知识点: 试题来源: 解析 A 反馈 收藏
Since Cos2 x + Sin2 x = 1, Sin2 x = 1-cos2 x. So you have 2 cos x -(1-cos2 x) = cos2 x 2cos x - 1 + cos2 x = cos2x 2cos x -1 = 0 cos x = 1/2 Since cosine is positive, you have an answer in the first and the other in the fourth quadrant. You c...
平时习惯用matlab,所以在使用Python时直接使用了cos,sin,exp等函数,运行发现报错未定义函数,用from math importpi,sin,cos,exp定义后会报以下错误。 只需要把定义改成from numpy.ma import exp,sin,cos即可。 MATLAB二维图绘制 MatLab的二维绘图功能一、plot函数作图 x=0:pi/200:2*pi;sin=sin(x);cos=cos(x...
函数y=cos[2(pi)t]的最小正周期：T1=2(pi)/2(pi)=1；函数y=sin10t 的最小正周期：T2=2(pi)/10=pi/5.T1,T2 的最小公倍数为：pi.所以复合函数f(x)=cos[2(pi)t]+(sin10t)的周期为：pi.