Question

Упростите выражение √2cosa-2cos(45°-a)/2sin(30°+a)-√3sina

Answer 1

Ответ:

$\boxed{\dfrac{\sqrt{2} \cos \alpha - 2 \cos(45^{\circ} - \alpha ) }{2 \sin (30^{\circ} + \alpha) - \sqrt{3} \sin \alpha }= -\sqrt{2}\ \rm tg \ \alpha}$

Формулы:

$\cos (\alpha - \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\rm tg \ \alpha = \dfrac{\sin\alpha }{\cos\alpha }$

Объяснение:

$\dfrac{\sqrt{2} \cos \alpha - 2 \cos(45^{\circ} - \alpha ) }{2 \sin (30^{\circ} + \alpha) - \sqrt{3} \sin \alpha }= \dfrac{\sqrt{2} \cos \alpha - ( \sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha ) }{ \cos \alpha + \sqrt{3} \sin \alpha - \sqrt{3} \sin \alpha} =$

$= \dfrac{\sqrt{2} \cos \alpha - \sqrt{2} \cos \alpha - \sqrt{2} \sin \alpha }{ \cos \alpha } = \dfrac{-\sqrt{2} \sin \alpha }{\cos \alpha } = -\sqrt{2}\ \rm tg \ \alpha$

а)

$2 \cos(45^{\circ} - \alpha ) = 2 ( \cos \alpha \cos 45^{\circ} + \sin \alpha \sin 45^{\circ} ) = 2 \bigg ( \dfrac{\sqrt{2} }{2} \cos \alpha + \dfrac{\sqrt{2} }{2} \sin \alpha \bigg ) =$

$= \dfrac{2\sqrt{2} }{2} \cos \alpha + \dfrac{2\sqrt{2} }{2} \sin \alpha = \sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha$

б)

$2 \sin (30^{\circ} + \alpha) = 2 ( \sin 30^{\circ} \cos \alpha + \cos 30^{\circ} \sin \alpha ) = 2 \bigg (\dfrac{1}{2} \cos \alpha + \dfrac{\sqrt{3}}{2} \sin \alpha \bigg ) =$

$= \dfrac{2 \cdot 1}{2} \cos \alpha + \dfrac{2\sqrt{3}}{2} \sin \alpha \bigg = \cos \alpha + \sqrt{3} \sin \alpha$