If x = cos@ + isin@ , y = cos€ + isin€ ' prove
that (x - y)/(x + y) = itan((@ - €))/2
step sir plz
Lat's just use u and v for ease of typing
x = cosu + i sinu
y = cosv + i sinv
(x-y)/(x+y) =
(cosu-cosv) + (sinu-sinv)i
-------------------------------
(cosu+cosy) + (sinu+siny)i
Now multiply top&bottom by the conjugate of the denominator, collect terms, and recall that
tan (u-v)/2 = sin(u-v)/(1+cos(u-v))
that will get you where you need to go.
To prove the given statement:
1. Express x and y in terms of their real and imaginary parts.
Let x = cos(@) + i*sin(@) = a + bi
Let y = cos(€) + i*sin(€) = c + di
where a = cos(@), b = sin(@), c = cos(€), and d = sin(€).
2. Calculate the numerator: (x - y).
(x - y) = (a + bi) - (c + di)
= (a - c) + (b - d)i
3. Calculate the denominator: (x + y).
(x + y) = (a + bi) + (c + di)
= (a + c) + (b + d)i
4. Simplify the expression (x - y)/(x + y):
[(a - c) + (b - d)i] / [(a + c) + (b + d)i]
To simplify further, we need to rationalize the denominator.
Multiply the numerator and denominator by the conjugate of the denominator:
[(a - c) + (b - d)i] * [(a + c) - (b + d)i] / [(a + c) + (b + d)i] * [(a + c) - (b + d)i]
Using the FOIL method (First, Outer, Inner, Last):
Numerator:
= [(a - c)(a + c) - (a - c)(b + d)i + (b - d)i(a + c) - (b - d)(b + d)i^2]
= [(a^2 - c^2) - (ab + ad - bc + cd)i]
Denominator:
= [(a + c)(a + c) - (b + d)(b + d)i^2]
= [(a^2 + 2ac + c^2) + (b^2 + 2bd + d^2)]
The imaginary part i^2 simplifies to -1:
Numerator:
= [(a^2 - c^2) - (ab + ad - bc + cd)i]
Denominator:
= [(a^2 + 2ac + c^2) + (b^2 + 2bd + d^2)]
Combine like terms for both numerator and denominator:
Numerator:
= [a^2 - c^2 - ab - ad + bc - cd + (-1)(ab + ad - bc + cd)i]
= [a^2 - c^2 - ab - ad + bc - cd - ab - ad + bc - cd)i]
= [a^2 - c^2 - 2ab - 2ad + 2bc - 2cd)i]
Denominator:
= [a^2 + 2ac + c^2 + b^2 + 2bd + d^2]
The expression (x - y)/(x + y) becomes:
[(a^2 - c^2 - 2ab - 2ad + 2bc - 2cd)i] / [a^2 + 2ac + c^2 + b^2 + 2bd + d^2]
5. Simplify the expression further:
Divide both the numerator and the denominator of the expression by a^2 + c^2 (the real part in the denominator):
[(a^2 - c^2 - 2ab - 2ad + 2bc - 2cd) / (a^2 + c^2)] * [i / 1]
= [(a^2 - c^2 - 2ab - 2ad + 2bc - 2cd) / (a^2 + c^2)] * i
6. Apply the double angle formula for tangent:
tan(2θ) = 2tan(θ) / (1 - tan^2(θ))
In our case, let θ = (@ - €) / 2:
tan((@ - €) / 2) = 2tan((@ - €) / 4) / (1 - tan^2((@ - €) / 4))
By substitution, let θ = (@ - €) / 4:
tan(θ) = 2tan(θ / 2) / (1 - tan^2(θ / 2))
Matching this form with our expression, we can conclude that:
[(a^2 - c^2 - 2ab - 2ad + 2bc - 2cd) / (a^2 + c^2)] * i = tan((@ - €) / 2)
Hence, we have proved that (x - y)/(x + y) = itan((@ - €))/2.