Question

${(1+\cos \theta +i\sin \theta )}^3+{(1-\cos \theta -i\sin \theta )}^3$ is equal to

• A. $2^4{\cos }^3\left(\frac{\theta }{2}\right)\cos \left(\frac{3\theta }{2}\right)$
• B. $2^3{\cos }^3\left(\frac{\theta }{2}\right)\cos \left(\frac{3\theta }{2}\right)$
• C. $2^3{\cos }^3\left(\frac{\theta }{2}\right)\sin \left(\frac{3\theta }{2}\right)$
• D. $2^4{\cos }^3\left(\frac{\theta }{2}\right)\sin \left(\frac{3\theta }{2}\right)$

Answer 1

Answer: (A)

${(1+\cos \theta +i\sin \theta )}^3+{(1+\cos \theta -i\sin \theta )}^3$
$={\left[1+2{\cos }^2\frac{\theta }{2}-1+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right]}^3$
$+{\left[1+2{\cos }^2\frac{\theta }{2}-1-i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right]}^3$
$={\left[2\cos \frac{\theta }{2}\left(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}\right)\right]}^3$
$+{\left[2\cos \frac{\theta }{2}\left(\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}\right)\right]}^3$
$=8{\cos }^3\frac{\theta }{2}\left[\cos \frac{3\theta }{2}+i\sin \frac{3\theta }{2}+\cos \frac{3\theta }{2}-i\sin \frac{3\theta }{2}\right]$
$=8{\cos }^3\frac{\theta }{2}\left[2\cos \frac{3\theta }{2}\right]$
$=16{\cos }^3\frac{\theta }{2}.\cos \frac{3\theta }{2}$
$=2^4{\cos }^3\frac{\theta }{2}.\cos \frac{3\theta }{2}$