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 $=option (4)$