Exercitiul din poza. Este cu sin si tg si nu ma pricep la astea ca sunt prea multe relatii.​

Răspuns:

folosesti formuala tangentei pe jumate

sinα=2t/(1+t²)  unde t=tgα/2

3/5=2t/(1+t²) Aplici proprietatea fundamentala a proportiilor

3(1+t²)=5*2t

3t²+3=10t

t²-10t+3=0

t1=(5-√22)/2

t2=5+√22)/2

Alegem solutia corecta

Daca α∈(π/2,π)

α/2=(π/4,π/2)=>tgα/2>tgπ/4  =1

Dar t1=(5-√22)/2<1   nu   corespunde   problemei

Solutie

t2=tgα/2=(5+√22)/2

b)aplici formula fundamentala  a  trigonometriei

sin²α+cos²α=1

(3/4)²+cos²α=1

9/16+cos²α=1

cos²α=1-9/16

cos²α=7/16

cosα=√7/116=±√7/4

Deoarece α∈cadranului 2 se ia solutiabnegativa

cosα= -√7/4

Explicație pas cu pas:

[tex]\it a)\ \alpha\in\Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow \dfrac{\alpha}{2}\in\Big(\dfrac{\pi}{4},\ \dfrac{\pi}{2}\Big) \Rightarrow tg\dfrac{\alpha}{2}>tg\dfrac{\pi}{4} \Rightarrow tg\dfrac{\alpha}{2}>1\ \ \ \ \ (*)[/tex]

[tex]\it sin\alpha=\dfrac{2t}{1+t^2},\ \ unde\ \ t=tg\alpha\\ \\ \\ \dfrac{2t}{1+t^2}=\dfrac{3}{5} \Rightarrow 3+3t^2=10t \Rightarrow 3t^2-10t+3=0 \Rightarrow 3t^2-9t-t+3=0 \Rightarrow \\ \\ \\ \Rightarrow 3t(t-3)-(t-3)=0 \Rightarrow (t-3)(3t-1)=0 \Rightarrow \\ \\ \\ \Rightarrow \begin{cases}\it t-3=0 \Rightarrow t=3\\ \\ \it 3t-1=0 \Rightarrow t=\dfrac{1}{3} \stackrel{(*)}{\Longrightarrow}\ nu\ convine\end{cases}\\ \\ \\ Deci,\ tg\dfrac{\alpha}{2}=3[/tex]

[tex]\it b)\ \alpha\in(0,\ \dfrac{\pi}{2}) \Rightarrow cos\alpha>0\\ \\ cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\Big(\dfrac{3}{5}\Big)^2}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\\ \\ \\ tg\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{5}\cdot\dfrac{5}{4}=\dfrac{3}{4}[/tex]