`sin^2(pi/8+A/2)-sin^2(pi/8-A/2)=1/(sqrt(2))sinA`

Prove that sin^2(pi/8+A/2)-sin^2(pi/8-A/2)=1/sqrt2sinA

Prove that sin^2(pi/8+A/2)-sin^2(pi/8-A/2)=1/sqrt(2) sin A .

Prove that: sin^2(A/2+pi/8) - sin^2(A/2-pi/8)= 1/sqrt(2)sinA

Prove that : cos^2 (pi/8+x/2)-sin^2 (pi/8-x/2)= 1/sqrt2 cos x .

Prove that cos^(2)(pi/8-A/2)-cos^(2)(pi/8+A/2) = sin(pi/4). sinA=1/sqrt(2)sinA

Prove that: sin^(2)((pi)/(8)+(A)/(2))-sin^(2)((pi)/(8)-(A)/(2))=(1)/(sqrt(2))sin A

Prove that cos^(2)(pi/8-A/2)-cos^(2)(pi/8+A/2) [1-sin^(2)(pi/8-A/2)]-[1-sin^(2)(pi/8+A/2)] =sin^(2)(pi/8+A/2)-sin^(2)(pi/8-A/2) =sin{(pi/8+A/2)+(pi/8-A/2)} sin{(pi/8+A/2)-(pi/8-A/2)} s=sinpi/4. sinA=1/sqrt(2)sinA =RHS Hence Proved.

sin^(2)((A)/(2)+(pi)/(8))-sin^(2)((A)/(2)-(pi)/(8))=(1)/(sqrt(2))sin A

If f(x)=sin^(2)((pi)/8 + (x)/2) -sin^2((pi)/8-(x)/(2)) , then the period of f is

Evaluate : sin^(2) (pi/8 +x/2) - sin^(2) (pi/8 - x/2)