`sin^2 (pi/8+A/2)-sin^2(pi/8-A/2)` is equal to :

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(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/sqrt(2)sinA

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.

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

The value of sin^(2)""(pi)/(8) + sin^(2)""(3pi)/(8) + sin^(2)""(5pi)/(8) + sin^(2) ""(7pi)/(8) is equal to a) (1)/(8) b) (1)/(4) c) (1)/(2) d)2

The value of ( sin theta )/( sin^(2) (pi//8 + theta //2)-sin^(2)(pi//8- theta //2))=

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