Question #d8c45

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Answer 1 · 2016-07-27T15:44:36

Given

#costheta+sintheta=sqrt2costheta#

Following the formula

#(a+b)^2+(a-b)^2=2(a^2+b^2)#

we can wite

#(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)#

Putting

#costheta+sintheta=sqrt2costheta" "# we get

#(sqrt2costheta)^2+(costheta-sintheta)^2=2*1#

#=>2cos^2theta+(costheta-sintheta)^2=2*1#

#=>(costheta-sintheta)^2=2-2cos^2theta#

#=>(costheta-sintheta)^2=2(1-cos^2theta)#

#=>(costheta-sintheta)^2=2sin^2theta#

#=>costheta-sintheta=sqrt2sintheta#

Proved

In other way when
#costheta-sintheta=sqrt2sintheta" "# we get

#(costheta+sintheta)^2+(sqrt2sintheta)^2=2*1#

#=>(costheta+sintheta)^2 + 2sin^2theta =2#

#=>(costheta+sintheta)^2=2-2sin^2theta#

#=>(costheta+sintheta)^2=2(1-sin^2theta)#

#=>(costheta+sintheta)^2=2cos^2theta#

#=>costheta+sintheta=sqrt2costheta#

Proved