यदि alpha = 22^circ 30' है,तो (1 + cos alpha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) दिया है $\alpha = 22^\circ 30' = \frac{\pi}{8}$.माना $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.यहाँ $3\alpha =

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(A) दिया है $\alpha = 22^\circ 30' = \frac{\pi}{8}$.माना $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.यहाँ $3\alpha = 67^\circ 30'$,$5\alpha = 112^\circ 30'$,और $7\alpha = 157^\circ 30'$ है।हम जानते हैं कि $\cos 5\alpha = \cos(180^\circ - 67^\circ 30') = -\cos 3\alpha$ और $\cos 7\alpha = \cos(180^\circ - 22^\circ 30') = -\cos \alpha$.अतः,$P = (1 + \cos \alpha)(1 - \cos \alpha)(1 + \cos 3\alpha)(1 - \cos 3\alpha)$.$P = (1 - \cos^2 \alpha)(1 - \cos^2 3\alpha) = \sin^2 \alpha \sin^2 3\alpha$.चूँकि $\alpha = 22.5^\circ$,$\sin^2 \alpha = \frac{1 - \cos 45^\circ}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$.चूँकि $3\alpha = 67.5^\circ$,$\sin^2 3\alpha = \frac{1 - \cos 135^\circ}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}}$.$P = \left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right) \left(\frac{\sqrt{2} + 1}{2\sqrt{2}}\right) = \frac{2 - 1}{8} = \frac{1}{8}$.

यदि $\alpha = 22^\circ 30'$ है,तो $(1 + \cos \alpha )(1 + \cos 3\alpha )(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ का मान ज्ञात कीजिए।

Similar Questions

$\cos^2 76^{\circ} + \cos^2 16^{\circ} - \cos 76^{\circ} \cos 16^{\circ}$ का मान ज्ञात कीजिए।

यदि $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$ है, तो $\sinh x$ का मान क्या है?

$\frac{\cos 12^\circ - \sin 12^\circ}{\cos 12^\circ + \sin 12^\circ} + \frac{\sin 147^\circ}{\cos 147^\circ} = $

$\sec {50^o} + \tan {50^o}$ का मान क्या है?

$0 \leq x \leq \pi$ के लिए,यदि $81^{\sin ^2 x}+81^{\cos ^2 x}=30$ है,तो $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module