Answer
$\sqrt 2 \sin (t-45^{\circ} )$
Work Step by Step
The given equation has form of: $A\sin (Bt+\phi)$ where, $A=\sqrt {a_1^2+a_2^2}$
We are given that $- \sin t + \cos t$ with $a_1=-1$ and $a_2=1$ and $B=1$
So, we have: $A=\sqrt {(-1)^2+(1)^2}=\sqrt 2$
Now, $\cos \phi =\dfrac{a_1}{A}=-\dfrac{1}{\sqrt 2}; \sin \phi =\dfrac{a_2}{A}=\dfrac{1}{\sqrt 2}$ and $\tan \phi =\dfrac{a_2}{a_1}=\dfrac{1}{-1}=-1$
and $\phi =\tan^{-1}(-1)= -45^{\circ}$
or, $\phi =-45^{\circ}$
Therefore, we have: $- \sin t+ \cos t=A\sin (Bt+\phi)=\sqrt 2 \sin (t-45^{\circ} )$
