Question Number 126701 by Eric002 last updated on 23/Dec/20
$${use}\:{right}\:{triangles}\:{to}\:{explain} \\ $$$${why}\:{cos}^{−\mathrm{1}} \left({x}\right)+{sin}^{−\mathrm{1}} \left({x}\right)=\pi/\mathrm{2} \\ $$
Answered by ebi last updated on 23/Dec/20
$$ \\ $$$${let}\:{sin}^{−\mathrm{1}} \left({x}\right)=\theta,\:{then}\:{x}={sin}\:\theta \\ $$$${sin}\:\theta={cos}\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$${cos}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{2}}−\theta \\ $$$${cos}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} \left({x}\right) \\ $$$${cos}^{−\mathrm{1}} \left({x}\right)+{sin}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 24/Dec/20
![let f(x)=arcosx +arcsinx−(π/2) ∀ x∈[−1,1] f^′ (x)=−(1/( (√(1−x^2 ))))+(1/( (√(1−x^2 ))))=0 ⇒f(x)=c f(0)=(π/2)−(π/2)=0 ⇒∀x∈[−1,1] f(x)=0 ⇒arcosx +arcsinx=(π/2)](Q126743.png)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arcosx}\:+\mathrm{arcsinx}−\frac{\pi}{\mathrm{2}} \\ $$$$\forall\:\mathrm{x}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\mathrm{f}^{'} \left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{c} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\forall\mathrm{x}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\:\Rightarrow\mathrm{arcosx}\:+\mathrm{arcsinx}=\frac{\pi}{\mathrm{2}} \\ $$