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# Get rid of fraction

0
94
3
+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!!..

Mar 3, 2023

#1
+118623
+1

$$\frac{\frac{\sqrt2}{2}cos\;x\;+\;sin\;x}{sin\;x}\\~\\ =\frac{\frac{\sqrt2cosx+2sinx}{2}}{sinx}\\ =\frac{\sqrt2cosx+2sinx}{2}\div sinx\\ =\frac{\sqrt2cosx+2sinx}{2sinx}\\ =\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\frac{2sinx}{\sqrt2}}\\ =\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\sqrt2sinx}\\ =\frac{cosx+\sqrt2sinx}{\sqrt2sinx}\\$$

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}\\

Mar 4, 2023
#2
+1124
0

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..

juriemagic  Mar 4, 2023
#3
+118623
+1

Yes that is correct and it is easier than what I did.