Question

4 и 5 задание (смотрите на изображение)

Answer 1

$1. 13-12sin^2x=13-12 - 12cos^2x=1-12*0.11=1-1.32=-0.32$
$2. ctga = -4 \\ tga = \frac{1}{-4} \\ tg^2a + 1 = \frac{1}{cos^2a} \\ \frac{17}{16} = \frac{1}{cos^2} \\ cosa= \frac{4}{ \sqrt{17} }\\ \frac{ \sqrt{17} }{ \frac{4}{ \sqrt{17} } }= \frac{17}{4}=4.25$
$3. \ \sqrt{6} *sin \frac{63pi}{4} * tg \frac{64pi}{3} = \sqrt{6} *sin \frac{7pi}{4} * tg \frac{4pi}{3}= \\ \sqrt{6} *sin ( \frac{3pi}{2} - \frac{pi}{4}) * tg (pi + \frac{pi}{3})= \sqrt{6} *(-cos \frac{pi}{4}) * tg \frac{pi}{3}=\\ \sqrt{6} *(-\frac{\sqrt{2} }{2}) * \sqrt{3}= -3$ 
$4 \ \frac{1- 2sin^254}{8tg9 * sin^299} = \frac{cos108}{8tg9 *cos9}= \frac{cos108}{8sin9}= \frac{-sin18}{sin9}= \frac{-2sin9cos9}{8sin9}= \frac{-cos9}{4}$$\sqrt{2} * \frac{sin40*cos5-cos40*sin5}{-(cos25cos35 - sin25sin35) }= \sqrt{2} * \frac{sin45}{cos60}= \sqrt{2}* \sqrt{2}=2$