Question

Ştiind că x € (0,pi/2) şi sin x =4/5 să se calculeze cos x, tgx, ctgx. Vă rog mult ajutați-mă

Answer 1

sin²x+cos²x=1, cos²x=1-sin²x, cosx=+-√(1-sin²x)=+-√(1-16/25)=
+-√9/5=+-3/5
deoarece x∈(+,π/2), cosx>0, cosx=3/5
tgx=sinx/cosx=4/3
ctgx=1/tgx=3/4

Answer 2

$\displaystyle x \in \left(0, \frac{\pi}{2} \right)~~~~~~~~~~~~~sin~x= \frac{4}{5} ~~~~~~~~~~~~~~~~cos~x,~tg~x,~ctg~x=? \\ \\ cos^2x+sin^2x=1 \Rightarrow cos^2x=1-sin^2x \Rightarrow cos~x=\pm \sqrt{1-sin^2x} \\ \\ x \in \left(0, \frac{\pi}{2} \right) \Rightarrow cos~x\ \textgreater \ 0 \Rightarrow cos~x= \sqrt{1-sin^2x} \\ \\ cos~x= \sqrt{1-\left( \frac{4}{5} \right)^2} = \sqrt{1- \frac{16}{25} } = \sqrt{ \frac{25-16}{25} }= \sqrt{ \frac{9}{25} } = \frac{ \sqrt{9} }{ \sqrt{25} } = \frac{3}{5} \\ \\ cos~x= \frac{3}{5}$
$\displaystyle tg~x= \frac{sin~x}{cos~x} = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{5} : \frac{3}{5} = \frac{4}{\not5} \cdot \frac{\not5}{3} = \frac{4}{3} \\ \\ tg~x= \frac{4}{3} \\ \\ ctg~x= \frac{cos~x}{sin~x} = \frac{ \frac{3}{5} }{ \frac{4}{5} } = \frac{3}{5} : \frac{4}{5} = \frac{3}{\not5} \cdot \frac{\not5}{4} = \frac{3}{4} \\ \\ ctg~x= \frac{3}{4}$