Sin^2(π\8+A/4)-sin^2(π\8-A/4)=1/√2sinA/2

Answer 1

If your question means:

Prove sin² ( π / 8 + A / 4 ) - sin² ( π / 8 - A / 4 ) = ( 1 / √2 ) ∙ sin ( A / 2 )

then use identity:

sin² α - sin² β = sin ( α + β ) ∙ sin ( α - β )

In this case:

α = π / 8 + A / 4 , β = π / 8 - A / 4

so

sin² ( π / 8 + A / 4 ) - sin² ( π / 8 - A / 4 ) =

sin ( π / 8 + A / 4 + π / 8 - A / 4 ) ∙ sin [ π / 8 + A / 4 - ( π / 8 - A / 4 ) ] =

sin ( π / 8 + π / 8 ) ∙ sin ( π / 8 + A / 4 - π / 8 + A / 4 ) =

sin ( 2 π / 8 ) ∙ sin ( A / 4 + A / 4 ) = sin ( π / 4 ) ∙ sin ( 2 A / 4 ) =

sin ( π / 4 ) ∙ sin ( A / 2 ) = ( 1 / √2 ) ∙ sin ( A / 2 )

Answer 2

Remark:

π / 4 rad = 45°

sin ( π / 4 ) = sin 45° = √2 / 2 = √2 ∙ 1 / √2 ∙ 2 = 1 / √2

This means √2 / 2 is the same as 1 / √2