+1124 Hi friends....please a last question for today:
The sum goes like this:
√22Cosx+SinxSinx
becomes
Sinx+Cosx√2Sinx
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!!.. 
+118696
√22cosx+sinxsinx =√2cosx+2sinx2sinx=√2cosx+2sinx2÷sinx=√2cosx+2sinx2sinx=√2cosx+2sinx√22sinx√2=√2cosx+2sinx√2√2sinx=cosx+√2sinx√2sinx
BUT If I put brackets into you question, this is what I get.
√22[cosx+sinx]sinx=√22[cosx+sinx]×1sinx=√22[cosx+sinx]×1sinx×√2√2=22√2[cosx+sinx]×1sinx=1√2[cosx+sinx]×1sinx=cosx+sinx√2sinx
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:?
√22(SinxCosx)Sinx
√22(SinxCosx)Sinx∗√2√2
which really means:
√2∗√22(SinxCosx)√2Sinx
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..
