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Question Number 136679 by mnjuly1970 last updated on 24/Mar/21
         arcsin(√x) +arcsin((√(1−x)) )=t        sin(α)=(√x)    sin(β)=(√(1−x))       α+β=t      sin(α)cos(β)+cos(α)sin(β)=sin(t)    (√(x )) .(√(1−1+x)) +(√(1−x)) .(√(1−x))   =x+1−x=1     t=(π/2)...
$$\:\:\:\:\:\:\:\:\:{arcsin}\sqrt{{x}}\:+{arcsin}\left(\sqrt{\mathrm{1}−{x}}\:\right)={t} \\ $$$$\:\:\:\:\:\:{sin}\left(\alpha\right)=\sqrt{{x}}\:\:\:\:{sin}\left(\beta\right)=\sqrt{\mathrm{1}−{x}}\: \\ $$$$\:\:\:\:\alpha+\beta={t} \\ $$$$\:\:\:\:{sin}\left(\alpha\right){cos}\left(\beta\right)+{cos}\left(\alpha\right){sin}\left(\beta\right)={sin}\left({t}\right) \\ $$$$\:\:\sqrt{{x}\:}\:.\sqrt{\mathrm{1}−\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}\:.\sqrt{\mathrm{1}−{x}}\: \\ $$$$={x}+\mathrm{1}−{x}=\mathrm{1} \\ $$$$\:\:\:{t}=\frac{\pi}{\mathrm{2}}… \\ $$$$\:\:\: \\ $$
Answered by Olaf last updated on 25/Mar/21
Let f(x) = arcsin(√x)+arcsin(√(1−x))  f′(x) = (1/(2(√x))).(1/( (√(1−x))))−(1/(2(√(1−x)))).(1/( (√(1−(1−x)))))  f′(x) = (1/(2(√x))).(1/( (√(1−x))))−(1/(2(√(1−x)))).(1/( (√x)))  f′(x) = 0  ⇒ f(x) = cste = f(0) = arcsin(0)+arcsin(1)  = 0+(π/2) = (π/2)
$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\mathrm{arcsin}\sqrt{{x}}+\mathrm{arcsin}\sqrt{\mathrm{1}−{x}} \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−{x}\right)}} \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}}.\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{f}\left({x}\right)\:=\:\mathrm{cste}\:=\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{arcsin}\left(\mathrm{0}\right)+\mathrm{arcsin}\left(\mathrm{1}\right) \\ $$$$=\:\mathrm{0}+\frac{\pi}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{2}} \\ $$

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