回答:
说明:
回答:
另一种方法……
说明:
鉴于: -
#sintheta cdot costheta = 1/2#
#=> 2 cdot sintheta cdot costheta = 1#
#“所以,”#
#sintheta + costheta#
#= SQRT((sintheta + costheta)^ 2)#
#= sqrt(sin ^ 2theta + 2 cdot sintheta cdot costheta + cos ^ 2theta#
#= sqrt((sin ^ 2theta + cos ^ 2theta)+2 cdot sintheta cdot costheta#
#= SQRT(1 + 1)#
#=#SQRT2 希望能帮助到你…
谢谢…
:-)
找到theta的值,if,Cos(theta)/ 1 - sin(theta)+ cos(theta)/ 1 + sin(theta)= 4?
Theta = pi / 3或60 ^ @好的。我们得到:costheta /(1-sintheta)+ costheta /(1 + sintheta)= 4我们暂时忽略RHS。 costheta /(1-sintheta)+ costheta /(1 + sintheta)(costheta(1 + sintheta)+ costheta(1-sintheta))/((1-sintheta)(1 + sintheta))(costheta((1-sintheta) )+(1 + sintheta)))/(1-sin ^ 2theta)(costheta(1-sintheta + 1 + sintheta))/(1-sin ^ 2theta)(2costheta)/(1-sin ^ 2theta)根据毕达哥拉斯的身份,罪^ 2theta + cos ^ 2theta = 1。所以:cos ^ 2theta = 1-sin ^ 2theta现在我们知道了,我们可以写:(2costheta)/ cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ - 1(1/2)θ= pi / 3,当0 <=θ<= pi时。以度为单位,θ= 60 ^ @当0 ^ @ <= theta <= 180 ^ @
简化(1- cos theta + sin theta)/(1+ cos theta + sin theta)?
= sin(theta)/(1 + cos(theta))(1-cos(theta)+ sin(theta))/(1 + cos(theta)+ sin(theta))=(1-cos(theta)+ sin(theta))*(1 + cos(theta)+ sin(theta))/(1 + cos(theta)+ sin(theta))^ 2 =((1 + sin(θ))^ 2-cos ^ 2(theta))/(1 + cos ^ 2(θ)+ sin ^ 2(theta)+2 sin(theta)+2 cos(theta)+ 2 sin(theta)cos(theta))=((1+ sin(theta))^ 2-cos ^ 2(theta))/(2 + 2 sin(theta)+2 cos(theta)+ 2 sin(theta)cos(theta))=((1 + sin(theta) )^ 2-cos ^ 2(theta))/(2(1 + cos(theta))+ 2 sin(theta)(1 + cos(theta))=(1/2)((1 + sin(theta)) )^ 2-cos ^ 2(theta))/((1 + cos(theta))(1 + sin(theta))=(1/2)(1 + sin(theta))/(1 + cos(theta) )) - (1/2)(cos ^ 2(theta))/((1 + cos(theta))(
表明,(1 + cos theta + i * sin theta)^ n +(1 + cos theta - i * sin theta)^ n = 2 ^(n + 1)*(cos theta / 2)^ n * cos( n * theta / 2)?
请看下面。设1 + costheta + isintheta = r(cosalpha + isinalpha),这里r = sqrt((1 + costheta)^ 2 + sin ^ 2theta)= sqrt(2 + 2costheta)= sqrt(2 + 4cos ^ 2(theta / 2) )-2)= 2cos(theta / 2)和tanalpha = sintheta /(1 + costheta)==(2sin(theta / 2)cos(theta / 2))/(2cos ^ 2(theta / 2))= tan (theta / 2)或alpha = theta / 2然后1 + costheta-isintheta = r(cos(-alpha)+ isin(-alpha))= r(cosalpha-isinalpha)我们可以写(1 + costheta + isintheta) ^ n +(1 + costheta-isintheta)^ n使用DE MOivre定理为r ^ n(cosnalpha + isinnalpha + cosnalpha-isinnalpha)= 2r ^ ncosnalpha = 2 * 2 ^ ncos ^ n(theta / 2)cos((ntheta) / 2)= 2 ^(n + 1)cos ^ n(theta / 2)cos((nθ)/ 2)