if a=cos2x + sin2x and b=cos2y + isin2y.then find (a + b)
I have a sneaky suspicion that your first equation is
a = cos 2x + i sin 2x
a+b = (cos 2x + cos 2y) + i(sin 2x + sin 2y)
recall cosA + cosB = 2cos( (A+B)/2 )cos( (A-B)/2)
so cos 2x + cos 2y = 2 cos(x+y) cos(x-y)
recall sinA + sinB = 2sin( (A+B)/2)cos( (A+B)/2)
so sin 2x + sin 2y = 2sin(x+y)cos(x+y)
then a+b = (cos 2x + cos 2y) + i(sin 2x + sin 2y)
= 2 cos(x+y) cos(x-y) + 2i sin(x+y)cos(x+y)
To find the sum of a + b, we can simply add the corresponding terms of a and b.
Let's start by substituting the given expressions for a and b:
a = cos(2x) + sin(2x)
b = cos(2y) + i sin(2y)
Now we can add the real and imaginary parts separately:
Real part:
a_r = cos(2x)
b_r = cos(2y)
Imaginary part:
a_i = sin(2x)
b_i = i sin(2y)
Summing up the real parts and imaginary parts separately, we get:
(a + b) = (cos(2x) + cos(2y)) + (sin(2x) + i sin(2y))
So, the sum of a + b is:
(a + b) = cos(2x) + cos(2y) + sin(2x) + i sin(2y)
Please note that this expression cannot be simplified further unless you have additional information or constraints on the values of x and y.