Question

How do you simplify `sin(x + (3π)/2) cos x`?

Answer 1

`-cos^2x`

Explanation:

`sin(pi+(pi/2+x))cosx`
knowing that `sin(pi+alpha)=-sin(alpha)`
`=-sin(pi/2+x)cosx`
knowing that `sin(pi/2+alpha)=cos(alpha)`
`=-cosxcosx`
`=-cos^2x`

Answer 2

`-cos^2x`

Explanation:

Expand `sin(x+(3pi)/2)" using "color(blue)"addition formula"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(sin(A+B)=sinAcosB+cosAsinB)color(white)(a/a)|)))`

`rArrsin(x+(3pi)/2)=sinxcos((3pi)/2)+cosxsin((3pi)/2)`

`color(orange)"Reminder"`

`color(red)(bar(ul(|color(white)(a/a)color(black)(cos((3pi)/2)=0" and " sin((3pi)/2)=-1)color(white)(a/a)|)))`

`rArrsinxcos((3pi)/2)+cosxsin((3pi)/2)`

`=0-cosx=-cosx`

`rArrsin(x+(3pi)/2)cosx=-cosx(cosx)=-cos^2x`