Question

cos2 a=? stiind ca tgx=4

Answer 1

Salut,

$cos(2a)=cos^2a-sin^2a=\dfrac{cos^2a-sin^2a}{1}=\dfrac{cos^2a-sin^2a}{cos^2a+sin^2a}=\\\\\\=\dfrac{\dfrac{cos^2a-sin^2a}{cos^2a}}{\dfrac{cos^2a+sin^2a}{cos^2a}}=\dfrac{1-tg^2a}{1+tg^2a}=\dfrac{1-16}{1+16}=-\dfrac{15}{17}.$

Green eyes.

Answer 2

[tex]\it tga = 4 \Longrightarrow tg^2a = 16 \ \ \ \ (*) \\\;\\ \\\;\\ tg^2a= \dfrac{sin^2a}{cos^2a} =\dfrac{1-cos^2a}{cos^2a} = \dfrac{1}{cos^2a} -1 \ \ \ \ \ (**) \\\;\\ \\\;\\ (*),\ (**) \Rightarrow \dfrac{1}{cos^2 a} - 1 = 16 \Rightarrow \dfrac{1}{cos^2a}=17 \Rightarrow cos^2a=\dfrac{1}{17} [/tex]

$\it cos2a=2cos^2a-1 = \dfrac{2}{17} -1 = -\dfrac{15}{17}$