Question

Express $\frac{1}{1+\cos 2\theta +i\sin 2\theta }$ in the form of $a+ib$

Answer 1

Solution:-

$\frac{1}{(1+\cos 2\theta )+i\sin 2\theta }$

$=\frac{1\times ((1+\cos 2\theta )-i\sin 2\theta )}{((1+\cos 2\theta )+i\sin 2\theta )\times ((1+\cos 2\theta )-i\sin 2\theta )}\{\because \text{Rationalising the denominator}\}$

$=\frac{((1+\cos 2\theta )-i\sin 2\theta )}{{(1+\cos 2\theta )}^2-{\left(i\sin 2\theta \right)}^2}$

$=\frac{((1+\cos 2\theta )-i\sin 2\theta )}{(1^2+{\cos }^{2}2\theta +2\times 1\times \cos 2\theta )-\left(i^2{\sin }^{2}2\theta \right)}$

$=\frac{((1+\cos 2\theta )-i\sin 2\theta )}{1+{\cos }^{2}2\theta +2\cos 2\theta -{(-1)}^2{\sin }^{2}2\theta }$

$=\frac{((1+\cos 2\theta )-i\sin 2\theta )}{1+2\cos 2\theta +{\cos }^{2}2\theta +{\sin }^{2}2\theta }$

$\Rightarrow =\frac{((1+\cos 2\theta )-i\sin 2\theta )}{1+2\cos 2\theta +1}\{\because {\sin }^2\theta +{\cos }^2\theta =1\}$

$\Rightarrow =\frac{((1+\cos 2\theta )-i\sin 2\theta )}{2(1+\cos 2\theta )}$

$=\frac{(1+\cos 2\theta )}{2(1+\cos 2\theta )}-\frac{i\sin 2\theta }{2(1+\cos 2\theta )}$

$\Rightarrow =\frac{1}{2}-\frac{i2\sin \theta \cos \theta }{2{\cos }^2\theta }\{\because \sin 2\theta =2\sin \theta \cos \theta \&\cos 2\theta =2{\cos }^2\theta -1\}$

$\Rightarrow =\frac{1}{2}-i\tan \theta \{\because \frac{\sin \theta }{\cos \theta }=\tan \theta \}$

$\Rightarrow =\frac{1}{2}+(-\tan \theta )i\{\text{In the form of a + ib; where,}a=\frac{1}{2}\text{and}b=-\tan \theta \}$