How do you simplify $\frac{4 {cos}^{3} x}{{sin}^{2} x} \cdot {(\frac{sin x}{4 cos x})}^{2}$ $(4cos^3x)/(sin^2x)*(sinx/(4cosx))^2$ ?

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Answer 1 · 2016-08-28T13:07:57

$= \frac{cos x}{4}$ $=cosx/4$

$\frac{4 {cos}^{3} x}{{sin}^{2} x} \cdot {(\frac{sin x}{4 cos x})}^{2}$ $(4cos^3x)/sin^2x*(sinx/(4cosx))^2$

$= \frac{4 {cos}^{3} x}{{sin}^{2} x} \cdot (\frac{{sin}^{2} x}{16 {cos}^{2} x})$ $=(4cos^3x)/sin^2x*(sin^2x/(16cos^2x))$

$= \frac{cos x}{4}$ $=cosx/4$

How do you simplify 4cos3xsin2x⋅(sinx4cosx)2(4cos^3x)/(sin^2x)*(sinx/(4cosx))^2?