+1124 Hi friends....please a last question for today:
The sum goes like this:
\({{\sqrt{2} \over 2}Cos x +Sinx} \over Sinx\)
becomes
\(Sin x +Cos x \over \sqrt{2}Sinx\)
I understand the sqrt of 2 going to the bottom, but what happenms to the denominator "2"??. Thanks a lot for your patience today. All the best. Untill next time!!.. 
+118609
\(\frac{\frac{\sqrt2}{2}cos\;x\;+\;sin\;x}{sin\;x}\\~\\ =\frac{\frac{\sqrt2cosx+2sinx}{2}}{sinx}\\ =\frac{\sqrt2cosx+2sinx}{2}\div sinx\\ =\frac{\sqrt2cosx+2sinx}{2sinx}\\ =\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\frac{2sinx}{\sqrt2}}\\ =\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\sqrt2sinx}\\ =\frac{cosx+\sqrt2sinx}{\sqrt2sinx}\\\)
BUT If I put brackets into you question, this is what I get.
\(\frac{\frac{\sqrt2}{2}[cosx+sinx]}{sinx}\\ =\frac{\sqrt2}{2}[cosx+sinx]\times\frac{1}{sinx} \\ =\frac{\sqrt2}{2}[cosx+sinx]\times\frac{1}{sinx} \times \frac{\sqrt2}{\sqrt2}\\ =\frac{2}{2\sqrt2}[cosx+sinx]\times\frac{1}{sinx} \\ =\frac{1}{\sqrt2}[cosx+sinx]\times\frac{1}{sinx} \\ =\frac{cosx+sinx}{\sqrt2sinx} \\\)
LaTex:
\frac{\frac{\sqrt2}{2}cos\;x\;+\;sin\;x}{sin\;x}\\~\\
=\frac{\frac{\sqrt2cosx+2sinx}{2}}{sinx}\\
=\frac{\sqrt2cosx+2sinx}{2}\div sinx\\
=\frac{\sqrt2cosx+2sinx}{2sinx}\\
=\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\frac{2sinx}{\sqrt2}}\\
=\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\sqrt2sinx}\\
=\frac{cosx+\sqrt2sinx}{\sqrt2sinx}\\
+1124 Melody, once again, ..Thank you. So it's all about rasionalizing the numerator. I could do it like this:?
\({{\sqrt2 \over2}(SinxCosx)} \over Sinx\)
\({{{\sqrt2 \over2}(SinxCosx)} \over Sinx}*{ \sqrt2 \over \sqrt2}\)
which really means:
\({{{\sqrt2* {\sqrt2} \over2}(SinxCosx)} \over \sqrt2Sinx}\)
which will then also produce the final correct answer?
By the way, the brackets were needed...I have forgotten to add them in the original sum...oopsie..sorry..
