If a= 2cosx+ isinx and b = 2cosy +isiny then prove that a+b = 2cos(x-y)(cos(x+y)+isin(x+y))
just plug and chug
a+b = 2cosx+i sinx + 2cosy + i siny
= 2(cosx+cosy) + (sinx+siny)i
Now,
(sinx+cosx)^2 = sin^2x+2sinx cosx+cos^2x
= 1+2sinx cosx
similarly, (siny+cosy)^2 = 1+2siny cosy
Since 2^2-b^2 = (a+b)(a-b), we have
(sinx+cosx)^2 - (siny+cosy)^2 = ((sinx+cosx)+(siny+cosy))((sinx+cosx)(siny+cosy))
Try expanding that out, and use your sum/difference formulas to get to the right side.
To prove the given equation (a + b) = 2cos(x - y)(cos(x + y) + isin(x + y)), we will start by expanding both a and b.
Given:
a = 2cosx + isinx
b = 2cosy + isiny
Adding a and b:
a + b = (2cosx + isinx) + (2cosy + isiny)
Next, we will distribute the addition across the terms:
a + b = 2cosx + 2cosy + isinx + isiny
Now, let's regroup the real and imaginary terms separately:
a + b = 2cosx + 2cosy + i(sinx + siny)
To simplify this expression further, we will transform the cosine and sine terms using trigonometric identities. Using the cosine sum and difference formulas, we have:
a + b = 2cosx + 2cosy + i(sin(x)cos(y) + cos(x)sin(y))
Now, let's express the sine and cosine terms in terms of their sum and difference:
a + b = 2cosx + 2cosy + i(sin(x + y) + sin(x - y))
Next, we can distribute the 2 across both cosine terms:
a + b = 2(cosx + cosy) + i(sin(x + y) + sin(x - y))
Now, we will use the double angle identity for cosine, which states that cos(2θ) = 2cos^2(θ) - 1:
a + b = 2(2cos(x - y)cos(x + y)/2) + i(sin(x + y) + sin(x - y))
Simplifying this expression:
a + b = 2cos(x - y)cos(x + y) + i(sin(x + y) + sin(x - y))
Finally, we can re-arrange the terms to match the right-hand side of the given equation:
a + b = 2cos(x - y)(cos(x + y) + isin(x + y))
Thus, we have proven that a + b = 2cos(x - y)(cos(x + y) + isin(x + y)).