Question

How do you find arcsin x = arccos (x/2)?

Answer 1

`x=2/sqrt5" is the Soln."`

Explanation:

Let, `arc sinx=arc cos (x/2)= theta rArr sintheta=x, costheta=x/2.`

But, `sin^2theta+cos^2theta=1 :. x^2+(x/2)^2=1, or, (5x^2)/4=1.`

`:. x=+-2/sqrt5`

Verification :

`x=-2/sqrt5`

`rArr arcsin(x)=arc sin(-2/sqrt5)=theta in [-pi/2,pi/2].`

`rArr sintheta=-2/sqrt5 lt0 rArr theta in [-pi/2,0] :.costheta gt0,` and,

`costheta=+sqrt(1-sin^2theta)=sqrt(1-4/5)=1/sqrt5.`

Now,`-pi/2lethetale0:.-pi/2+piletheta+pile0+pi`, i.e.,

`pi/2letheta+pilepi, or, (pi+theta) in [0,pi]`

Thus,

`cos(pi+theta)=-costheta=-1/sqrt5=x/2, &, (pi+theta) in [0,pi].`

Therefore, `arc cos(x/2)=theta+pinetheta=arc (sinx).`

Hence, `x=-2/sqrt5" is not the Soln."`

On the above lines, we can verify that `x=2/sqrt5" is the Soln."`