Question

algebra ii h - kling - 02 (lms)
alg. iih: 7 - 2 evaluating trigonometric functions: day 1: mathxl (lms graded)
part 1 of 4
identify the quadrant and reference angle for the given angle. then find the sine, cosine, and tangent.
\\( \frac { 25 \pi } { 4 } \\)
the angle is in and the reference angle is
(type an exact answer, using radicals as needed. use integers or fractions for any numbers in the expression.)
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Explanation:

Step 1: Find the coterminal angle

To find the coterminal angle of \(\frac{25\pi}{4}\), we use the formula \(\theta - 2k\pi\) (\(k\in\mathbb{Z}\)).
Let \(k = 3\), then \(\frac{25\pi}{4}-2\times3\pi=\frac{25\pi}{4}-6\pi=\frac{25\pi - 24\pi}{4}=\frac{\pi}{4}\).
Since \(\frac{\pi}{4}\) is in the first - quadrant, and \(\frac{25\pi}{4}\) and \(\frac{\pi}{4}\) are coterminal.

Step 2: Determine the quadrant

The angle \(\frac{25\pi}{4}\) is coterminal with \(\frac{\pi}{4}\), so \(\frac{25\pi}{4}\) is in the first quadrant.

Step 3: Find the reference angle

For an angle \(\theta\) in the first quadrant, the reference angle \(\theta_{r}=\theta\). So the reference angle of \(\frac{25\pi}{4}\) is \(\frac{\pi}{4}\).

Step 4: Calculate trigonometric functions

We know that \(\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\), \(\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\), \(\tan\frac{\pi}{4} = 1\).
Since \(\frac{25\pi}{4}\) and \(\frac{\pi}{4}\) are coterminal and in the first quadrant:
\(\sin\frac{25\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\), \(\cos\frac{25\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\), \(\tan\frac{25\pi}{4}=\tan\frac{\pi}{4}=1\).

Answer:

The angle is in the first quadrant and the reference angle is \(\frac{\pi}{4}\). \(\sin\frac{25\pi}{4}=\frac{\sqrt{2}}{2}\), \(\cos\frac{25\pi}{4}=\frac{\sqrt{2}}{2}\), \(\tan\frac{25\pi}{4}=1\).