Question

How do you prove `sin ((3pi)/2 - x) = -cos x`?

Answer 1

`sin( (3pi)/2)cosx-cos((3 pi)/2)sinx=-1*cosx-0*sinx = -cosx`

Explanation:

Use the composite argument property for sin(A-B)=sinAcosB-cosAsinB to disolve the left hand side and simplify to show that it equals the right hand side

Answer 2

Make use of the compound angle formula and standard trig ratios.

Explanation:

We may use the following compound angle formula to simplify the left hand side in this case :

`sin(A+-B)=sinAcosB+-cosAsinB`.

So in this case, we get :

`sin((3pi)/2-x)=sin((3pi)/2)cosx-cos((3pi)/2)sinx`

`=(-1)cosx-(0)sinx`

`=-cosx`