माना alpha ,beta इस प्रकार है कि pi

Question

(d) $\sin \alpha + \sin \beta = - \frac{{21}}{{65}},\;\cos \alpha + \cos \beta = - \frac{{27}}{{65}}$
अब ${(\sin \alpha + \sin \beta )^2} + {(\cos \a

Vedclass · Accepted Answer

$ - \frac{3}{{\sqrt {130} }}$

Vedclass · Answer

(d) $\sin \alpha + \sin \beta = - \frac{{21}}{{65}},\;\cos \alpha + \cos \beta = - \frac{{27}}{{65}}$
अब ${(\sin \alpha + \sin \beta )^2} + {(\cos \alpha + \cos \beta )^2}$
$ = {\left( {\frac{{ - 21}}{{65}}} \right)^2} + {\left( {\frac{{ - 27}}{{65}}} \right)^2}$
==> $2 + 2\sin \alpha \sin \beta + 2\cos \alpha \cos \beta = \frac{{441}}{{{{65}^2}}} + \frac{{729}}{{{{65}^2}}}$
==> $2 + 2\left[ {\cos (\alpha - \beta )} \right] = \frac{{1170}}{{{{(65)}^2}}} \Rightarrow 2.2{\cos ^2}\left( {\frac{{\alpha + \beta }}{2}} \right) = \frac{{1170}}{{{{(65)}^2}}}$
==> $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = \frac{{3\sqrt {130} }}{{130}} = \frac{3}{{\sqrt {130} }}$
अत: $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = \frac{{ - 3}}{{\sqrt {130} }}$, .

माना $\alpha ,\beta $ इस प्रकार है कि $\pi < (\alpha - \beta ) < 3\pi $. यदि $\sin \alpha + \sin \beta = - \frac{{21}}{{65}}$ तथा $\cos \alpha + \cos \beta = - \frac{{27}}{{65}},$ तो $\cos \frac{{\alpha - \beta }}{2}$ का मान है

Similar Questions

यदि $\tan \,(A + B) = p,\,\,\tan \,(A - B) = q,$ तो $\tan \,2A$ का मान $p$ तथा $q$ के पदों में है

यदि $\frac{{2\sin \alpha }}{{\{ 1 + \cos \alpha + \sin \alpha \} }} = y,$ हो, तो $\frac{{\{ 1 - \cos \alpha + \sin \alpha \} }}{{1 + \sin \alpha }} $ बराबर है

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi $ का मान है

$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = $

व्यंजक $\frac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$ बराबर है