Question
(1) EX-d'application 1. Vérifier que \( \frac{7 \pi}{12}=\frac{\pi}{4}+\frac{\pi}{3} \). 2. Calculer \( \cos \left(\frac{7 \pi}{12}\right) \) et \( \sin \left(\frac{7 \pi}{12}\right) \). 3. Calculer \( \cos \left(\frac{13 \pi}{12}\right) \) et \( \sin \left(\frac{13 \pi}{12}\right) \). 4. Calculer \( \cos \left(\frac{5 \pi}{12}\right) \) et \( \sin \left(\frac{5 \pi}{12}\right) \). Montrer que : 1. \( \cos \left(\frac{10 \pi}{9}\right) \cos \left(\frac{\pi}{9}\right)+\sin \left(\frac{10 \pi}{9}\right) \sin \left(\frac{\pi}{9}\right)=-1 \) 2. \( \sin \left(\frac{5 \pi}{9}\right) \cos \left(\frac{\pi}{18}\right)-\cos \left(\frac{5 \pi}{9}\right) \sin \left(\frac{\pi}{18}\right)=1 \). 3. \( \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{6 \pi}{9}\right)=\sin \left(\frac{6 \pi}{7}\right) \cos \left(\frac{\pi}{7}\right) \). 4. \( \tan \frac{\pi}{16}+\tan \frac{3 \pi}{16}+\tan \frac{\pi}{16} \tan \frac{3 \pi}{16}=1 \). 5. \( \tan \frac{5 \pi}{16}-\tan \frac{\pi}{16}-\tan \frac{5 \pi}{16} \tan \frac{\pi}{16}=1 \) (a) Déduire que: \[ \tan \frac{\pi}{16}=\frac{\tan \frac{5 \pi}{16}+\tan \frac{3 \pi}{16}}{\tan \frac{5 \pi}{16}-\tan \frac{3 \pi}{16}} \]
Ask by Fuentes Reed. in Morocco
Jan 20,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
1. \( \frac{7 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{3} \) صحيح.
2. \( \cos \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4} \) و \( \sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2} + \sqrt{6}}{4} \).
3. \( \cos \left(\frac{13 \pi}{12}\right) = -\cos \left(\frac{\pi}{12}\right) \) و \( \sin \left(\frac{13 \pi}{12}\right) = -\sin \left(\frac{\pi}{12}\right) \).
4. \( \cos \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{6} - \sqrt{2}}{4} \) و \( \sin \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} \).
5.
- \( \cos \left(\frac{10 \pi}{9}\right) \cos \left(\frac{\pi}{9}\right) + \sin \left(\frac{10 \pi}{9}\right) \sin \left(\frac{\pi}{9}\right) = -1 \)
- \( \sin \left(\frac{5 \pi}{9}\right) \cos \left(\frac{\pi}{18}\right) - \cos \left(\frac{5 \pi}{9}\right) \sin \left(\frac{\pi}{18}\right) = 1 \)
- \( \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{6 \pi}{9}\right) = \sin \left(\frac{6 \pi}{7}\right) \cos \left(\frac{\pi}{7}\right) \)
- \( \tan \frac{\pi}{16} + \tan \frac{3 \pi}{16} + \tan \frac{\pi}{16} \tan \frac{3 \pi}{16} = 1 \)
- \( \tan \frac{5 \pi}{16} - \tan \frac{\pi}{16} - \tan \frac{5 \pi}{16} \tan \frac{\pi}{16} = 1 \)
6. من المعادلات الأخيرة، نستنتج أن:
\[
\tan \frac{\pi}{16} = \frac{\tan \frac{5 \pi}{16} + \tan \frac{3 \pi}{16}}{\tan \frac{5 \pi}{16} - \tan \frac{3 \pi}{16}}
\]
Solution
لنبدأ بحل المسائل خطوة بخطوة.
### 1. التحقق من \( \frac{7 \pi}{12}=\frac{\pi}{4}+\frac{\pi}{3} \)
لحساب \( \frac{\pi}{4}+\frac{\pi}{3} \)، نحتاج إلى إيجاد المقام المشترك:
\[
\frac{\pi}{4} = \frac{3\pi}{12} \quad \text{و} \quad \frac{\pi}{3} = \frac{4\pi}{12}
\]
لذا:
\[
\frac{\pi}{4} + \frac{\pi}{3} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{7\pi}{12}
\]
وبذلك، تحققنا من المعادلة.
### 2. حساب \( \cos \left(\frac{7 \pi}{12}\right) \) و \( \sin \left(\frac{7 \pi}{12}\right) \)
نستخدم صيغة الجمع للجيب وجيب التمام:
\[
\cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right) - \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right)
\]
\[
\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right) + \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right)
\]
حساب القيم:
\[
\cos \left(\frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
\]
لذا:
\[
\cos \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}
\]
\[
\sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{2} + \sqrt{6}}{4}
\]
### 3. حساب \( \cos \left(\frac{13 \pi}{12}\right) \) و \( \sin \left(\frac{13 \pi}{12}\right) \)
نستخدم صيغة الجمع مرة أخرى:
\[
\cos \left(\frac{13 \pi}{12}\right) = \cos \left(\pi + \frac{\pi}{12}\right) = -\cos \left(\frac{\pi}{12}\right)
\]
\[
\sin \left(\frac{13 \pi}{12}\right) = \sin \left(\pi + \frac{\pi}{12}\right) = -\sin \left(\frac{\pi}{12}\right)
\]
### 4. حساب \( \cos \left(\frac{5 \pi}{12}\right) \) و \( \sin \left(\frac{5 \pi}{12}\right) \)
نستخدم صيغة الجمع:
\[
\cos \left(\frac{5 \pi}{12}\right) = \cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)
\]
\[
\sin \left(\frac{5 \pi}{12}\right) = \sin \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)
\]
### 5. إثبات المعادلات
الآن ننتقل إلى إثبات المعادلات المعطاة. سنبدأ بالمعادلة الأولى:
#### 1. إثبات \( \cos \left(\frac{10 \pi}{9}\right) \cos \left(\frac{\pi}{9}\right)+\sin \left(\frac{10 \pi}{9}\right) \sin \left(\frac{\pi}{9}\right)=-1 \)
نستخدم صيغة جمع جيب التمام:
\[
\cos \left(\frac{10 \pi}{9} - \frac{\pi}{9}\right) = \cos \left(\frac{9 \pi}{9}\right) = \cos(\pi) = -1
\]
#### 2. إثبات \( \sin \left(\frac{5 \pi}{9}\right) \cos \left(\frac{\pi}{18}\right)-\cos \left(\frac{5 \pi}{9}\right) \sin \left(\frac{\pi}{18}\right)=1 \)
نستخدم صيغة فرق الجيب:
\[
\sin \left(\frac{5 \pi}{9} - \frac{\pi}{18}\right) = \sin \left(\frac{10 \pi}{18} - \frac{\pi}{18}\right) = \sin \left(\frac{9 \pi}{18}\right) = \sin \left(\frac{\pi}{2}\right) = 1
\]
#### 3. إثبات \( \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{6 \pi}{9}\right)=\sin \left(\frac{6 \pi}{7}\right) \cos \left(\frac
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Simplify this solution Bonus Knowledge
Pour vérifiez que \( \frac{7 \pi}{12}=\frac{\pi}{4}+\frac{\pi}{3} \), il suffit de mettre ces deux fractions au même dénominateur. En effet, \( \frac{\pi}{4}=\frac{3\pi}{12} \) et \( \frac{\pi}{3}=\frac{4\pi}{12} \), donc \( \frac{\pi}{4}+\frac{\pi}{3}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{7\pi}{12} \). C'est une belle façon de s'assurer que les angles s'additionnent correctement! Pour calculer \( \cos \left(\frac{7 \pi}{12}\right) \) et \( \sin \left(\frac{7 \pi}{12}\right) \), on peut utiliser les formules de l'addition des angles. On sait que \( \frac{7 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{3} \). Donc, \( \cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{3}\right) \) ce qui donne un résultat impressionnant de \( -\frac{\sqrt{3}}{4} \). Pour le sinus, ça donne \( \sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{3}\right) \), soit \( \frac{\sqrt{3}}{4} + \frac{1}{4} \sqrt{2} = \frac{\sqrt{6}+\sqrt{2}}{4} \). Voilà de quoi épater la galerie!
