Question

Помогите кто-нибудь пожалуйста​

Answer 1

1.

Угол принадлежит 2 четверти, косинус отрицательный.

$\sin( \alpha ) = \frac{7}{9}$

$\cos( \alpha ) = \sqrt{1 - \sin {}^{2} ( \alpha ) } \\ \cos( \alpha ) = - \sqrt{1 - \frac{49}{81} } = \\ = - \sqrt{ \frac{32}{81} } = - \frac{ 4\sqrt{2} }{9}$

$tg( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{7}{9} \times ( - \frac{9}{4 \sqrt{2} } ) = \\ = - \frac{7}{4 \sqrt{2} } = - \frac{7 \sqrt{2} }{8}$

$ctg( \alpha ) = \frac{1}{tg( \alpha )} = - \frac{4 \sqrt{2} }{7}$

2.

а

$\cos {}^{2} ( \alpha ) - \cos(2 \alpha ) = \\ = \cos {}^{2} ( \alpha ) - ( \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) ) = \\ = \cos {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha ) + \sin {}^{2} ( \alpha ) = \sin {}^{2} ( \alpha )$

б

$\sin( \frac{\pi}{2} + \alpha ) - \cos(\pi - \alpha ) + tg(\pi - \alpha ) + ctg(\pi - \alpha ) = \\ = \cos( \alpha ) + \cos( \alpha ) - tg (\alpha ) - ctg( \alpha ) = \\ = 2 \cos( \alpha ) - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } = \\ = \frac{2 \cos {}^{2} ( \alpha ) \sin( \alpha ) - \sin {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha ) }{ \sin( \alpha ) \cos( \alpha ) } = \\ = \frac{2 \cos {}^{2} ( \alpha ) \sin( \alpha ) - 1 }{ \sin( \alpha ) \cos( \alpha ) } = \frac{ \sin( 2\alpha ) \cos( \alpha ) - 1 }{ \sin( \alpha ) \cos( \alpha ) }$

3

$\sin( \frac{\pi}{12} ) \cos( \frac{11\pi}{12} ) + \cos( \frac{\pi}{12} ) \sin( \frac{11\pi}{12} ) = \\ = \sin( \frac{\pi}{12} + \frac{11\pi}{12} ) = \sin(\pi) = 0$

4

a

$2 \cos(x) - \sqrt{2} = 0 \\ \cos(x) = \frac{ \sqrt{2} }{2} \\ x = \pm \frac{\pi}{4} + 2\pi \: n$

n принадлежит Z.

б

$\sin( \frac{x}{2} - \frac{\pi}{6} ) - 1 = 0 \\ \sin( \frac{x}{2} - \frac{\pi}{6} ) = 1 \\ \frac{x}{2} - \frac{\pi}{6} = \frac{\pi}{2} + 2\pi \: n \\ \frac{x}{2} = \frac{\pi}{2} + \frac{\pi}{6} + 2\pi \: n \\ \frac{x}{2} = \frac{2\pi}{3} + 2\pi \: n \\ x = \frac{4\pi}{3} + 4\pi \: n$

n принадлежит Z.

в

$\cos {}^{2} (x) + 2 \sin(x) + 2 = 0 \\ 1 - \sin {}^{2} (x) + 2\sin(x) + 2 = 0 \\ \sin {}^{2} (x) - 2\sin(x) - 3 = 0 \\ \\ \sin(x) = t \\ \\ {t}^{2} - 2t - 3 = 0 \\ D = 4 + 12 = 16 \\ t_1 = \frac{2 + 4}{2} = 3 \\ t_2 = - 1 \\ \\ \sin(x) = 3$

нет корней

$\sin(x) = - 1 \\ x = - \frac{\pi}{2} + 2 \pi \: n$

n принадлежит Z.

г

${tg}^{2} x - tgx +1 = 0 \\ \\ tgx = t \\ \\ {t}^{2} - t + 1 = 0 \\ D = 1 - 4 < 0$

нет корней

Answer 2

Ответ:

$1)\ \ sina=\dfrac{7}{9}\\\\sin^2a+cos^2a=1\ \ \to \ \ cos^2a=1-sin^2a\ \ ,\ \ cosa=\pm \sqrt{1-sin^2a}\\\\\dfrac{\pi}{2}<a<\pi \ \ \to \ \ cosa<0\ \ ,\ \ cosa=-\sqrt{1-sin^2a}=-\sqrt{1-\dfrac{49}{81}}=-\dfrac{\sqrt{32}}{9}=\\\\=-\dfrac{4\sqrt2}{9}\\\\tga=\dfrac{sina}{cosa}=\dfrac{7/9}{-4\sqrt2/9}=-\dfrac{7}{4\sqrt2}=-\dfrac{7\sqrt2}{8}\\\\ctga=\dfrac{cosa}{sina}=-\dfrac{4\sqrt2}{7}$

$2a)\ \ cos2a-cos^2a=(2cos^2a-1)-cos^2a=cos^2a-1=-sin^2a\\\\b)\ \ sin(\dfrac{\pi}{2}-a)-cos(\pi -a)+tg(\pi -a)+ctg(\pi -a)=cosa+cosa-tga-ctga=\\\\=2cosa-tga-ctga=2cosa-\dfrac{sina}{cosa}-\dfrac{cosa}{sina}=2cosa-\dfrac{sin^2a+cos^2a}{sina\cdot cosa}=\\\\=2cosa-\dfrac{1}{\frac{1}{2}\, sin2a}=2cosa-\dfrac{2}{sin2a}=\dfrac{2\, (cosa\cdot sin2a-1)}{sin2a}$

$3)\ \ sin\dfrac{\pi}{12}\cdot cos\dfrac{11\pi}{12}+cos\dfrac{\pi}{12}\cdot sin\dfrac{11\pi }{12}=sin\Big(\dfrac{\pi}{12}+\dfrac{11\pi }{12}\Big)=sin\pi =0$

$4a)\ \ 2cosx-\sqrt2=0\ \ \ \to \ \ \ \sqrt2\cdot (\sqrt2cosx-1)=0\ \ ,\\\\cosx=\dfrac{1}{\sqrt2}\ \ ,\ \ \ x=\pm \dfrac{\pi}{4}+2\pi n\ ,\ n\in Z\\\\Otvet:\ x=\pm \dfrac{\pi }{4}+2\pi n\ ,\ n\in Z$

$b)\ \ sin\Big(\dfrac{x}{2}-\dfrac{\pi}{6}\Big)-1=0\ \ \ \to \ \ \ sin\Big(\dfrac{x}{2}-\dfrac{\pi}{6}\Big)=1\ \ ,\\\\\Big(\dfrac{x}{2}-\dfrac{\pi}{6}\Big)=\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\\\\\dfrac{x}{2}=\dfrac{\pi}{6}+\dfrac{\pi}{2}+2\pi n\ \ \ ,\ \ \ \dfrac{x}{2}=\dfrac{4\pi}{6}+2\pi n\ \ ,\ \ \ \dfrac{x}{2}=\dfrac{2\pi}{3}+2\pi n\ ,\ n\in Z\ ,$

$Otvet:\ x=\dfrac{4\pi}{3}+4\pi n\ ,\ n\in Z$

$c)\ \ cos^2x+2sinx+2=0\\\\(1-sin^2x)+2sinx+2=0\ \ ,\ \ \ -sin^2x+2sinx+3=0\ \ ,\\\\sin^2x-2sinx-3=0\ \ ,\ \ (sinx)_1=3\ ,\ (sinx)_2=-1\ \ (teorema\ Vieta)\\\\sina=3\ \ \to \ \ x\in \varnothing \ ,\ tak\ kak\ \ |sinx|\leq 1\\\\sina=-1\ \ \to \ \ \ x=-\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\\\\Otvet:\ x=-\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\ .$

$d)\ \ tg^2x-tgx+1=0\\\\tgx=t\ ,\ \ t^2-t+1=0\ \ ,\ \ D=1-4=-3<0\ \ \to \ \ t\in \varnothing \\\\Otvet:\ x\in \varnothing \ .$