Adjacent: \(\sqrt{17^2 – 8^2} = \sqrt{289 – 64} = \sqrt{225} = 15\)

• \(\sin\theta = \frac{8}{17}\)    \(\csc\theta = \frac{17}{8}\)
• \(\cos\theta = \frac{15}{17}\)    \(\sec\theta = \frac{17}{15}\)
• \(\tan\theta = \frac{8}{15}\)    \(\cot\theta = \frac{15}{8}\)

Opposite: \(\sqrt{25^2 – 7^2} = \sqrt{625 – 49} = \sqrt{576} = 24\)

• \(\sin\theta = \frac{24}{25}\)    \(\csc\theta = \frac{25}{24}\)
• \(\cos\theta = \frac{7}{25}\)    \(\sec\theta = \frac{25}{7}\)
• \(\tan\theta = \frac{24}{7}\)    \(\cot\theta = \frac{7}{24}\)

In Q IV, sine is negative.

$$\sin\theta = -\sqrt{1 – \frac{64}{289}} = -\sqrt{\frac{225}{289}} = -\frac{15}{17}$$

$$\tan\theta = \frac{-15/17}{8/17} = -\frac{15}{8}$$

Hypotenuse: \(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\)

In Q II: sine is positive, cosine is negative.

$$\sin\theta = \frac{5}{13} \qquad \cos\theta = -\frac{12}{13}$$

\(\sin\theta = \frac{5}{13}\), so opposite = 5, hypotenuse = 13, adjacent = 12

• \(\sin\theta = \frac{5}{13}\)    \(\csc\theta = \frac{13}{5}\)
• \(\cos\theta = \frac{12}{13}\)    \(\sec\theta = \frac{13}{12}\)
• \(\tan\theta = \frac{5}{12}\)    \(\cot\theta = \frac{12}{5}\)

Q II, reference angle = 30°, sine is positive in Q II

$$\sin(150°) = \frac{1}{2}$$

Q III, reference angle = \(\frac{\pi}{6}\), cosine is negative in Q III

$$\cos\!\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$

Q IV, reference angle = 45°, tangent is negative in Q IV

$$\tan(315°) = -1$$

Q III, reference angle = \(\frac{\pi}{3}\), sine is negative in Q III

$$\sin\!\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$

Q III, reference angle = 30°, cosine is negative in Q III

$$\cos(210°) = -\frac{\sqrt{3}}{2}$$

Q IV, reference angle = \(\frac{\pi}{3}\), tangent is negative in Q IV

$$\tan\!\left(\frac{5\pi}{3}\right) = -\sqrt{3}$$

Amplitude = 3, midline = −2

Maximum: \(-2 + 3 = 1\)    Minimum: \(-2 – 3 = -5\)

Range: \([-5,\ 1]\)

$$x = 0,\quad x = \pi,\quad x = 2\pi$$

Odd, because \(\sin(-x) = -\sin(x)\) for all x.

The graph has 180° rotational symmetry about the origin.

Amplitude = 2, midline = 5. Since A is negative, the graph is reflected.

Maximum: \(5 + 2 = 7\)

Both have period \(2\pi\) — the graph completes one full cycle every \(2\pi\) units.

\(A = 4,\ B = 2,\ C = \pi,\ D = 3\)

• Amplitude: 4
• Period: \(\frac{2\pi}{2} = \pi\)
• Phase shift: \(\frac{\pi}{2}\) to the right
• Vertical shift: up 3 (midline \(y = 3\))

Rewrite: \(y = -3\sin\!\left(\frac{1}{2}x – \left(-\frac{\pi}{4}\right)\right) – 1\), so \(A = -3,\ B = \frac{1}{2},\ C = -\frac{\pi}{4},\ D = -1\)

• Amplitude: \(|-3| = 3\)
• Period: \(\frac{2\pi}{1/2} = 4\pi\)
• Phase shift: \(\frac{-\pi/4}{1/2} = -\frac{\pi}{2}\) → left by \(\frac{\pi}{2}\)
• Vertical shift: down 1 (midline \(y = -1\))
• Note: reflected over midline (negative A)

$$A = \frac{6-(-2)}{2} = 4 \qquad D = \frac{6+(-2)}{2} = 2$$

$$\frac{2\pi}{B} = 3\pi \Rightarrow B = \frac{2}{3}$$

$$y = 4\sin\!\left(\frac{2x}{3}\right) + 2$$

$$A = \frac{10-2}{2} = 4 \qquad D = \frac{10+2}{2} = 6$$

$$\frac{2\pi}{B} = \pi \Rightarrow B = 2$$

$$y = 4\cos(2x) + 6$$

Amplitude = 5, midline = 2

Maximum: \(2 + 5 = 7\)    Minimum: \(2 – 5 = -3\)

Range: \([-3,\ 7]\)

\(A = 6,\ B = 4,\ C = \frac{\pi}{3},\ D = 5\)

• Amplitude: 6
• Period: \(\frac{2\pi}{4} = \frac{\pi}{2}\)
• Phase shift: \(\frac{\pi/3}{4} = \frac{\pi}{12}\) to the right
• Vertical shift: up 5 (midline \(y = 5\))

Period: \(\frac{\pi}{4}\)

Asymptotes: where \(4x = \frac{\pi}{2} + n\pi\), so \(x = \frac{\pi}{8} + \frac{n\pi}{4}\)

Period: \(\frac{\pi}{1/3} = 3\pi\)

Asymptotes: where \(\frac{x}{3} = n\pi\), so \(x = 3n\pi\)

Range: \((-\infty,\ \infty)\) — tangent has no maximum or minimum value.

Tangent equals zero where sine equals zero:

$$x = 0,\quad x = \pi,\quad x = 2\pi$$

Period: \(\frac{\pi}{3}\)

Asymptotes: where \(3x = \frac{\pi}{2} + n\pi\), so \(x = \frac{\pi}{6} + \frac{n\pi}{3}\)

Note: the vertical shift −1 does not affect asymptotes or period.

Asymptotes where \(\sin(3x) = 0\), i.e., \(3x = n\pi\)

$$x = \frac{n\pi}{3}, \quad n \in \mathbb{Z}$$

$$(-\infty,\ -1] \cup [1,\ \infty)$$

Secant never takes values between −1 and 1.

Asymptotes where \(\cos\!\left(\frac{x}{2}\right) = 0\), i.e., \(\frac{x}{2} = \frac{\pi}{2} + n\pi\)

$$x = \pi + 2n\pi, \quad n \in \mathbb{Z}$$

Period: \(2\pi\) — same as \(\sin(x)\) since csc is its reciprocal.

\(\sec(x) = \frac{1}{\cos(x)}\). For an x-intercept we need \(\sec(x) = 0\), which means \(\frac{1}{\cos(x)} = 0\). A fraction can only equal zero if the numerator is zero — but the numerator is always 1. So there are no x-intercepts.

Q IV, reference angle = \(\frac{\pi}{6}\), cosine is positive in Q IV

$$\cos\!\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

In Q III, cosine is also negative. Adjacent: \(\sqrt{5^2 – 4^2} = 3\)

• \(\sin\theta = -\frac{4}{5}\)    \(\csc\theta = -\frac{5}{4}\)
• \(\cos\theta = -\frac{3}{5}\)    \(\sec\theta = -\frac{5}{3}\)
• \(\tan\theta = \frac{4}{3}\)    \(\cot\theta = \frac{3}{4}\)

Note: tan and cot are positive in Q III ✅

\(A = -\frac{1}{2},\ B = 1,\ C = -\frac{\pi}{6},\ D = 4\)

• Amplitude: \(\frac{1}{2}\)
• Period: \(2\pi\)
• Phase shift: \(\frac{\pi}{6}\) to the left
• Vertical shift: up 4 (midline \(y = 4\))
• Note: reflected over midline (negative A)

$$A = \frac{3-(-7)}{2} = 5 \qquad D = \frac{3+(-7)}{2} = -2$$

$$\frac{2\pi}{B} = 8\pi \Rightarrow B = \frac{1}{4}$$

$$y = 5\cos\!\left(\frac{x}{4}\right) – 2$$

Period: \(\frac{\pi}{1/4} = 4\pi\)

Asymptotes: where \(\frac{x}{4} = \frac{\pi}{2} + n\pi\), so \(x = 2\pi + 4n\pi\)

Base range of \(\sec(x)\) is \((-\infty, -1] \cup [1, \infty)\).

Multiply by 3: \((-\infty, -3] \cup [3, \infty)\)

Shift down 1: \((-\infty, -4] \cup [2, \infty)\)

Range: \((-\infty,\ -4] \cup [2,\ \infty)\)

Q IV, reference angle = 30°, sine is negative in Q IV

$$\sin(330°) = -\frac{1}{2}$$

\(\cos\theta = -\frac{5}{13}\), so adjacent = −5, hypotenuse = 13

Opposite: \(\sqrt{13^2 – 5^2} = 12\). In Q II, sine is positive.

$$\sin\theta = \frac{12}{13} \qquad \tan\theta = \frac{12/13}{-5/13} = -\frac{12}{5}$$