Could someone please explain to me how to solve this?

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Answer 1 · 2018-08-04T19:15:34

#cos[sin^-1(2/3)+2sin^-1(-1/3)]=(7sqrt5+8sqrt2)/27#

Let ,

#A=cos[sin^-1(2/3)+2sin^-1(-1/3)]#

Substitute , #x=sin^-1(2/3)tox in[-pi/2,pi/2]#

# and y=2sin^-1(-1/3)toy/2 in [-pi/4,pi/4]#

Now , #x=sin^-1(2/3)#

#=>color(red)(sinx=2/3 >0=>x in [0,pi/2]to1^(st)Quadrant#

#=>cosx=sqrt(1-sin^2x)=sqrt(1-4/9)=>color(red)(cosx=sqrt5/3#

Again ,

#y=2sin^-1(-1/3)#

#=>y/2=sin^-1(-1/3)#

#=>color(blue)(sin(y/2)=-1/3 <0=>y/2 in [-pi/4,0]tocolor(red)(4^(th)Quadrant#

#=>color(blue)(cos(y/2)=sqrt(1-sin^2(y/2))=sqrt(1-1/9)=sqrt8/3=(2sqrt2)/3#

So, #siny=2sin(y/2)cos(y/2)=2(-1/3)(2sqrt2/3)#

#=>color(red)(siny=-(4sqrt2)/9#

#and cosy=cos^2(y/2)-sin^2(y/2)#

#=>cosy=((2sqrt2)/3)^2-(-1/3)^2=8/9-1/9#

#=>color(red)cosy=7/9#

Hence ,

#A=cos(x+y)#

#A=cosxcosy-sinxsiny#

#:.A=sqrt5/3*7/9-2/3(-(4sqrt2)/9)#

#=>A=(7sqrt5)/27+(8sqrt2)/27#

#=>A=(7sqrt5+8sqrt2)/27#