I have already figured out the answer: [1+sinxcosx-(sinx+cosx)]/2. I did that by looking how to calculate what was already in the squared circle and then squared the result!!!
This is a straightforward calculation, yielding E=[sin^2 x cos^x]/[2[1+sinx +cosx+sinxcosx] = sin^xcos^x/2(1+sinx)(1+cosx) = 1/2[1-sin x-cos x +sinx cost].
I have already figured out the answer: [1+sinxcosx-(sinx+cosx)]/2. I did that by looking how to calculate what was already in the squared circle and then squared the result!!!
This is a straightforward calculation, yielding E=[sin^2 x cos^x]/[2[1+sinx +cosx+sinxcosx] = sin^xcos^x/2(1+sinx)(1+cosx) = 1/2[1-sin x-cos x +sinx cost].
Αν θυμαμαι καλα tanχ=εφχ; cotχ=σφχ; secχ= τεμνουσα χ=1/cosχ =1/συνχ; και cscχ=συντεμνουσα χ=1/sinχ=1/ημχ. Αρα η σχεση γινεται : παρονομαστης=εφχ+σφχ+1/συνχ+1/ημχ=ημχ/συνχ+ συνχ/ημχ +1/συνχ+1/ημχ=(1+ημχ+συνχ)/ημχ συνχ. Αρα Ε=[ημ(χ/2)[συν(χ/2)-ημ(χ/2)]]^2=(ημ^2(χ/2)×(1-ημχ)=[(1-συνχ)(1-ημχ)]/2=[(συν0-συνχ)(ημ(π/2)-ημχ)]/2....