Please, explain this problem to me, I got lost at the step where they turned the denominator into 1

Question

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I'm sorry I just don't understand this question, and I can't quite figure out why you would swap the denominator to the numerator, then make the denominator a 1.

Answer 1

In general, for $a,b\in\Bbb{R}$ with $a^2+b^2>0$, we have $$\frac{1}{a-ib} = \frac{a+ib}{(a+ib)(a-ib)} = \frac{a+ib}{a^2+b^2}.$$ This formula serves to make a complex denominator real. If we rewrite this equation using $z\in \Bbb{C}$, we get a more compact version. $$\frac1z = \frac{\bar{z}}{|z|^2} \quad\forall\,z\in\Bbb{C}\setminus\{0\}$$ Applying this formula to $z = \sin\dfrac{x}2 + i\cos\dfrac{x}2$ answers your question.

Answer 2

They are multiplying the top and the bottom of the fraction conjugate of $$ sin (x/2) -i cos(x/2)$$ which is $$ sin(x/2) +i cos(x/2)$$

What happened here is that the denominator simplified to $1$ because $$ (sin ( x/2) -i cos(x/2))((sin (x/2) -i cos(x/2)) = sin^2 (x/2) + cos^2(x/2)=1$$