How do you find the exact value sin(x-y) if #sinx=4/9, siny=1/4#?

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Answer 1 · 2016-12-29T06:03:54

The answer is #=(4sqrt15-sqrt65)/36=0.21#

We need

#sin(A-B)=sinAcosB-sinBcosA#

#sin^2A+cos^2A=1#

Here, we have

#sinx=4/9#

#cos^2x=1-sin^2x=1-16/81=65/81#

#cosx=sqrt65/9#

#siny=1/4#

#cos^2y=1-sin^2y=1-1/16=15/16#

#cosy=sqrt15/4#

Therefore,

#sin(x-y)=sinxcosy-sinycosx#

#=4/9*sqrt15/4-1/4*sqrt65/9#

#=sqrt15/9-sqrt65/36#

#=(4sqrt15-sqrt65)/36=0.21#