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Question-161668




Question Number 161668 by mnjuly1970 last updated on 21/Dec/21
Answered by mr W last updated on 21/Dec/21
((sin^2  x)/(1−sin^2  x))=1+sin^2  x  sin^4  x+sin^2  x−1=0  sin^2  x=((−1+(√5))/2)=φ  4 sin^2  (x/2)(1−sin^2  (x/2))=φ  sin^4  (x/2)−sin^2  (x/2)+(φ/4)=0  sin^2  (x/2)=((1±(√(1−φ)))/2)  (sin^2  (x/2))_(max) =((1+(√(1−φ)))/2)  =((1+(√((3−(√5))/2)))/2)=(1/2)+((√(6−2(√5)))/4)=(1/2)+(((√5)−1)/4)=(((√5)+1)/2)    ((√(6−2(√5)))/4)=((√(6−(√(20))))/4)  ⇒answer (3) is correct:
$$\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}=\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{x}+\mathrm{sin}^{\mathrm{2}} \:{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\phi \\ $$$$\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\right)=\phi \\ $$$$\mathrm{sin}^{\mathrm{4}} \:\frac{{x}}{\mathrm{2}}−\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}+\frac{\phi}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\phi}}{\mathrm{2}} \\ $$$$\left(\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\right)_{{max}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}−\phi}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}+\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\frac{\sqrt{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{6}−\sqrt{\mathrm{20}}}}{\mathrm{4}} \\ $$$$\Rightarrow{answer}\:\left(\mathrm{3}\right)\:{is}\:{correct}: \\ $$
Commented by mnjuly1970 last updated on 21/Dec/21
geateful sir W...very excellent
$${geateful}\:{sir}\:{W}…{very}\:{excellent} \\ $$

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