sum_{k=0}^4 sin^2 left( (2k+1) frac{pi}{20} r | vedclass

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

Question

(B) माना $S = \sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} \right)$.योगफल का विस्तार करने पर:$S = \sin^2 \frac{\pi}{20} + \sin^2 \frac{3\pi}{20} +

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(B) माना $S = \sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} ight)$.योगफल का विस्तार करने पर:$S = \sin^2 \frac{\pi}{20} + \sin^2 \frac{3\pi}{20} + \sin^2 \frac{5\pi}{20} + \sin^2 \frac{7\pi}{20} + \sin^2 \frac{9\pi}{20}$.सर्वसमिका $\sin^2 	heta = \cos^2 \left( \frac{\pi}{2} - 	heta ight)$ का उपयोग करने पर:$\sin^2 \frac{\pi}{20} = \cos^2 \frac{9\pi}{20}$ और $\sin^2 \frac{3\pi}{20} = \cos^2 \frac{7\pi}{20}$.अतः,$S = (\sin^2 \frac{9\pi}{20} + \cos^2 \frac{9\pi}{20}) + (\sin^2 \frac{7\pi}{20} + \cos^2 \frac{7\pi}{20}) + \sin^2 \frac{\pi}{4}$.$S = 1 + 1 + \frac{1}{2} = \frac{5}{2}$.

$\sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} \right) =$

Similar Questions

यदि $\sin x \cdot \cosh y = \cos \theta$ और $\cos x \cdot \sinh y = \sin \theta$ है,तो $\sin^2 x + \cosh^2 y = $

$\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right)$ का मान ज्ञात कीजिए।

$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ का मान है

यदि $\theta \in \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ है, तो $\sqrt{4 \cos^{4} \theta + \sin^{2} 2 \theta} + 4 \cot \theta \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module