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Question Number 141383 by Willson last updated on 18/May/21

sin^2 (((f(x))/2))=1−(1/2)(√(1−x^2 )) find f(x)

$$\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$

Answered by mr W last updated on 18/May/21

sin^2 (((f(x))/2))=1−(1/2)(√(1−x^2 )) 1−sin^2 (((f(x))/2))=(1/2)(√(1−x^2 )) cos^2 (((f(x))/2))=(1/2)(√(1−x^2 )) 2cos^2 (((f(x))/2))−1=(√(1−x^2 ))−1 cos (f(x))=(√(1−x^2 ))−1 ⇒f(x)=cos^(−1) ((√(1−x^2 ))−1)

$$\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{2cos}^{\mathrm{2}} \:\left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\right)−\mathrm{1}=\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }−\mathrm{1} \\ $$$$\mathrm{cos}\:\left({f}\left({x}\right)\right)=\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }−\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$

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