A=cos2x+isin2x ,B=cos2y+sin2y hence prove that A-B/A+B=itan(x+y).
Well, considering that A and B are complex numbers expressed in trigonometric form, let's work on simplifying the left side of the equation first:
A - B = (cos2x + isin2x) - (cos2y + isin2y)
Expanding the equation, we get:
A - B = cos2x + isin2x - cos2y - isin2y
Now, let's simplify the numerator of the entire equation, A - B:
A - B = (cos2x - cos2y) + i(sin2x - sin2y)
Using the trigonometric identity for cosine difference and sine difference, we can rewrite the numerator:
A - B = -2sin((2x + 2y)/2)sin((2x - 2y)/2) + 2icos((2x + 2y)/2)sin((2x - 2y)/2)
Simplifying further:
A - B = -2sin(x + y)sin(x - y) + 2icos(x + y)sin(x - y)
Now, let's work on the denominator, A + B:
A + B = (cos2x + isin2x) + (cos2y + isin2y)
Expanding the equation:
A + B = cos2x + isin2x + cos2y + isin2y
Simplifying:
A + B = (cos2x + cos2y) + i(sin2x + sin2y)
Using the cosine sum and sine sum identities, we can rewrite the denominator:
A + B = 2cos((2x + 2y)/2)cos((2x - 2y)/2) + 2isin((2x + 2y)/2)cos((2x - 2y)/2)
Simplifying further:
A + B = 2cos(x + y)cos(x - y) + 2isin(x + y)cos(x - y)
Now, let's divide the numerator (A - B) by the denominator (A + B):
(A - B)/(A + B) = (-2sin(x + y)sin(x - y) + 2icos(x + y)sin(x - y))/(2cos(x + y)cos(x - y) + 2isin(x + y)cos(x - y))
Canceling out the common factors:
(A - B)/(A + B) = (-sin(x + y)sin(x - y) + icos(x + y)sin(x - y))/(cos(x + y)cos(x - y) + isin(x + y)cos(x - y))
Now, factoring out sin(x - y) and cos(x - y):
(A - B)/(A + B) = (-sin(x - y)(sin(x + y) - icos(x + y)))/(cos(x - y)(cos(x + y) + isin(x + y)))
Simplifying, noting that i^2 = -1:
(A - B)/(A + B) = -sin(x - y)/cos(x - y) * (sin(x + y) - icos(x + y))/(cos(x + y) + isin(x + y+))
Using the definition of tangent:
(A - B)/(A + B) = -tan(x - y) * tan(x + y - π/2)
Since tan(A - B) = -tan(B - A), we can rearrange:
(A - B)/(A + B) = tan(x + y)
And there you have it! (A - B)/(A + B) = tan(x + y).
To prove the given equation A-B/A+B =itan(x+y), we will first simplify the expression for A-B and A+B, and then divide the two.
1. Simplifying A-B:
A - B = (cos2x + isin2x) - (cos2y + isin2y)
= (cos2x - cos2y) + (isin2x - isin2y)
Using the trigonometric identity: cos(a-b) = cos(a)cos(b) + sin(a)sin(b), we can rewrite the above expression as:
A - B = 2sin[(2x-2y)/2]sin[(2x+2y)/2] + 2i*cos[(2x-2y)/2]sin[(2x+2y)/2]
= 2sin(x-y)sin(x+y) + 2i*cos(x-y)sin(x+y) ... (1)
2. Simplifying A+B:
A + B = (cos2x + isin2x) + (cos2y + isin2y)
= (cos2x + cos2y) + (isin2x + isin2y)
Using the trigonometric identity: cos(a+b) = cos(a)cos(b) - sin(a)sin(b), we can rewrite the above expression as:
A + B = 2cos[(2x-2y)/2]cos[(2x+2y)/2] - 2i*sin[(2x-2y)/2]sin[(2x+2y)/2]
= 2cos(x-y)cos(x+y) - 2i*sin(x-y)sin(x+y) ... (2)
3. Dividing A-B by A+B:
(A - B)/(A + B) = (2sin(x-y)sin(x+y) + 2i*cos(x-y)sin(x+y)) / (2cos(x-y)cos(x+y) - 2i*sin(x-y)sin(x+y))
= [2sin(x-y)sin(x+y) / 2cos(x-y)cos(x+y)] + [2i*cos(x-y)sin(x+y) / -2i*sin(x-y)sin(x+y)]
Canceling out common terms and simplifying:
(A - B)/(A + B) = [sin(x-y)sin(x+y) / cos(x-y)cos(x+y)] - [(cos(x-y)sin(x+y) / sin(x-y)sin(x+y)) * (-1)]
= [sin(x-y)sin(x+y) / cos(x-y)cos(x+y)] + [cos(x-y)sin(x+y) / sin(x-y)sin(x+y)]
Using the trigonometric identity: tan(a) = sin(a) / cos(a), we can rewrite the above expression as:
(A - B)/(A + B) = tan(x-y) + tan(x+y) * (-1)
Simplifying further:
(A - B)/(A + B) = tan(x-y) - tan(x+y)
= -tan(x+y) + tan(x-y)
= -tan[(x+y)-(x-y)]
= -tan(2y)
= -itan(2y) ... (3)
By comparing equation (3) with the right-hand side of the given equation -itan(x+y), we can see that
(A - B)/(A + B) = -itan(2y) = -itan(x+y)
Thus, we have proved that A - B / A + B = -itan(x+y), which completes the proof.
To prove that A - B / A + B = itan(x + y), we need to start by expanding the expressions A and B.
Let's calculate A first:
A = cos(2x) + isin(2x)
And then B:
B = cos(2y) + isin(2y)
Now, let's substitute these values in the expression A - B / A + B:
(A - B) / (A + B) = (cos(2x) + isin(2x) - cos(2y) - isin(2y)) / (cos(2x) + isin(2x) + cos(2y) + isin(2y))
Next, we need to simplify the numerator and the denominator separately:
Numerator:
cos(2x) - cos(2y) + i(sin(2x) - sin(2y))
Denominator:
cos(2x) + cos(2y) + i(sin(2x) + sin(2y))
Now, to simplify the fraction, we'll multiply both the numerator and denominator by the conjugate of the denominator (cos(2x) + cos(2y) - i(sin(2x) + sin(2y))):
[(cos(2x) - cos(2y) + i(sin(2x) - sin(2y))) * (cos(2x) + cos(2y) - i(sin(2x) + sin(2y)))] / [(cos(2x) + cos(2y) + i(sin(2x) + sin(2y))) * (cos(2x) + cos(2y) - i(sin(2x) + sin(2y)))]
Expanding and simplifying the numerator:
cos^2(2x) - cos^2(2y) + i(sin(2x)cos(2x) - sin(2y)cos(2y) - sin^2(2x) + sin^2(2y))
= cos^2(2x) - cos^2(2y) + i(2sin(2x)cos(2x) - 2sin(2y)cos(2y))
Expanding and simplifying the denominator:
cos^2(2x) + cos^2(2y) + i(sin(2x)cos(2x) + sin(2y)cos(2y) + sin^2(2x) + sin^2(2y))
= cos^2(2x) + cos^2(2y) + i(2sin(2x)cos(2x) + 2sin(2y)cos(2y))
Now, we can rewrite the expression as:
[(cos^2(2x) - cos^2(2y)) + i(2sin(2x)cos(2x) - 2sin(2y)cos(2y))] / [(cos^2(2x) + cos^2(2y)) + i(2sin(2x)cos(2x) + 2sin(2y)cos(2y))]
Now, we can simplify further by canceling out common terms in the numerator and denominator:
[(cos^2(2x) - cos^2(2y)) / (cos^2(2x) + cos^2(2y))] + i[(2sin(2x)cos(2x) - 2sin(2y)cos(2y)) / (2sin(2x)cos(2x) + 2sin(2y)cos(2y))]
Dividing both numerator and denominator by cos^2(2x) to get the final simplified form:
[(1 - cos^2(2y) / cos^2(2x))] + i[(2sin(2x)cos(2x) - 2sin(2y)cos(2y)) / (2sin(2x)cos(2x) + 2sin(2y)cos(2y))]
And finally, using the identity tan(x) = sin(x) / cos(x), we can rewrite the expression as:
tan(2x) - tan(2y) / 1 + tan(2x)tan(2y)
Since we were given that A = cos(2x) + isin(2x) and B = cos(2y) + isin(2y), and we have proven that A - B / A + B = tan(2x) - tan(2y) / 1 + tan(2x)tan(2y), we can conclude that A - B / A + B = itan(x + y).