A reference angle is the smallest possible angle between a given angle measurement and the x-axis.

In this case, our angle \(\displaystyle 257^{\circ}\) lies in Quadrant III, so the angle is found by the formula \(\displaystyle \angle A_r = \angle A - 180^{\circ}\).

\(\displaystyle \angle A_r = \angle A - 180^{\circ}\) \(\displaystyle =\) \(\displaystyle 257^{\circ} - 180^{\circ} = 77^{\circ}\)

Thus, the reference angle for \(\displaystyle 257^{\circ}\) is \(\displaystyle 77^{\circ}\).

A reference angle is the smallest possible angle between a given angle measurement and the x-axis.

In this case, our angle \(\displaystyle 125^{\circ}\) lies in Quadrant II, so we can find our reference angle using the formula

\(\displaystyle \angle A_r = 180^{\circ}- \angle A\).

\(\displaystyle \angle A_r = 180^{\circ}- \angle A\) \(\displaystyle =\) \(\displaystyle 180^{\circ} - 125^{\circ} = 55^{\circ}\)

Thus, the reference angle for \(\displaystyle 125^{\circ}\) is \(\displaystyle 55^{\circ}\).

 The period of sinΘ is 2Π, so we set the new angle equal to the base period of 2Π and solve for Θ.

So 4Θ = 2Π and Θ = Π/2.

For the function

\(\displaystyle sin\, (Ax)\)

the period is equal to

\(\displaystyle \frac{2\pi}{A}\)

in this case

\(\displaystyle \frac{2\pi}{\pi}\)

which reduces to \(\displaystyle 2\).

For the function

\(\displaystyle sin\, (Ax)\)

the period is equal to

\(\displaystyle \frac{2\pi}{A}\)

in this case

\(\displaystyle \frac{2\pi}{\frac{1}{2}}=2\pi \cdot \frac{2}{1}\)

which reduces to \(\displaystyle 4\pi\).

To find the period of Sine and Cosine functions you use the formula:
\(\displaystyle \frac{2\pi}{\left | b\right |}\) where \(\displaystyle b\) comes from \(\displaystyle f(t) = a\sin(bt)\). Looking at our formula you see b is 4 so 
\(\displaystyle \frac{2\pi}{\left | 4\right |}=\frac{\pi}{2}\)

To find the period of a sine, cosine, cosecant, or secant funciton use the formula:

\(\displaystyle \omega = \frac{2\pi}{\left|b\right|}\) where \(\displaystyle b\) comes from the general formula: \(\displaystyle c(t)=a\sin(bt)\). We see that for our equation \(\displaystyle \left|b\right| = 2\) and so the period is \(\displaystyle \pi\) when you reduce the fraction.

To find period, simply remember the following formula:

\(\displaystyle \textup{period}=\frac{2\pi}{B}\)

where B is the coefficient in front of x. Thus,

\(\displaystyle \textup{period}=\frac{2\pi}{4}=\frac{\pi}{2}\)

The function \(\displaystyle y=-9\sin(5\theta-3)-3\) is related to the parent function \(\displaystyle y=\sin(\theta)\), which has a domain of \(\displaystyle (-\infty.\infty)\).

The value of theta for \(\displaystyle y=-9\sin(5\theta-3)-3\) has no restriction and is valid for all real numbers.  

The answer is \(\displaystyle (-\infty,\infty)\).

For both Sine and Cosine, since there are no asymptotes like Tangent and Cotangent functions, the function can take in any value for \(\displaystyle t\). Thus the domain is:

\(\displaystyle (-\infty,\infty)\)