Draw the graph of `y=sin^(-1)(2xsqrt(1-x^(2)))`

Explanation

We have `y=f(x)=sin^(-1)(2xsqrt(1-x^(2)))`
Clearly, the domain of the function is [-1, 1].
Let `x=sintheta, theta in[-pi//2,pi//2]`
`rArr" "theta=sin^(-1)x`

Now `sin(2xsqrt(1-x^(2)))=sin^(-1)(2 sinthetacostheta)`
`=sin^(-1)(sin2theta)`
`=sin^(-1)(sinalpha)," where "alpha in [-pi,pi]`
Now consider the graph of `y=sin^(-1)(sin alpha)," where "alpha in [-pi, pi]`
From the grph,
`y=f(x)=sin^(-1)(2xsqrt(1-x^(2)))=sin^(-1)(sin alpha)`
`={{:(-alpha-pi,-pilealphalt-pi//2),(alpha, -pi//2alphalepi//2),(-alpha+pi,pi//2ltalphalepi):}`
`={{:(-2sin^(-1)x-pi,-pile2sin^(-1)xlt-pi//2),(2sin^(-1)x,-pi//2le2sin^(-1)xlepi//2),(-2sin^(-1)x+pi,pi//2lt2sin^(-1)xltpi):}`
`={{:(-2sin^(-1)x-pi,-pile2sin^(-1)xlt-pi//4),(2sin^(-1)x,-pi//pile4sin^(-1)xlepi//4),(-2sin^(-1)x+pi,pi//4ltsin^(-1)xltpi//2):}`
`={{:(-2sin^(-1)x-pi,-1lexlt-1/sqrt2),(2sin^(-1)x,-1/sqrt2lexle1/sqrt2),(-2sin^(-1)x+pi,-1/sqrt2lexle1):}`
`={{:(-2/sqrt(1-x^(2)),-1ltxlt-1/sqrt(2)),(2/sqrt(1-x^(2)),-1ltxlt-1/sqrt(2)),(-2/sqrt(1-x^(2)),-1ltxlt1):}`
Clearly, y = f(x) is non-differentiable at `x=+-1/sqrt2`
y=f(x) inreases for `x in (-1/sqrt2,-1/sqrt2)` and decreases for `x in (-1,-1/sqrt2)cup(1/sqrt2,1)`
For f(x) `= -2 sin^(-1)x-pi`
`f(-1)=-2sin^(-1)(-1)-pi=0,f(-1/sqrt2)=-2(-pi/4)-pi=-pi/2`
Thus, `sin^(-1)(2xsqrt(1-x^(2)))` decreases from 0 to `-pi/2` when x increases from -1 to `-1/sqrt2`.
For `f(x)=2sin^(-2)x`
`f(-1/sqrt2)=2(-pi/4)=-pi/2`
`f(1/sqrt2)=2(pi/4)=pi/2`
Thus, `sin^(-1)(2xsqrt(1-x^(2)))" increases from "=pi/2"to"pi/2" when x in creases from "-1/sqrt2"to"1/sqrt2`
For `f(x)=-2sin^(-1)x+pi`
`f(1/sqrt2)=-2(pi/4)+pi=pi/2,f(1)=-2sin^(-1)(1)+pi=0`
Thus, `sin^(-1)(2xsqrt(1-x^(2)))" decreases from "pi/2"to"0`
When x increases from `1/sqrt2"to1`.
From this information, we can draw the graph of `sin^(-1)(2xsqrt(1-x^(2)))` as shown in the adjecnt figure.