cos left(frac{3 pi}{4}+xright)-sin left(frac{ | vedclass

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Question

(A) हम त्रिकोणमितीय सर्वसमिकाओं का उपयोग करते हैं: $\cos(A+B) = \cos A \cos B - \sin A \sin B$ और $\sin(A-B) = \sin A \cos B - \cos A \sin B$.$\cos \l

Vedclass · Accepted Answer

$-\sqrt{2} \cos x$

Vedclass · Answer

(A) हम त्रिकोणमितीय सर्वसमिकाओं का उपयोग करते हैं: $\cos(A+B) = \cos A \cos B - \sin A \sin B$ और $\sin(A-B) = \sin A \cos B - \cos A \sin B$.$\cos \left(\frac{3 \pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right)$$= \left(\cos \frac{3 \pi}{4} \cos x - \sin \frac{3 \pi}{4} \sin x\right) - \left(\sin \frac{\pi}{4} \cos x - \cos \frac{\pi}{4} \sin x\right)$चूंकि $\cos \frac{3 \pi}{4} = -\frac{1}{\sqrt{2}}$,$\sin \frac{3 \pi}{4} = \frac{1}{\sqrt{2}}$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,और $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:$= \left(-\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x\right) - \left(\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x\right)$$= -\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x$$= -\frac{2}{\sqrt{2}} \cos x = -\sqrt{2} \cos x$.

$\cos \left(\frac{3 \pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right) = $

Similar Questions

$\tan 15^\circ = $

मान ज्ञात कीजिए: $\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma )$

यदि $\sin \theta = \frac{-12}{13}$,$\cos \phi = \frac{-4}{5}$ और $\theta, \phi$ तीसरे चतुर्थांश में स्थित हैं,तो $\tan(\theta - \phi) =$

सिद्ध कीजिए कि $\cos \left( \frac{\pi}{4} + x \right) + \cos \left( \frac{\pi}{4} - x \right) = \sqrt{2} \cos x$.

यदि $\cos \alpha + \cos \beta = \frac{1}{3}$ और $\sin \alpha + \sin \beta = \frac{1}{4}$ है,तो $\cos (\alpha + \beta) = $

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