Question

 

6) tan x=3. 180 độ < x < 270 độ

Tính sin2x ; tan2x ;cos4x

Answer 1

\(180^o< x< 270^o\)

\(1+tan^2x=\dfrac{1}{cos^2x}\Leftrightarrow1+3^2=\dfrac{1}{cos^2x}\Leftrightarrow cosx=-\dfrac{\sqrt{10}}{10}\)

\(sinx=tanx.cosx=-\dfrac{3\sqrt{10}}{10}\)

\(sin2x=2sinxcosx=2.\left(-\dfrac{3\sqrt{10}}{10}\right).\left(-\dfrac{\sqrt{10}}{10}\right)=\dfrac{3}{5}\)

\(tan2x=\dfrac{2tanx}{1-tan^2x}=\dfrac{2.3}{1-3^2}=-\dfrac{3}{4}\)

\(cos4x=8cos^4x-8cos^2x+1=8.\left(-\dfrac{\sqrt{10}}{10}\right)^4-8.\left(-\dfrac{\sqrt{10}}{10}\right)^2+1=\dfrac{7}{25}\)

Answer 2

tới đây rồi làm tiếp nhé, bận=)