Question

a)\(\cot\alpha=-\alpha,a>0\Rightarrow\cos\alpha=?\)

b)\(\cos x=\frac{1}{2}\Rightarrow3\sin^2x+4\cos^2x=?\)

c) \(\sin\alpha=\frac{4}{5}\Rightarrow2\sin\alpha-\cos\alpha=?\) ( góc tù)

Answer 1

a, \(cota=-a\Rightarrow sina=-a.cosa\)

Khi đó \(sin^2a+cos^2a=a^2.cos^2a+cos^2a=\left(a^2+1\right)cos^2a=1\Rightarrow cosa=\sqrt{\frac{1}{a^2+1}}\)

b, \(sin^2x+cos^2x=sin^2x+\frac{1}{4}=1\Rightarrow sin^2x=\frac{3}{4}\)

\(\Rightarrow3sin^2x+4cos^2x=3.\frac{3}{4}+4.\frac{1}{4}=\frac{13}{4}\)

c, \(sin^2a+cos^2a=\frac{16}{25}+cos^2a=1\Rightarrow cosa=-\frac{3}{5}\)

\(\Rightarrow2sina-cosa=2.\frac{4}{5}+\frac{3}{5}=\frac{11}{5}\)