Trig expression help?

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Answer 1 · 2018-08-06T09:18:45

The answer is #=option (4)#

Let

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

And

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

Then,

#cos(x+2y)=cosxcos2y-sinxsin2y#

#sinx=2/3#

#cosx=sqrt(1-sin^2x)=sqrt(1-4/9)#

#=sqrt5/3#

#siny=-1/3#

#cosy=sqrt(1-sin^2y)=sqrt(1-1/9)=(2sqrt2)/3#

#sin(2y)=2sinycosy=2*-1/3*(2sqrt2)/3#

#=-(4sqrt2)/9#

#cos(2y)=1-2sin^2y=1-2*1/9=7/9#

Finally,

#cos(x+2y)=sqrt5/3*7/9-2/3*-(4sqrt2)/9#

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

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

The answer is #=option (4)#