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Diketahui sudut lancip A dengan cos 2A = 1/3. Nilai tan A = …

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Jawab:
Cos 2A = 1 – 2 sin² A
1/3 = 1 – 2 sin² A
2 sin² A = 1 – 1/3
2 sin² A = 2/3
sin² A = 1/3/2
sin² A = 2/3 × 2 = 1/3
cos² A = 1 – sin² A
cos² A = 1 – 1/3
cos² A = 2/3
cos A = √2/3

tan A = sin A/cos A
           = √1/3/√2/3 × √2/3/√2/3
           = √2/9/2/3
tan A = 1/3√2 × 3/2 = 1/2√2
  Nilai limit x mendekati 3 x² – 2x – 3/2x – 6 = ....