Algebra 2 Graphs of Trigonometric Functions and Identities Quiz: 10 Questions with Explanations

A -6B 6C 12D 2π

Amplitude is the absolute value of A, so 6. The negative sign reflects the graph but does not change the amplitude.

A \(\frac{\pi}{3}\)B \(6\pi\)C \(2\pi\)D \(\frac{\pi}{6}\)

Period = 2π / 6 = π/3.

A \(y = 0\)B \(y = 1\)C \(y = 2\)D \(y = 3\)

The midline is halfway between the maximum 3 and the minimum -1, so y = 1.

A 3B 12C 7D 1

Maximum = midline + amplitude = 4 + 3 = 7.

A \(\sin^{2}x + \cos^{2}x = 1\)B \(\sin x + \cos x = 1\)C \(\sin^{2}x – \cos^{2}x = 1\)D \(\sin x\cos x = 1\)

The Pythagorean identity follows from the unit circle having radius 1.

A \(-\frac{3}{5}\)B \(\frac{1}{5}\)C \(\frac{3}{5}\)D \(\frac{9}{25}\)

cos² θ = 1 – 16/25 = 9/25, and cosine is positive in quadrant I, so cos θ = 3/5.

A \(y = 5\cos x + 2\)B \(y = 2\sin x + 5\)C \(y = 2\cos(2x) + 5\)D \(y = 2\cos x + 5\)

Cosine starts at its maximum; A = 2 and D = 5; B = 1 gives period 2π.

A Cosine is sine shifted right by πB Cosine is sine reflected over the x-axisC They are the same graphD Cosine is sine shifted left by π/2

cos x = sin(x + π/2), a left shift of π/2.

A \(\frac{15}{17}\)B \(-\frac{15}{17}\)C \(\frac{9}{17}\)D \(\frac{8}{15}\)

sin² θ = 1 – 64/289 = 225/289; sine is positive in quadrant II.

A 7 feetB 9 feetC 2 feetD 14 feet

Maximum = midline + amplitude = 7 + 2 = 9.