If a=cos2x+isin2x ,b=cos2y +isin2y prove thAt a+b=2Cos(x-y)(cos(x+y)+isin(x+y))

see "neha"'s post -- the one after yours...

To prove that a + b = 2cos(x-y)(cos(x+y) + isin(x+y)), we can start by simplifying the left-hand side (LHS) and right-hand side (RHS) separately.

Let's begin with the LHS:
a + b = (cos2x + isin2x) + (cos2y + isin2y)

Using the properties of addition for complex numbers, we can combine the real and imaginary parts separately:
= (cos2x + cos2y) + i(sin2x + sin2y)

Next, we can use the trigonometric identity cos(A + B) = cos(A)cos(B) - sin(A)sin(B) to rewrite the terms in terms of cosine and sine sums:
= 2cos(x)cos(y) - 2sin(x)sin(y) + i(2sin(x)cos(y) + 2cos(x)sin(y))

Let's simplify further:
= 2[cos(x)cos(y) - sin(x)sin(y)] + i[2sin(x)cos(y) + 2cos(x)sin(y)]

Now, let's rewrite the terms inside the brackets using the trigonometric identity cos(A - B) = cos(A)cos(B) + sin(A)sin(B) to replace the cosine and sine differences:
= 2cos(x-y) + i2sin(x+y)

Finally, multiply the entire expression by (cos(x+y) + isin(x+y)) to get the desired form:
= 2cos(x-y)(cos(x+y) + isin(x+y))

Hence, we have shown that a + b = 2cos(x-y)(cos(x+y) + isin(x+y)).