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Question Number 117101 by harckinwunmy last updated on 09/Oct/20

$$\mathrm{If}{\sin }^2\theta \mathrm{and}{\cos }^2\theta \mathrm{are}\mathrm{the}\mathrm{roots}$$$$\mathrm{of}\mathrm{quadratic}\mathrm{equation},\mathrm{find}\mathrm{the}\mathrm{equation}.$$

Answered by bemath last updated on 09/Oct/20

$$\Rightarrow (\mathrm{x}-\sin {}^2\theta )(\mathrm{x}-\cos {}^2\theta )=0$$$$\Rightarrow {\mathrm{x}}^2-(\sin {}^2\theta +\cos {}^2\theta )\mathrm{x}+\sin {}^2\theta \cos {}^2\theta =0$$$$\Rightarrow {\mathrm{x}}^2-\mathrm{x}+\frac{1}{4}\sin {}^2\left(2\theta \right)=0$$$$\Rightarrow {4\mathrm{x}}^2-4\mathrm{x}+\sin {}^2\left(2\theta \right)=0$$

Commented by harckinwunmy last updated on 09/Oct/20

$$$$$$\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{how}\mathrm{you}\mathrm{got}\frac{1}{4}\sin {}^2\left(2\theta \right)=$$$$\mathrm{from}\sin {}^2\theta \cos {}^2\theta$$

Commented by bobhans last updated on 09/Oct/20

$$\Rightarrow \sin {}^2\theta \cos {}^2\theta ={(\frac{1}{2}.2\sin \theta \cos \theta )}^2$$$$\Rightarrow \frac{1}{4}.\sin {}^2\left(2\theta \right)$$

Answered by Olaf last updated on 09/Oct/20

$$x^2-\mathrm{S}x+\mathrm{P}=0$$$$\Rightarrow x^2-x+\frac{1}{4}{\sin }^{2}2\theta =0$$$$\mathrm{The}\mathrm{result}\mathrm{is}\mathrm{immediate}.$$

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