We know that:

\(\displaystyle{\sin{{\left(\alpha-\beta\right)}}}={\sin{\alpha}}{\cos{\beta}}-{\cos{\alpha}}{\sin{\beta}}\)

Therefore,

\(\displaystyle{\sin{{\left({n}{w}{t}-{90}^{{\circ}}\right)}}}={\sin{{\left({n}{w}{t}\right)}}}{\cos{{\left({90}^{{\circ}}\right)}}}-{\cos{{\left({n}{w}{t}\right)}}}{\sin{{\left({90}^{{\circ}}\right)}}}={\sin{{\left({n}{w}{t}\right)}}}{\left({0}\right)}-{\cos{{\left({n}{w}{t}\right)}}}{\left({1}\right)}=-{\cos{{\left({n}{w}{t}\right)}}}\)

\(\displaystyle{\sin{{\left(\alpha-\beta\right)}}}={\sin{\alpha}}{\cos{\beta}}-{\cos{\alpha}}{\sin{\beta}}\)

Therefore,

\(\displaystyle{\sin{{\left({n}{w}{t}-{90}^{{\circ}}\right)}}}={\sin{{\left({n}{w}{t}\right)}}}{\cos{{\left({90}^{{\circ}}\right)}}}-{\cos{{\left({n}{w}{t}\right)}}}{\sin{{\left({90}^{{\circ}}\right)}}}={\sin{{\left({n}{w}{t}\right)}}}{\left({0}\right)}-{\cos{{\left({n}{w}{t}\right)}}}{\left({1}\right)}=-{\cos{{\left({n}{w}{t}\right)}}}\)