Question

cosx cosx/2- cos3x cos 9x/2=sin4x sin7x/2​

Answer 1

Answer:

we know:-

cosa.cosb = 1/2[cos(a + b) + cos(a - b)]

cos(-∅) = cos∅

cosa - cosb = 2sin(a + b)/2 * sin(b - a)/2

now,

L.H.S:- cosx.cos(x/2) - cos3x.cos(9x/2)

≈» 1/2[{cos(3x/2) + cos(x/2)} - {cos(15x/2) + cos(3x/2)}]

≈» 1/2[cos(3x/2) + cos(x/2) - cos(15x/2) - cos(3x/2)]

≈» 1/2[cos(x/2) - cos(15x/2)]

≈> 1/2*2[sin{(x/2) + (15x/2)}/2 * sin{(15x/2) - (x/2)}/2]

≈» [sin(8x/2) * sin(7x/2)]

≈» sin4x * sin7x/2

I think some misprints in questions may be but according to solution we can say that sin4x.sin7x/2 is the correct answer