用π/2=1+1!/3+2!/(3*5)+3!/(3*5*7)+4!/(3*5*7*9)+.+n!/(3*5*...*(2n+1))公式求出π的近似值
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用π/2=1+1!/3+2!/(3*5)+3!/(3*5*7)+4!/(3*5*7*9)+.+n!/(3*5*...*(2n+1))公式求出π的近似值
用π/2=1+1!/3+2!/(3*5)+3!/(3*5*7)+4!/(3*5*7*9)+.+n!/(3*5*...*(2n+1))公式求出π的近似值
用π/2=1+1!/3+2!/(3*5)+3!/(3*5*7)+4!/(3*5*7*9)+.+n!/(3*5*...*(2n+1))公式求出π的近似值
求近似值,给一个较大的n值直接算就行.差不多小数点后第四五位不变了就可以不用算了
sin0=0 sin(1/6π)=1/2sin(1/3π)=√3/2sin(1/2π)=1sin(2/3π)=√3/2sin(5/6π)=1/2sin(π)=0sin(-π)=0sin(-5/6π)=-1/2sin(-2/3π)=-√3/2sin(-1/2π)=-1sin(-1/3π)=-√3/2sin(-1/6π)=-1/2sin(7/6π)=-1/2sin(4/3π)=-√3/2sin(5/3π)=-√3/2sin(11/6π)=-1/
已知sin(π/2+α)-cos(3π/2-α)=1/5(π/2
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cos5π/3为什么等于1/2不是-1/2?cos5π/3=cos(5π/3-2π)=cos(-π/3)=-cosπ/3=-1/2不是这样的吗
用π/2=1+1!/3+2!/(3*5)+3!/(3*5*7)+4!/(3*5*7*9)+.+n!/(3*5*...*(2n+1))公式求出π的近似值
已知sina+cosa=1/5,且2π/3
函数y=tan(1/2x-π/3)的一个单调递增区间A、(-π/3,5π/3)B、(π/6,7π/6)C、(-2π/3,4π/3)D、(-π/6,5π/6)
用五点法作出 y=2sin(3x+π/5)-1的函数图像
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vb 根据公式计算π的近似值π/2=1+1/3+1/3*2/5+1/3*2/5*3/7
Matlab计算π/2=1+(1/3)+(1/3*2/5)+(1/3*2/5*3/7)+...怎么计算π的值?.
cos(π+α)=-1/3,sin(3π/2-α)=sin(π/3+α)=1/3,cos(5π/6+α)=sin(α-π/4)=1/3,cos(π/4+α)=
若cos(π+α)=-1/2,3π/2
已知cos(π+a)=-1/2,3π/2
若cos(π+a)=-1/2 ,3π/2
若cos(π+α)=-1/2,3/2π
若cos(π+α)=-1/2,3/2π
若cos(π+a)=-1/2,3/2π