(i*x)*asin(sin(12))

1.	(ix)asin(sin(12))
Answer» sin^-1{sinx}=x; iff x is in proper limits we have to check wether x is in its proper limit i.e. -π/2<=x<=π/2Now,let nπ<x<(n+1)π let,x=(n+1)π-y {iff x is more near to (n+1)π}x=nπ+y {iff x is more near to nπ} since difference b/w max and min value of x is π so range of y is given by: 0<y<π/2 So, (n+1)π-x = y; is clearly in its proper limits[ since : 0<y<π/2] Now, 3π<12<4π sin12=-sin(4π-12)=sin(12–4π) {sin(2nπ-x)=-sinx} Therefore, sin^-1{(sin12)} = sin^-1{sin(12–4π)}=12–4π [ sin^-1{sinx}=x ; iff x is in proper limits] sin^-1(sin12)=Sin^-1sin12=12

(ix)asin(sin(12))

Answer»

sin^-1{sinx}=x; iff x is in proper limits

we have to check wether x is in its proper limit i.e. -π/2<=x<=π/2Now,let nπ<x<(n+1)π

let,x=(n+1)π-y {iff x is more near to (n+1)π}x=nπ+y {iff x is more near to nπ}

since difference b/w max and min value of x is π so range of y is given by: 0<y<π/2

So, (n+1)π-x = y; is clearly in its proper limits[ since : 0<y<π/2]

Now, 3π<12<4π

sin12=-sin(4π-12)=sin(12–4π) {sin(2nπ-x)=-sinx}

Therefore, sin^-1{(sin12)} = sin^-1{sin(12–4π)}=12–4π

[ sin^-1{sinx}=x ; iff x is in proper limits]

sin^-1(sin12)=Sin^-1sin12=12