If (sinx + siny) = a and (cosx + cosy) = b, then find the value of (sinx siny + cosx cosy).
[A] (a²+b²-2)/ 2
[B] (a2-b2-2)/ 2
[C] (a²+b²-1)/ 2
[D](a²+b²+2)/2
(sinx + siny) = a (both sides squaring). a2 =sin2x +sin2y +2sinx.siny … (1) (cosx + cosy) = b (both sides squaring). b2 =cos2x +cos2y+2cosx.cosy … (2) Now EQ (1) +EQ (2) … (sin2x +sin2y +2sinx.siny) + cos2x +cos2y+2cosx.cosy =a2+b2 =1+1+2(sinx. siny+cosx. cosy) = a2+b2 2(sinx. siny+cosx. Cosy) = a2+b2-2 sinx. siny+cosx. Cosy = (a2+b2-2)/2.
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