Ответ:
[tex]cos\beta =0,4[/tex]
Так как [tex]\beta \in 4[/tex] четверти, то [tex]sin\beta < 0\ ,\ tg\beta < 0\ ,\ ctg\beta < 0[/tex] .
Воспользуемся основным тригонометрическим тождеством:
[tex]sin^3\beta +cos^2\beta =1\ \ \ \Rightarrow \ \ \ sin^2\beta =1-cos^2\beta =1-0,16=0,84\ ,\\\\sin\beta =-\sqrt{0,84}=-\sqrt{\dfrac{84}{100}}=-\dfrac{2\sqrt{21}}{10}=-\dfrac{\sqrt{21}}{5}\\\\tg\beta =\dfrac{sin\beta }{cos\beta }=-\dfrac{\sqrt{21}}{5\cdot 0,4}=-\dfrac{\sqrt{21}}{2}\\\\ctg\beta =\dfrac{1}{tg\beta }=-\dfrac{5}{\sqrt{21}}[/tex]
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Ответ:
[tex]cos\beta =0,4[/tex]
Так как [tex]\beta \in 4[/tex] четверти, то [tex]sin\beta < 0\ ,\ tg\beta < 0\ ,\ ctg\beta < 0[/tex] .
Воспользуемся основным тригонометрическим тождеством:
[tex]sin^3\beta +cos^2\beta =1\ \ \ \Rightarrow \ \ \ sin^2\beta =1-cos^2\beta =1-0,16=0,84\ ,\\\\sin\beta =-\sqrt{0,84}=-\sqrt{\dfrac{84}{100}}=-\dfrac{2\sqrt{21}}{10}=-\dfrac{\sqrt{21}}{5}\\\\tg\beta =\dfrac{sin\beta }{cos\beta }=-\dfrac{\sqrt{21}}{5\cdot 0,4}=-\dfrac{\sqrt{21}}{2}\\\\ctg\beta =\dfrac{1}{tg\beta }=-\dfrac{5}{\sqrt{21}}[/tex]