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

2 Answers
Mar 1, 2016

#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

Mar 1, 2016

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#