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  4. (1+ cos θ +i sin θ )3+(1- cos θ -i sin θ )3 is equal to

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

Rajasthan PETRajasthan PET 2004

Solution:

$ {{(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} $