Question

Sigma sin (a+b )sin (a-b)/cos2acos2b

Answer 1

Method I:

if you know the formula;
★cosC + cosD = 2 cos((C+D)/2) cos((C-D)/2)
★ cosC - cosD = 2 sin((C+D)/2) sin((C-D)/2)

Now from this 2nd formula, you can see:
→ 2 sin((C+D)/2) sin((C-D)/2) = cosC - cosD
→ sin((C+D)/2) sin((C-D)/2) = (cosC - cosD)/2

On comparing the Left Hand side of this equation with the Numerator of the question, you might be thinking the term 2 is missing in the Numerator, or you can also say:
C = 2a , & D = 2b

So, the Numerator of the question can be simplified to:
sin((2a + 2b)/2) sin((2a - 2b)/2)
which will be equal to:
= [cos(2a) - cos(2b)]/2

Now solve it.
ok...
This question is very easy, looking little lengthy, because to explain you I've written a lot extra...

Method II:
if you know the formula;
★sin(A+B) = sinA cosB + cosA sinB
★sin(A-B) = sinA cosB - cosA sinB

then put it in the equation, we get Numerator as :
(sinA cosB + cosA sinB)(sinA cosB - cosA sinB)
here if we assume (sinA cosB) as "x" and (cosA sinB) as "y", then
→ (x + y) (x - y) = x² - y²

→ [(sinA cosB)² - (cosA sinB)²]
and then performing these steps, your can solve it.

Thankyou!!!