In a triangle ABC, the real part of the equation (acosB+bcosA + i(asinB-bsinA)) ^n

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If A B C are the angles of a triangle, and a, b, c are the corresponding sides, then the real part of $ (acosB+bcosA + i(asinB-bsinA))^n $is?

My book says that the real part of this is $(acosB+bcosA) ^ n = c^n$

But if you expand this expression using the binomial theorem , won't there be more terms not containing 'i'?

Please explain.

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Hint: Using sine Law of triangle, $a\cos B+b\cos A =2R\sin(A+B)=c$

and $a\sin B=b\sin A$