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Question Number 190993 by mathlove last updated on 16/Apr/23

$${sin}^{2}x\cdot {cos}^{2}x=?$$

Answered by cortano12 last updated on 16/Apr/23

$$=\frac{1}{4}\sin {}^{2}2\mathrm{x}$$

Answered by mustafazaheen last updated on 16/Apr/23

$$=[\frac{2}{2}\sin \left(\mathrm{x}\right)\cdot \cos \left(\mathrm{x}\right)][\frac{2}{2}\sin \left(\mathrm{x}\right)\cdot \cos \left(\mathrm{x}\right)]$$$$=\frac{\sin \left(2\mathrm{x}\right)}{2}\times \frac{\sin \left(2\mathrm{x}\right)}{2}$$$$=\frac{1}{4}{\sin }^2\left(2\mathrm{x}\right)$$

Answered by manxsol last updated on 16/Apr/23

$$y={sin}^{2}x\cdot {cos}^{2}x$$$$4y=4{sin}^{2}x\cdot {cos}^{2}x$$$$4y={\left(2sinxcosx\right)}^2$$$$4y={\left(sin2x\right)}^2$$$$y=\frac{{\left(sin2x\right)}^2}{4}$$

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