# Given x+y=pi, evaluate the ratio (cosx+i*sinx)/(cosy-i*siny)?

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### 2 Answers

We have to find the value of (cos x + i*sin x)/(cos y - i*sin y) given that x + y = pi

(cos x + i*sin x)/(cos y - i*sin y)

=> (cos x + i*sin x)(cos y + i*sin y)/(cos y - i*sin y)(cos y + i*sin y)

=> (cos x + i*sin x)(cos y + i*sin y)/((cos y)^2 + (sin y)^2)

=> cos x*cos y + i*sin x*cos y + i*cos x*sin y +i^2*sin x*sin y

=> cos x*cos y - sin x*sin y + i*(sin x*cos y + cos x*sin y)

=> cos (x + y) + i*(sin x + y)

=> cos pi + i*sin pi

=> -1 + i*0

=> -1

**The ratio is equal to -1.**

We'll replace y by y = pi-x and we'll get:

(cosx+i*sinx)/(cosy-i*siny) = (cosx + i*sinx)/[cos(pi-x) - i*sin(pi-x)]

By definition, cos(pi-x) = - cos x and sin(pi-x) = sin x.

(cosx+i*sinx)/[cos(pi-x)-i*sin(pi-x)] = (cosx + i*sinx)/(-cos x - i*sin x)

(cosx + i*sinx)/(-cos x - i*sin x) = (cosx + i*sinx)/-(cos x + i*sin x)

(cosx+i*sinx)/(cosy-i*siny) = -1

**The requested value of the ratio is: (cosx+i*sinx)/(cosy-i*siny) = -1.**