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Question Number 140497 by benjo_mathlover last updated on 08/May/21

sin^(−1) (sin x)=x sin^(−1) (cos x)=? sin^(−1) (tan x)=?

$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)=\mathrm{x} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)=? \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{x}\right)=? \\ $$

Commented by mr W last updated on 08/May/21

sin^(−1) (sin x)=x is not always true.

$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)=\mathrm{x}\:{is}\:{not}\:{always}\:{true}. \\ $$

Commented by liberty last updated on 09/May/21

for what value of x this not true?

$$\mathrm{for}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{this}\:\mathrm{not}\:\mathrm{true}? \\ $$

Commented by mr W last updated on 09/May/21

example sin^(−1) (sin ((5π)/4))=−(π/4)≠((5π)/4)

$${example} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)=−\frac{\pi}{\mathrm{4}}\neq\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$

Commented by liberty last updated on 09/May/21

((5π)/4)∉D since not satisfy the equation

$$\frac{\mathrm{5}\pi}{\mathrm{4}}\notin\mathrm{D}\:\mathrm{since}\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$

Commented by liberty last updated on 09/May/21

y=sin^(−1) (x) has domain −(π/2)≤x≤(π/2)

$$\mathrm{y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{domain}\:−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$

Commented by mr W last updated on 09/May/21

y=sin^(−1) (x) has domain −1≤x≤1 y=sin^(−1) (x) has range −(π/2)≤x≤(π/2) y=sin^(−1) (sin (x)) has domain −∞≤x≤∞ y=sin^(−1) (sin (x)) has range −(π/2)≤x≤(π/2)

$$\mathrm{y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{domain}\:−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{1} \\ $$$$\mathrm{y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{has}\:{range}\:−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\left(\mathrm{x}\right)\right)\:\mathrm{has}\:\mathrm{domain}\:−\infty\leqslant\mathrm{x}\leqslant\infty \\ $$$$\mathrm{y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\left(\mathrm{x}\right)\right)\:\mathrm{has}\:{range}\:−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$

Commented by mr W last updated on 09/May/21

Answered by EDWIN88 last updated on 08/May/21

let sin^(−1) (cos x) = y cos x = sin y ; sin ((π/2)−x)=sin y we get y = (π/2)−x so sin^(−1) (cos x) = (π/2)−x

$$\mathrm{let}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{y} \\ $$$$\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{sin}\:\mathrm{y}\:;\:\cancel{\mathrm{sin}}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)=\cancel{\mathrm{sin}}\:\mathrm{y}\: \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{y}\:=\:\frac{\pi}{\mathrm{2}}−\mathrm{x}\:\mathrm{so}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\:=\:\frac{\pi}{\mathrm{2}}−\mathrm{x} \\ $$

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