3234363840 Simplify Assume that all radicands are nonnegat s
Solution
32) sqrt(cos2x sinx)*sqrt(sinx)=cosx*sqrt(sinx)*sqrt(sinx)=cosx*sinx since sqrt(cos2x)=cosxsqrt(sin2x)=sinx
sqrt(cos2x sinx)*sqrt(sinx)=cosx*sinx
34) sqrt(tan2x-2tanx sinx+sin2x)
using the formula a^2-2ab+b^2=(a-b)^2 where a=tanx,b=sinx
sqrt(tan2x-2tanx sinx+sin2x)=sqrt((tanx-sinx)2)=tanx-sinx
sqrt(tan2x-2tanx sinx+sin2x)=tanx-sinx
36) sqrt(cos(theta))*[sqrt(2cos(theta))+sqrt(cos(theta) sin(theta))]
=[sqrt(2cos2(theta))+sqrt(cos2(theta) sin(theta))]
=sqrt(2)*cos(theta)+cos(theta)*sqrt(sin(theta))
=cos(theta)*[sqrt(2)+sqrt(sin(theta))]
sqrt(cos(theta))*[sqrt(2cos(theta))+sqrt(cos(theta) sin(theta))]=cos(theta)*[sqrt(2)+sqrt(sin(theta))]
38) sqrt(cosx/tanx)
Rationalizing the denominator
sqrt(cosx/tanx)*sqrt(tanx/tanx)
=sqrt((cosx tanx)/tan2x)
=sqrt(cosx tanx)/tanx
substitute tanx=sinx/cosx
=sqrt(cosx sinx/cosx)/tanx
=sqrt(sinx)/tanx
sqrt(cosx/tanx) =sqrt(sinx)/tanx
40) sqrt((1-cos)/(1+cos))
Rationalizing the denominator
sqrt((1-cos)/(1+cos))=sqrt((1-cos)/(1+cos))*sqrt(1-cos)/sqrt(1-cos)
=sqrt((1-cos)2/(1-cos2))
use 1-cos2=sin2
=sqrt((1-cos)2/(sin2))
=(1-cos)/sin
sqrt((1-cos)/(1+cos))=(1-cos)/sin
