Question

What is the real part of $(\sin x+i\cos x{)}^3$ where $i=\sqrt{-1}$?

• A. $-\cos 3x$
• B. $-\sin 3x$
• C. $\sin 3x$
• D. $\cos 3x$

Answer 1

The correct option is B $-\sin 3x$

${(\sin x+i\cos x)}^3={\sin }^{3}x-i{\cos }^{3}x+3i\sin x\cos x(\sin x+i\cos x)$

$={\sin }^{3}x-i{\cos }^{3}x+3i{\sin }^{2}x\cos x-3\sin x{\cos }^{2}x$

$={\sin }^{3}x-3\sin x{\cos }^{2}x+i(3{\sin }^{2}x\cos x-{\cos }^{3}x)$

Real part is ${\sin }^{3}x-3\sin x{\cos }^{2}x$

$=\sin x({\sin }^{2}x-3{\cos }^{2}x)$

$=\sin x(-3+3{\sin }^{2}x+{\sin }^{2}x)$

$=\sin x(-3+4{\sin }^{2}x)$

$=-(3\sin x-4{\sin }^{3}x)$

$=-\sin 3x$