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

Question

7292 views around the world

You can reuse this answer
Creative Commons License

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."#