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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)...
arcsinx+arcsin(1x)=tsin(α)=xsin(β)=1xα+β=tsin(α)cos(β)+cos(α)sin(β)=sin(t)x.11+x+1x.1x=x+1x=1t=π2
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)
Letf(x)=arcsinx+arcsin1xf(x)=12x.11x121x.11(1x)f(x)=12x.11x121x.1xf(x)=0f(x)=cste=f(0)=arcsin(0)+arcsin(1)=0+π2=π2

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