Sqrt(sin^8xcos^5x)sqrt(cosx) What is the answer?

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sqrt(sin^8xcos^5x)sqrt(cosx)

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Answer 1 · 2018-03-12T00:37:19

The simplified expression is ${sin}^{4} x \cdot {cos}^{3} x$ $sin^4x*cos^3x$ .

Use these radical rules:

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$ $sqrta*sqrtb=sqrt(ab)$

$\sqrt{a^{2}} = a$ $color(red)sqrt(color(black)a^2)=a$

Now here's the problem:

$= \sqrt{{sin}^{8} x \cdot {cos}^{5} x} \cdot \sqrt{cos x}$ $color(white)=sqrt(sin^8x*cos^5x)*sqrt(cosx)$

$= \sqrt{{sin}^{8} x \cdot {cos}^{5} x \cdot cos x}$ $=sqrt(sin^8x*cos^5x*cosx)$

$= \sqrt{{sin}^{8} x \cdot {cos}^{6} x}$ $=sqrt(sin^8x*cos^6x)$

$= \sqrt{{sin}^{8} x} \cdot \sqrt{{cos}^{6} x}$ $=sqrt(sin^8x)*sqrt(cos^6x)$

$= \sqrt{{({sin}^{4} x)}^{2}} \cdot \sqrt{{({cos}^{3} x)}^{2}}$ $=color(red)sqrt(color(black)((sin^4x))^2)*color(red)sqrt(color(black)((cos^3x))^2)$

$= {sin}^{4} x \cdot {cos}^{3} x$ $=sin^4x*cos^3x$