In triangle ABC, let a,b,c be the lengths of the sides opposite to the angles A,B,C respectively. Let the value of a3cos3B+3a2bcos(A−2B)+3ab2cos(2A−B)+b3cos3A be l. Then the value of lc3 is equal to
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Solution
We have c=acosB+bcosA (By projection rule)
and asinB−bsinA=0. (By sine rule)
Thus, we can write c=a(cosB−isinB)+b(cosA+isinA) where i=√−1 ⇒c3=(ae−iB+beiA)3⇒c3=a3e−i3B+3a2be−i2BeiA+3ab2e−iBei2A+b3ei3A⇒c3=l
(Consider real part only as in LHS, we have real value only) ∴lc3=1