Question

помогите с математикой пожалуйста​

Answer 1

1-ое задание.

$sin(75^{\circ}) = \: ?$

Представим число $75$ как сумму двух табличных углов.

$75 = 30 + 45$ ⇒ $sin(75^{\circ}) = sin(30^{\circ} + 45^{\circ})$ ;

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

$sin(30^{\circ}+45^{\circ}) = sin 30^{\circ} * cos 45^{\circ} + cos 30^{\circ} * sin 45^{\circ}$

$sin(30^{\circ}+45^{\circ}) = \frac{1}{2} * \frac{\sqrt{2} }{2} + \frac{\sqrt{3} }{2} * \frac{\sqrt{2} }{2}$ $= \frac{\sqrt{2} }{4}+ \frac{\sqrt{6} }{4}$ $= \frac{\sqrt{2} + \sqrt{6} }{4}$

$cos(75^{\circ}) = \: ?$

$cos(75^{\circ}) = cos(30^{\circ} + 45^{\circ})$

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

$cos(30^{\circ}+45^{\circ}) = cos30^{\circ} * cos45^{\circ} - sin30^{\circ} * sin45^{\circ}$

$cos(30^{\circ}+45^{\circ}) = \frac{\sqrt{3} }{2} * \frac{\sqrt{2} }{2} - \frac{1}{2} * \frac{\sqrt{2} }{2}$ $= \frac{\sqrt{6} }{4} - \frac{\sqrt{2} }{4}$ $= \frac{\sqrt{6} - \sqrt{2} }{4}$

2-ое задание.

а) $sin(\pi+x) = sin\pi * cosx + cos\pi * sin x$ $= 0 * cosx + (-1) * sinx$ $= -sinx$

б) $cos(\pi+x) = cos\pi * cosx - sin\pi * sinx$ $= -1 * cosx - 0 * sinx$ $= -cosx$

в) $sin(\frac{\pi}{2} -x) = sin\frac{\pi}{2} * cos x - cos\frac{\pi}{2} * sinx$ $= 1 * cosx - 0 * sinx = cosx$

г) $cos(\frac{\pi}{2} - x) = cos\frac{\pi}{2} * cosx + sin\frac{\pi}{2} * sinx = 0 * cosx + 1 * sinx = sinx$

3-е задание.

$sin255^{\circ} = sin(210^{\circ} + 45^{\circ}) = sin210^{\circ} * cos45^{\circ} + cos210^{\circ} * sin45^{\circ} = -\frac{1}{2} * \frac{\sqrt{2} }{2} + (-\frac{\sqrt{3} }{2}) * \frac{\sqrt{2} }{2} = - \frac{\sqrt{2} }{4} - \frac{\sqrt{6} }{4} = - \frac{\sqrt{2} + \sqrt{6} }{4}$

$cos255^{\circ} = cos(210^{\circ} + 45^{\circ}) = cos 210^{\circ} * cos45^{\circ} - sin210^{\circ} * sin45^{\circ} = -\frac{\sqrt{3} }{2} * \frac{\sqrt{2} }{2} + \frac{1}{2} * \frac{\sqrt{2}}{2} = - \frac{\sqrt{6} }{4} + \frac{\sqrt{2} }{4} = \frac{\sqrt{2}-\sqrt{6} }{4}$