1.
Item 2. Determine the reference angle for \(240^\circ\) in standard position.
The angle \(240^\circ\) lies in Quadrant III. In Quadrant III, the reference angle is found by subtracting \(180^\circ\). \(240^\circ-180^\circ=60^\circ\) Therefore the reference angle is \(60^\circ\). Answer: A
2.
Item 1. Find the measure of the unknown angle \(x\). [Diagram will be inserted later.]
Use the sine ratio because the opposite side and hypotenuse are given. \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\) \(\sin x=\frac{20}{30}\) \(x=\sin^{-1}\left(\frac{20}{30}\right)\) \(x\approx 41.8^\circ\) Therefore the unknown angle is approximately \(41.8^\circ\). Answer: A
3.
Item 5. Visualize the graph of \(\sin\theta\) for \(-2\pi\leq \theta \leq 2\pi\). For what values of \(\theta\) does the graph intersect the \(\theta\)-axis?
The graph intersects the \(\theta\)-axis where \(\sin\theta=0\). For the sine function, zeros occur at integer multiples of \(\pi\). Within \(-2\pi\leq \theta\leq 2\pi\), the multiples of \(\pi\) are: \(\theta=-2\pi, -\pi, 0, \pi, 2\pi\) Therefore the graph intersects the \(\theta\)-axis at these five values. Answer: C
4.
Item 6. Visualize the graph of \(\cos\theta\) for \(-2\pi\leq \theta \leq 2\pi\). For what values of \(\theta\) is the graph at the minimum value?
The minimum value of \(\cos\theta\) is \(-1\). On the interval \(-2\pi\leq \theta\leq 2\pi\), \(\cos\theta=-1\) at: \(\theta=-\pi\) and \(\theta=\pi\) Therefore the graph is at its minimum value at \(-\pi\) and \(\pi\). Answer: C
5.
Item 4. Determine the exact value of \(\sin\left(\frac{\pi}{2}\right)+\cos\pi\).
Evaluate each trigonometric value separately. \(\sin\left(\frac{\pi}{2}\right)=1\) \(\cos\pi=-1\) Add the two values. \(\sin\left(\frac{\pi}{2}\right)+\cos\pi=1+(-1)=0\) Therefore the exact value is \(0\). Answer: A
6.
Item 3. An angle of \(240^\circ\) expressed in radians is:
Convert degrees to radians using \(180^\circ=\pi\) radians. \(240^\circ\times \frac{\pi}{180^\circ}=\frac{240\pi}{180}\) Simplify the fraction. \(\frac{240\pi}{180}=\frac{4\pi}{3}\) Therefore \(240^\circ=\frac{4\pi}{3}\). Answer: A
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