Answer:
[tex]-\cos nw_0t [/tex]
Step-by-step explanation:
We can use the double angle identity for sine.
[tex]\sin (A+B)=\sin A \cos B+\sin B\cos A[/tex]
The given expresion is
[tex]\sin (nw_0t-90) [/tex]
We apply the identity to get:
[tex]\sin (nw_0t-90)=\sin nw_0t \cos 90-\cos nw_0t \sin 90[/tex]
We simplify to obtain;
[tex]\sin (nw_0t-90)=\sin nw_0t (0)-\cos nw_0t (1)[/tex]
This simplifies to:
[tex]\sin (nw_0t-90)=-\cos nw_0t [/tex] The correct answer is D
