plz.. tell me is this correct?

Using the sum and difference identities for the sine, we have......

[sin(pi/4)cosA + sin(A) cos(pi/4)] * [sin(pi/4)cosA - sinAcos(pi/4)] =

[sin(pi/4)cosA]² - (sin(A)cos(pi/4)]² =

[(1/2)cos²A] - [(1/2)sin²(A)] =

(1/2)[cos²A - sin²A] =

(1/2)cos2A

Which is the same as your answer, Sasini .....  !!!!

Hi Sasini,

$$sin[\frac{\pi}{4}+A]*sin[\frac{\pi}{4}-A]\\\\
=[sin\frac{\pi}{4}cosA+cos\frac{\pi}{4}sinA][sin\frac{\pi}{4}cosA-cos\frac{\pi}{4}sinA]\\\\
\mbox{The two brackets are the same except one has a minus and the other a plus so this simplifies to the difference of 2 squares}\\\\
=[sin\frac{\pi}{4}cosA]^2-[cos\frac{\pi}{4}sinA]^2\\\\
=[sin^2\frac{\pi}{4}*cos^2A]-[cos^2\frac{\pi}{4}*sin^2A]\\\\
=[\frac{1}{2}cos^2A]-[\frac{1}{2}sin^2A]\\\\
=\frac{cos^2A-sin^2A}{2}\\\\
or\qquad
=\frac{1-2sin^2A}{2}\\\\
or\qquad
=\frac{cos2A}{2}$$

I got the same as well