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

If sin^2 θ and cos^2 θ are the roots   of quadratic equation, find the equation.

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

Answered by bemath last updated on 09/Oct/20

⇒(x−sin^2 θ)(x−cos^2 θ)=0  ⇒x^2 −(sin^2 θ+cos^2 θ)x+sin^2 θcos^2 θ=0  ⇒x^2 −x+(1/4)sin^2 (2θ)=0  ⇒4x^2 −4x+sin^2 (2θ)=0

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

Commented by harckinwunmy last updated on 09/Oct/20

  can you explain how you got (1/4)sin^2 (2θ)=  from sin^2 θcos^2 θ

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

Commented by bobhans last updated on 09/Oct/20

⇒sin^2 θ cos^2 θ = ((1/2).2sin θ cos θ)^2   ⇒ (1/4).sin^2 (2θ)

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

Answered by Olaf last updated on 09/Oct/20

x^2 −Sx+P = 0  ⇒ x^2 −x+(1/4)sin^2 2θ = 0  The result is immediate.

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

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