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,

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

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

$=sqrt15/9-sqrt65/36$

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

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