Answer
$10 \sin (t-0.664)$
Work Step by Step
The given equation has the form: $A\sin (Bt+\phi)$
$A=\sqrt {a_1^2+a_2^2}$
We are given that $8 \sin t-6 \cos t$ with $a_1=8$, $a_2=-6$ and $B=1$
So, we have: $A=\sqrt {(8)^2+(-6)^2}=10$
Now, $\cos \phi =\dfrac{a_1}{A}=\dfrac{8}{10}$ and $\tan \phi =\dfrac{a_1}{a_1}=\dfrac{-6}{8}=\dfrac{-3}{4}$
Since $\cos \phi$ is positive and $\sin \phi$ is negative, this implies that $\phi$ must be in the fourth quadrant.
We notice that $\tan^{1}(\dfrac{-3}{4}) \approx -0.644 $ is in the fourth quadrant. Thus, $\phi=\tan^{1}(\dfrac{-3}{4}) \approx -0.644 $
Therefore, we have: $8 \sin t-6 \cos t=A\sin (Bt+\phi)=10 \sin (t-0.664)$
