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

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Answer 1 · 2016-10-04T18:33:18

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

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