Question

_Cho góc nhọn x biết :

a) sin x = 0,6. Tính 2tan²x - cot x

b) cos x = 0,8. Tính 2tan x - 3cot² x

c) tan x = \(\frac{4}{3}\) . Tính sin x - cos² x

Answer 1

a/ \(sinx=0,6\Rightarrow cosx=\sqrt{1-sin^2x}=0,8\)

\(\Rightarrow tanx=\frac{sinx}{cosx}=\frac{0,6}{0,8}=\frac{3}{4}\) ; \(cotx=\frac{1}{tanx}=\frac{4}{3}\)

\(\Rightarrow2tan^2x-cotx=-\frac{5}{24}\)

b/ Tương tự \(sinx=\sqrt{1-cos^2x}=0,6\Rightarrow\left\{{}\begin{matrix}tanx=\frac{3}{4}\\cotx=\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow...\)

c/ \(\frac{16}{9}=tan^2x=\frac{sin^2x}{cos^2x}=\frac{1-cos^2x}{cos^2x}\)

\(\Rightarrow16cos^2x=9-9cos^2x\Rightarrow cos^2x=\frac{9}{25}\)

\(\Rightarrow sin^2x=1-cos^2x=\frac{16}{25}\Rightarrow sinx=\frac{4}{5}\)

\(\Rightarrow sinx-cos^2x=...\)