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Question: Express the following in the form A + iB . \(\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}\)...

Express the following in the form A + iB .
11−cos⁡θ+2isin⁡θ\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}

Explanation

Solution

Start by rationalizing the term . Use trigonometric identities and conversion formulas to simplify the fraction. Separate the real and imaginary part after simplification , the result obtained will be in the form of A + iB.

Complete step-by-step answer:
Given,
11−cos⁡θ+2isin⁡θ\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}
Let us start by rationalizing this equation, that is multiplying 1−cos⁡θ−2isin⁡θ1 - \cos \theta - 2i\sin \theta with the numerator and the denominator. We get
1×(1−cos⁡θ−2isin⁡θ)(1−cos⁡θ+2isin⁡θ)×(1−cos⁡θ−2isin⁡θ)\dfrac{{1 \times \left( {1 - \cos \theta - 2i\sin \theta } \right)}}{{\left( {1 - \cos \theta + 2i\sin \theta } \right) \times \left( {1 - \cos \theta - 2i\sin \theta } \right)}}
As we know (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}, applying this formula in denominator , we get
=1−cos⁡θ−2isin⁡θ(1−cos⁡θ)2−(2i)2(sin⁡2θ)= \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{{{\left( {1 - \cos \theta } \right)}^2} - {{\left( {2i} \right)}^2}\left( {{{\sin }^2}\theta } \right)}}
We know i2=−1&(a−b)2=a2+b2−2ab{i^2} = - 1\& {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, Applying this formula in denominator , we get
⇒1−cos⁡θ−2isin⁡θ1+cos⁡2θ−2cos⁡θ+4sin⁡2θ\Rightarrow \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{1 + {{\cos }^2}\theta - 2\cos \theta + 4{{\sin }^2}\theta }}
We know that ,sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
⇒1−cos⁡θ−2isin⁡θ2−2cos⁡θ+3sin⁡2θ\Rightarrow \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}
⇒1−cos⁡θ−2isin⁡θ2(1−cos⁡θ)+3sin⁡2θ\Rightarrow \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{2(1 - \cos \theta ) + 3{{\sin }^2}\theta }}
We know, 1−cos⁡θ=2sin⁡2θ21 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}
⇒2sin⁡2θ2−i2⋅2sin⁡θ2cos⁡θ22⋅2sin⁡2θ2+3⋅22sin⁡2θ2cos⁡2θ2\Rightarrow \dfrac{{2{{\sin }^2}\dfrac{\theta }{2} - i2 \cdot 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{2 \cdot 2{{\sin }^2}\dfrac{\theta }{2} + 3 \cdot {2^2}{{\sin }^2}\dfrac{\theta }{2}{{\cos }^2}\dfrac{\theta }{2}}}
Taking 4sin⁡2θ24{\sin ^2}\dfrac{\theta }{2}in denominator and 4sin⁡θ24\sin \dfrac{\theta }{2}in numerator as common , we get
⇒4sin⁡θ2(sin⁡θ2−icos⁡θ2)4sin⁡2θ2(1+3cos⁡2θ2)\Rightarrow \dfrac{{4\sin \dfrac{\theta }{2}(\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2})}}{{4{{\sin }^2}\dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}
On simplification , we get
⇒(sin⁡θ2−icos⁡θ2)sin⁡θ2(1+3cos⁡2θ2)\Rightarrow \dfrac{{(\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2})}}{{\sin \dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}
Now separating the real and imaginary part , we get
sin⁡θ2sin⁡θ2(1+3cos⁡2θ2)−icos⁡θ2sin⁡θ2(1+3cos⁡2θ2)\dfrac{{\sin \dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}} - i\dfrac{{\cos \dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}
On further simplification ,we get
1(1+3cos⁡2θ2)−icot⁡θ2(1+3cos⁡2θ2)\dfrac{1}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}} - i\dfrac{{\cot \dfrac{\theta }{2}}}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}
So this is in the form of A + iB , where A=1(1+3cos⁡2θ2) = \dfrac{1}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}and B=−cot⁡θ2(1+3cos⁡2θ2) = - \dfrac{{\cot \dfrac{\theta }{2}}}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}

Note: Similar questions can be solved by using the procedure , sometimes we just need to solve the exponent part first and sometimes we need to rationalize the terms. Students must know all the formulas related to trigonometric conversions. Attention must be given while substituting the values as it might lead to wrong answers.