sinx+ siny/ (cosx+cosy)= tan 1/2 (x+y)
prove this identity
To prove the identity sinx + siny / (cosx + cosy) = tan(1/2)(x+y), we will use the double-angle formula for tangent.
First, let's rewrite tan(1/2)(x+y) using the half-angle formula for tangent:
tan(1/2)(x+y) = (sin((x+y)/2))/(cos((x+y)/2))
Next, let's simplify sinx + siny / (cosx + cosy) using a common denominator for the numerator:
sinx + siny / (cosx + cosy) = (sinx * (cosy) + (siny) * (cosx))/(cosx + cosy)
Now, let's simplify the numerator of the expression:
sinx * cosy + siny * cosx = (sinx * cosy + sinx * cosx) + (siny * cosy - sinx * cosx)
= sinx * (cosy + cosx) + siny * (cosy - cosx)
Next, let's rewrite sinx * (cosy + cosx) and siny * (cosy - cosx) as products of sines and cosines:
sinx * (cosy + cosx) = sinx * (2*cos((x+y)/2)*cos((x-y)/2))
siny * (cosy - cosx) = siny * (2*sin((x+y)/2)*sin((x-y)/2))
Now, let's substitute these expressions back into our initial expression:
(sin((x+y)/2))/(cos((x+y)/2)) = (sinx * (2*cos((x+y)/2)*cos((x-y)/2)) + siny * (2*sin((x+y)/2)*sin((x-y)/2)))/(cosx + cosy)
Finally, let's simplify the numerator:
2(sinxcos((x+y)/2)cos((x-y)/2) + siny sin((x+y)/2)sin((x-y)/2)) = 2(sinx*cos((x+y)/2)[cos((x-y)/2) + sin((x-y)/2)])
= 2sinx*cos((x+y)/2)
Now, we can see that the numerator and denominator of our expression are equal:
2sinx*cos((x+y)/2) / (cosx + cosy) = 2sinx*cos((x+y)/2) / (cosx + cosy)
Therefore, we have proven that sinx + siny / (cosx + cosy) = tan(1/2)(x+y) using the double-angle formula for tangent.