Math  /  Algebra

QuestionNow consider h(x)=tan(π4(x+π/6))+3h(x)=\tan \left(\frac{\pi}{4}(x+\pi / 6)\right)+3 (a) (2 points) Compute h(0),h(π/6),h(π/2)h(0), h(\pi / 6), h(\pi / 2).

Studdy Solution
Simplify the expression:
h(π/2)=tan(π42π+π6)+3 h(\pi / 2) = \tan \left(\frac{\pi}{4} \cdot \frac{2\pi + \pi}{6}\right) + 3 h(π/2)=tan(π43π6)+3 h(\pi / 2) = \tan \left(\frac{\pi}{4} \cdot \frac{3\pi}{6}\right) + 3 h(π/2)=tan(π28)+3 h(\pi / 2) = \tan \left(\frac{\pi^2}{8}\right) + 3
Since π28\frac{\pi^2}{8} is not a standard angle, we approximate tan(π28)\tan\left(\frac{\pi^2}{8}\right) as a small angle, which is approximately π28\frac{\pi^2}{8}.
h(π/2)π28+3 h(\pi / 2) \approx \frac{\pi^2}{8} + 3
The values are:
h(0)π224+3 h(0) \approx \frac{\pi^2}{24} + 3 h(π/6)π212+3 h(\pi / 6) \approx \frac{\pi^2}{12} + 3 h(π/2)π28+3 h(\pi / 2) \approx \frac{\pi^2}{8} + 3

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