Expanding twice: 



T(x,y,t)=T+f C n cos(K ln X + K 2n y)cos(-^^- + S n ) + C n sin(K |n X+K 2n y)sin(^pi+8 n ) 



=T+f cos(K |n X + K 2n y)C n [cos-^cosS n -sin^PtsinS n ]+sin(K in X + K 2n y)C n 



[sin ^- cos S n + cos -^Pl sin S n ] 

 Substituting a n = C n cosS n b n = C n sinS n 



T(x,y,t)=T+S cos(K in X + K 2n y)[a n cos-^Pi-b n sin^^]+sin(K m X + K 2n y) 



L -:- 27rnt . .. _„. 27rnt l 

 I an sin — = — + bn cos — =- 



Example; 3 thermometers in right triangle 



Let L= ^ shortest wave length to be measured 



(0,0) (L,o) 



T(o,o,t) = ^ M |E^lTU R E= T +fa n cos^-b n sin2int 

 Solving for a specific value ajj of a n by multiplying both sides by cos 27 L*' dt and integrating 

 from t=o to t=T the term with T is zero at both limits of integration since A is a specific integer, 

 the integrand from the right hand portion of the summation becomes -b n sin ^ cos ^ 

 which can also be expressed as - 4?- sin 27rnt+27rlt _ &n_ s j n 27rnt ~ 27r ^ f which when integrated 

 with limits of t=o and t=T becomes zero for every value of n, and the integrand of the left 

 hand term in the summation gives the only contribution and at that only for n = { : 

 j^Ko.o.tJcos^Mdt^f f Q a n cos ^cos^dt 



= 2 jf ^[cos^n+jm+cos^n-JOtJdt 



All integrals of this summation are zero except for n=J£: 

 4([ ^ tcK0 ] dt ,|T 



ajj= ^-^Ko^tjcos^^dt 

 Similiarly solving for bj£ by multiplying both sides by sin 2 y^ dt and integrating from t=o to t=T 

 b r-f / T T(o,o,t)sin^- f dt 



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