2.
Item 2. Simplify \(\cos(\pi+\theta)\).
Solution: Use the cosine sum identity \(\cos(x+y)=\cos x\cos y-\sin x\sin y\). \(\cos(\pi+\theta)=\cos\pi\cos\theta-\sin\pi\sin\theta\). Since \(\cos\pi=-1\) and \(\sin\pi=0\), \(\cos(\pi+\theta)=(-1)\cos\theta-(0)\sin\theta=-\cos\theta\). Answer: C
3.
Item 3. Identify which step shows the first mistake within this proof. Prove: \(\cos(\alpha+\beta)\times\cos(\alpha-\beta)=\cos^2\alpha-\sin^2\beta\).
Solution: The first error is in Step 7. From Step 6, the expression should simplify to \(-1+\cos^2\beta+\cos^2\alpha\), not \(-1+\cos^2\beta-\cos^2\alpha\). Therefore the first incorrect step is Step 7. Answer: G
4.
Item 4. Simplify \(\frac{\sin 2x}{1-\cos 2x}\).
Solution: Use the double-angle identities \(\sin 2x=2\sin x\cos x\) and \(\cos 2x=1-2\sin^2 x\). \(\frac{\sin 2x}{1-\cos 2x}=\frac{2\sin x\cos x}{1-(1-2\sin^2 x)}=\frac{2\sin x\cos x}{2\sin^2 x}\). Cancel \(2\sin x\): \(\frac{\cos x}{\sin x}=\cot x\). Answer: A
5.
Item 5. Prove \(\cos^2(2x)=1-\sin^2(2x)\). Identify the first step in which an error is made.
Solution: The first error is in Step 6. Step 5 is \(2(\cos^2x+\sin^2x)-1-4\sin^2x\cos^2x\). Since \(\cos^2x+\sin^2x=1\), Step 6 should be \(2(1)-1-4\sin^2x\cos^2x\), not \(2(1)+1-4\sin^2x\cos^2x\). Therefore the first incorrect step is Step 6. Answer: F
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