Question

Много ! основное тригонометрическое тождество. найти: 1) cosx, если sinx=12/13; π/2

Answer 1

1) $Cosx = - \sqrt{1 - Sin ^{2} x} = - \sqrt{1-( \frac{12}{13}) ^{2} }= -\sqrt{1- \frac{144}{169} } = - \sqrt{ \frac{25}{169} }=- \frac{5}{13}$
2) $Sinx = \sqrt{1 -Cos ^{2}x } = \sqrt{1-( \frac{2}{5}) ^{2} }= \sqrt{1- \frac{4}{25} }= \sqrt{ \frac{21}{25} }= \frac{ \sqrt{21} }{5}$
$tgx = \frac{Sinx}{Cosx}= \frac{ \frac{ \sqrt{21} }{5} }{ \frac{2}{5} } = \frac{ \sqrt{21}*5 }{5*2} = \frac{ \sqrt{21} }{2}$
3) $Sinx = - \sqrt{1 - (- \frac{15}{17}) ^{2} } = - \sqrt{1- \frac{225}{289} } =- \sqrt{ \frac{64}{289} } =- \frac{8}{17}$
4) $1 + Ctg ^{2} x = \frac{1}{Sin ^{2} x}$
$Sin ^{2}x= \frac{1}{1+Ctg x^{2}x } = \frac{1}{1+(- \sqrt{2}) ^{2} } = \frac{1}{1+2} = \frac{1}{3}$
$Cosx = - \sqrt{1 - \frac{1}{3} }= - \sqrt{ \frac{2}{3} }= - \frac{ \sqrt{2} }{ \sqrt{3} }=- \frac{ \sqrt{2}* \sqrt{3} }{ \sqrt{3}* \sqrt{3} } =- \frac{ \sqrt{6} }{3}$

Answer 2

Решение в прикрепленном фото