Question

Question #ffc58

Answer 1

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)cos(A+B)=cosAcosB-sinAsinB`

`•color(white)(x)sin(A-B)=sinAcosB-cosAsinB`

`•color(white)(x)cos2A=cos^2A-sin^2A`

`"consider left hand side"`

`sinx(cosxcosy-sinxsiny)-cosx(sinxcosy-cosxsiny)`

`=cancel(sinxcosxcosy)-sin^2xsinycancel(-sinxcosxcosy)+cos^2xsiny`

`=siny(cos^2x-sin^2x)`

`=sinycos2x=" right hand side "rArr" proved"`

Answer 2

We have: `sin(x) cos(x + y) - cos(x) sin(x - y)`

Let's apply the compound angle identities for `sin(x)` and `cos(x)`:

`= sin(x) cdot (cos(x) cos(y) + sin(x) sin(y)) - cos(x) cdot (sin(x) cos(y) - cos(x) sin(y))`

`= sin(x) cos(x) cos(y) + sin^(2)(x) sin(y) - sin(x) cos(x) cos(y) - cos^(2)(x) sin(y)`

`= sin^(2)(x) sin(y) - cos^(2)(x) sin(y)`

`= sin(y) (sin^(2)(x) - cos^(2)(x))`

`= sin(y) (- (cos^(2)(x) - sin^(2)(x)))`

`= - sin(y) (cos^(2)(x) - sin^(2)(x))`

Then, let's apply the double angle identity for `cos(x)`; `cos(2 x) = cos^(2)(x) - sin^(2)(x)`:

`= - sin(y) cdot cos(2 x)`

`= - cos(2 x) sin(y)`