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Trigonometry · real student question

Evaluate tan(4.2 pi), giving an exact value.

Question

Evaluate

tan⁡(4.2π)\tan(4.2\pi)

giving an exact value.

Step-by-step solution

  1. Use the period of tangent, which is π\pi and not 2π2\pi. Unlike sine and cosine, tangent repeats every half turn:

    tan⁡(θ+kπ)=tan⁡θfor any integer k\tan(\theta+k\pi)=\tan\theta\quad\text{for any integer }k

    Using 2π2\pi here would still work but would leave more to reduce.

  2. Split off whole multiples of π\pi. Write 4.2=4+0.24.2=4+0.2:

    tan⁡(4.2π)=tan⁡(4π+0.2π)=tan⁡(0.2π)\tan(4.2\pi)=\tan(4\pi+0.2\pi)=\tan(0.2\pi)

    since 4π4\pi is four whole periods.

  3. Convert the remaining angle to a familiar form.

    0.2π=π5=36∘0.2\pi=\frac{\pi}{5}=36^\circ

    so the problem reduces to finding tan⁡36∘\tan 36^\circ.

  4. Quote the exact value for the pentagon angle. The angle 36∘36^\circ is tied to the regular pentagon and the golden ratio, and its tangent has a closed radical form:

    tan⁡π5=5−25\tan\frac{\pi}{5}=\sqrt{5-2\sqrt5}

  5. Verify numerically. 5≈2.236068\sqrt5\approx 2.236068, so 5−25≈0.5278645-2\sqrt5\approx 0.527864 and its square root is ≈0.726543\approx 0.726543. A direct evaluation gives tan⁡(4.2π)≈0.7265425\tan(4.2\pi)\approx 0.7265425 ✓\checkmark. The value is positive, consistent with π5\tfrac{\pi}{5} lying in the first quadrant.

Answer

tan⁡(4.2π)=tan⁡π5=5−25≈0.72654\tan(4.2\pi)=\tan\frac{\pi}{5}=\sqrt{5-2\sqrt5}\approx 0.72654

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