f(x)=sinx

1/f(165°) এর মান কত? 

Another Explanation (5): প্রশ্ন: \(f(x) = \sin x \times \frac{1}{f(165^\circ)}\) এর মান কত? উত্তর: \( \sqrt{6} + \sqrt{2} \) --- সমাধান: প্রথমে, \(f(165^\circ)\) এর মান নির্ণয় করি। \(f(165^\circ) = \sin 165^\circ\) \(\sin 165^\circ = \sin (180^\circ - 15^\circ) = \sin 15^\circ\) এবং, \(\sin 15^\circ = \sin (45^\circ - 30^\circ)\) \[ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \] মূল্যমান: \[ \sin 45^\circ = \frac{\sqrt{2}}{2} \] \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \] \[ \cos 45^\circ = \frac{\sqrt{2}}{2} \] \[ \sin 30^\circ = \frac{1}{2} \] অতএব, \[ \sin 15^\circ = \left( \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{2}}{2} \times \frac{1}{2} \right) \] \[ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \] সুতরাং, \[ f(165^\circ) = \sin 165^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} \] এখন, \(f(x)\) এর মান নির্ণয় করি: \[ f(x) = \sin x \times \frac{1}{f(165^\circ)} = \sin x \div \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right) = \sin x \times \frac{4}{\sqrt{6} - \sqrt{2}} \] সাধারণত, ডিনোমিনেটর রেডিক্যাল থেকে মুক্ত করতে র্যাশনালাইজ করবো: \[ f(x) = \sin x \times \frac{4}{\sqrt{6} - \sqrt{2}} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} = \sin x \times \frac{4(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} \] নোট করুন, \[ (\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2}) = 6 - 2 = 4 \] অতএব, \[ f(x) = \sin x \times \frac{4(\sqrt{6} + \sqrt{2})}{4} = \sin x \times (\sqrt{6} + \sqrt{2}) \] --- সুতরাং, \(f(x)\) এর মান হলো: \[ \boxed{ f(x) = \sin x \times (\sqrt{6} + \sqrt{2}) } \] প্রশ্নে উল্লেখিত মানটি হলো, অর্থাৎ \(f(x) = \sqrt{6} + \sqrt{2}\) অতএব, \( \sin x = 1 \) যেহেতু, \[ \sin x = 1 \implies x = 90^\circ \] --- **উত্তর:** \[ \boxed{ \sin 90^\circ \times (\sqrt{6} + \sqrt{2}) = 1 \times (\sqrt{6} + \sqrt{2}) = \sqrt{6} + \sqrt{2} } \]