1.
Given \(\sin A=-\frac13\) with angle \(A\) in Quadrant III, determine the exact value of \(\sin(A+\frac{\pi}{6})\).
Solution: \(\cos A=-\frac{\sqrt8}{3}\). Use \(\sin(x+y)=\sin x\cos y+\cos x\sin y\). Compute \((-\frac13)(\frac{\sqrt3}{2})+(-\frac{\sqrt8}{3})(\frac12)=\frac{-\sqrt3-\sqrt8}{6}\). Answer: B
2.
Identify which step is incorrect in the proof of \(\sin75^\circ=\frac14(\sqrt6+\sqrt2)\).
Step 3 incorrectly substitutes angles. Answer: F
3.
Simplify \(\frac{\cos2x+\sin^2x}{\sin2x}\).
Use \(\cos2x=\cos^2x-\sin^2x\), \(\sin2x=2\sin x\cos x\). Numerator becomes \(\cos^2x\), giving \(\frac{\cos x}{2\sin x}=\frac12\cot x\). Answer: C
4.
Prove \(\cos^2(2x)=1-\sin^2(2x)\). Identify the first incorrect step.
Step 6 changes -1 to +1 incorrectly. Answer: F
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