Question
What is the real part of $${\left( {\sin \,x + i\,\cos \,x} \right)^3}$$ where $$i = \sqrt { - 1} \,?$$
A.
$$ - \cos \,3x$$
B.
$$ - \sin \,3x$$
C.
$$ \sin \,3x$$
D.
$$ \cos \,3x$$
Answer :
$$ - \sin \,3x$$
Solution :
$$\eqalign{
& {\left( {\sin \,x + i\,\cos \,x} \right)^3} \cr
& = \,{\sin ^3}x + {\left( i \right)^3}\,{\cos ^3}x + 3i\,\left( {\sin \,x} \right)\,\left( {\cos \,x} \right)\left( {\sin \,x + i\,\cos \,x} \right) \cr
& = \,{\sin ^3}x - i\,{\cos ^3}x + 3i\,{\sin ^2}x\,\cos \,x - 3\,\sin \,x\,{\cos ^2}x \cr
& = \,\sin \,x\left( {{{\sin }^2}x - 3\,{{\cos }^2}x} \right) + i\,\cos \,x \cr
& {\text{Real}}\,{\text{part}}\,{\text{of}}\,{\left( {\sin \,x + i\,\cos \,x} \right)^3} \cr
& = \,\sin \;x\left( {{{\sin }^2}x - 3\,{{\cos }^2}x} \right) \cr
& = \,\sin \,x\left[ {{{\sin }^2}x - 3\left( {1 - {{\sin }^2}x} \right)} \right] \cr
& = \,\sin \,x\left[ {4{{\sin }^2}x - 3} \right] \cr
& = \,4{\sin ^3}x - 3\sin \,x \cr
& = \, - \left( {3\,\sin \,x - 4\,{{\sin }^3}x} \right) \cr
& = \, - \sin \,3x \cr} $$