Question

Trig expression help?

Answer 1

The answer is `=option (4)`

Explanation:

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)`