Step 1: Expand the expression using algebraic identity

\[ f(x) = (\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \]

Step 2: Apply trigonometric identities

We know that: \[ \sin^2 x + \cos^2 x = 1 \quad \text{and} \quad 2\sin x \cos x = \sin 2x \]

Step 3: Combine the terms

\[ f(x) = (\sin^2 x + \cos^2 x) + 2\sin x \cos x = 1 + \sin 2x \quad \text{(Proved)} \]

Key Points:

• Full expansion earns M1 mark
• Correct simplification to required form earns A1 mark

Graph Characteristics:

• Amplitude: 1 (oscillates between 0 and 2)
• Period: \(2\pi\)
• Key Points:
  • Maximum at \(x=0\), \(y=2\)
  • Minimum at \(x=\pi\), \(y=0\)
  • Crosses y=1 at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\)

Marking Scheme:

• A1: Correct cosine shape with proper amplitude and period
• A1: Accurate positioning of minimum point at \((\pi, 0)\)

Step 1: Analyze the relationship between \(f(x)\) and \(g(x)\)

We have: \[ f(x) = 1 + \sin 2x \quad \text{and} \quad g(x) = 1 + \cos x \]

Step 2: Recognize the horizontal stretch

The period of \(f(x)\) is \(\pi\) (since \(\sin 2x\) has period \(\frac{2\pi}{2} = \pi\))
The period of \(g(x)\) is \(2\pi\)
Thus, the horizontal stretch factor \(p = 2\)

Step 3: Determine the phase shift

Using trigonometric identity: \[ \cos x = \sin\left(x + \frac{\pi}{2}\right) \] Therefore, the translation vector has horizontal component \(k = -\frac{\pi}{2}\) (shift left by \(\frac{\pi}{2}\))

Final Answer:

\[ p = 2, \quad k = -\frac{\pi}{2} \]

Marking Scheme:

• A1: Correct value of \(p = 2\)
• A1: Correct value of \(k = -\frac{\pi}{2}\)

Markscheme Summary