Question

Trigonometrie (photo) nu stiu care dintre solutii este buna

Answer 1

a∈($\pi ,3 \pi /2$) deci a apartine cadranului 3
sin a= -4/5
tg a/2=?
Folosind formula fundamentala a trigonometriei vom afla cos a.
sin²a+cos²a=1
⇒cos²a=1-sin²a
cos²a=1-(-4/5)²=1-16/25   (amplificam pe 1 cu 25)
cos²a=(25-16)/25=9/25
⇒cos a=√(9/25)
cos a= -3/5 deoarece cosinusul in cadranul 3 este negativ
tg a/2= sin a/(1+cos a)
tg a/2= $\frac{ -\frac{4}{5} }{1+(- \frac{3}{5}) } = \frac{ -\frac{4}{5} }{ \frac{2}{5} } =- \frac{4}{5} : \frac{2}{5} =- \frac{4}{5} * \frac{5}{2} =- 2$

Answer 2

[tex]\it a\in(\pi,\ \dfrac{3\pi}{4}) \Rightarrow cosa\ \textless \ 0 \\\;\\ \\\;\\ cosa=-\sqrt{1-sin^2a} =-\sqrt{1-\dfrac{16}{25}} =-\sqrt{\dfrac{9}{25}} =-\dfrac{3}{5} [/tex]

[tex]\it tg\dfrac{a}{2} =\dfrac{sin\dfrac{a}{2}}{cos\dfrac{a}{2}}= \dfrac{sin\dfrac{a}{2}}{cos\dfrac{a}{2}}\cdot \dfrac{2cos\dfrac{a}{2}}{2cos\dfrac{a}{2}} = \dfrac{sina}{2cos^2\dfrac{a}{2}} =\dfrac{sina}{1+cosa} = \\\;\\ \\\;\\ =\dfrac{-\dfrac{4}{5}}{1-\dfrac{3}{5}} =\dfrac{-\dfrac{4}{5}}{\dfrac{2}{5}} =-\dfrac{4}{2} = -2[/tex]