Prove that cos(pi/10) = (1/4) sqrt(10 + 2 sqrt(5)) given tan(pi/10) = sqrt((5 − 2 * sqrt(5))/5)

نعلم أن $\tan^2\left(\frac{\pi}{10}\right) = \frac{5-2\sqrt{5}}{5}$.

$$1 + \tan^2\left(\frac{\pi}{10}\right) = 1 + \frac{5-2\sqrt{5}}{5} = \frac{5}{5} + \frac{5-2\sqrt{5}}{5} = \frac{5+5-2\sqrt{5}}{5} = \frac{10-2\sqrt{5}}{5}$$

إذن:

$$\frac{1}{\cos^2\left(\frac{\pi}{10}\right)} = \frac{10-2\sqrt{5}}{5}$$ $$\cos^2\left(\frac{\pi}{10}\right) = \frac{5}{10-2\sqrt{5}}$$

$$\cos^2\left(\frac{\pi}{10}\right) = \frac{5}{10-2\sqrt{5}} \times \frac{10+2\sqrt{5}}{10+2\sqrt{5}}$$ $$= \frac{5(10+2\sqrt{5})}{10^2 - (2\sqrt{5})^2} = \frac{5(10+2\sqrt{5})}{100 - (4 \times 5)} = \frac{5(10+2\sqrt{5})}{100 - 20} = \frac{5(10+2\sqrt{5})}{80}$$ $$= \frac{10+2\sqrt{5}}{16}$$

بما أن $\frac{\pi}{10}$ تقع في الربع الأول ($0 < \frac{\pi}{10} < \frac{\pi}{2}$)، فإن $\cos\left(\frac{\pi}{10}\right) > 0$.

$$\cos\left(\frac{\pi}{10}\right) = \sqrt{\frac{10+2\sqrt{5}}{16}} = \frac{\sqrt{10+2\sqrt{5}}}{\sqrt{16}} = \frac{\sqrt{10+2\sqrt{5}}}{4}$$ $$\boxed{\cos\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{10+2\sqrt{5}}}$$

2) حساب قيمة $\sin\left(\frac{\pi}{10}\right)$

$$\sin\left(\frac{\pi}{10}\right) = \tan\left(\frac{\pi}{10}\right) \times \cos\left(\frac{\pi}{10}\right)$$

نعوض بالقيم المعطاة والمحسوبة:

$$\sin\left(\frac{\pi}{10}\right) = \sqrt{\frac{5-2\sqrt{5}}{5}} \times \frac{1}{4}\sqrt{10+2\sqrt{5}}$$ $$\sin\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{\frac{(5-2\sqrt{5})(10+2\sqrt{5})}{5}}$$

نضرب البسط داخل الجذر:

$$(5-2\sqrt{5})(10+2\sqrt{5}) = 5 \times 10 + 5 \times 2\sqrt{5} - 2\sqrt{5} \times 10 - 2\sqrt{5} \times 2\sqrt{5}$$ $$= 50 + 10\sqrt{5} - 20\sqrt{5} - 4 \times 5$$ $$= 50 - 10\sqrt{5} - 20$$ $$= 30 - 10\sqrt{5}$$

نعود إلى تعبير $\sin\left(\frac{\pi}{10}\right)$:

$$\sin\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{\frac{30-10\sqrt{5}}{5}}$$ $$\sin\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{\frac{10(3-\sqrt{5})}{5}}$$ $$\sin\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{2(3-\sqrt{5})}$$ $$\boxed{\sin\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{6-2\sqrt{5}}}$$

3) استنتاج القيمة الدقيقة لـ $\cos\left(\frac{9\pi}{10}\right)$ و $\sin\left(\frac{3\pi}{5}\right)$

• حساب $\cos\left(\frac{9\pi}{10}\right)$: نلاحظ أن $\frac{9\pi}{10} = \pi - \frac{\pi}{10}$. باستخدام المتطابقة $\cos(\pi - x) = -\cos x$:

  $$\cos\left(\frac{9\pi}{10}\right) = \cos\left(\pi - \frac{\pi}{10}\right) = -\cos\left(\frac{\pi}{10}\right)$$

  من الجزء 1، لدينا $\cos\left(\frac{\pi}{10}\right) = \frac{1}{4}\sqrt{10+2\sqrt{5}}$. إذن:

  $$\boxed{\cos\left(\frac{9\pi}{10}\right) = -\frac{1}{4}\sqrt{10+2\sqrt{5}}}$$

• حساب $\sin\left(\frac{3\pi}{5}\right)$: نلاحظ أن $\frac{3\pi}{5} = \frac{6\pi}{10}$. يمكننا كتابة $\frac{6\pi}{10}$ كـ $\frac{5\pi + \pi}{10} = \frac{5\pi}{10} + \frac{\pi}{10} = \frac{\pi}{2} + \frac{\pi}{10}$. باستخدام المتطابقة $\sin\left(\frac{\pi}{2} + x\right) = \cos x$:

  $$\sin\left(\frac{3\pi}{5}\right) = \sin\left(\frac{\pi}{2} + \frac{\pi}{10}\right) = \cos\left(\frac{\pi}{10}\right)$$

  من الجزء 1، لدينا $\cos\left(\frac{\pi}{10}\right) =