Question

Exercitiul 6. Cum il aflu pe sin?

Answer 1

Salut,

x ∈ CI
cos x = √2/2.

sin²x+cos²x = 1
sin²x+2/4 = 1
sin²x+1/2 = 1
sin²x = 2/2-1/2
sin²x = 1/2
sin x = 1/√2
sin x = √2/2.

tg x = sin x/cos x
tg x = √2/2·2/√2
tg x = 1.



Succes! :3

Answer 2

$\displaystyle \mathtt{x\in\left(0, \frac{\pi}{2}\right);~cos~x= \frac{ \sqrt{2}} {2}}\\ \\ \mathtt{sin^2x+cos^2x=1\Rightarrow sin^2x=1-cos^2x \Rightarrow sin~x=\pm \sqrt{1-cos^2x} }$

$\displaystyle \mathtt{x~este~in~cadranul~I,~sin~x>0,~cos~x>0\Rightarrow sin~x= \sqrt{1-cos^2x}} \\ \\ \mathtt{sin~x= \sqrt{1-\left( \frac{\sqrt{2} }{2} }\right)^2 }= \sqrt{1- \frac{2}{4} }= \sqrt{\frac{4-2}{4}}= \sqrt{ \frac{2}{4}} = \frac{\sqrt{2}} {\sqrt{4}} = \frac{\sqrt{2}}{2} }\\ \\ \mathtt{tg~x= \frac{sin~x}{cos~x} = \frac{ \frac{\sqrt{2}}{2}} {\frac{\sqrt{2}}{2}}= \frac{\sqrt{2}}{2} \cdot \frac{2}{\sqrt{2}}=1}$