# solve arcsin x + arccos 1/sq root 2=pie /2

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`sin^(-1)(x)+cos^(-1)(1/sqrt(2))=pi/2`

`sin^(-1)(x)=pi/2-cos^(-1)(1/sqrt(2))`

`x=sin(pi/2-cos^(-1)(1/sqrt(2)))`

`=sin(pi/2)cos(cos^(-1)(1/sqrt(2)))-cos(pi/2)sin(cos^(-1)(1/sqrt(2)))`

`=1/sqrt(2)-0xxsin(cos^(-1)(1sqrt(2)))`

`x=1/sqrt(2)`

`because`

`sin(pi/2)=1`

`cos(cos^(-1)(1/sqrt(2)))=1/sqrt(2)`

`cos(pi/2)=0`

`sin(A-B)=sin(A)cos(B)-cos(A)sin(B)`

Thus

x=1/sqrt(2)