Wiedząc, że alfa jest kątem ostrym, takim, że tg a = 2, oblicz wartość wyrażenia: a) sin^2 a - cos^2 a b)sin a + cos a c) sin a tg a + cos a ctg a d) 2 sin a cos a   W jaki sposób obliczyć to zadanie???  

  [latex]tg a =2\ frac{sin a}{cos a}=2\ sin a = 2cosa \[/latex]       [latex]sin^2 a + cos^2 a = 1\ (2cos a)^2 + cos^2 a = 1\ 4cos^2 a + cos^2 a = 1\ 5cos^2 a = 1[/latex]     [latex]cos^2 a = frac{1}{5}\ cos a = frac{sqrt5}{5}\ sina = 2cdot frac{sqrt5}{5}=frac{2sqrt5}{5}[/latex]     a)   [latex]sin^2 a - cos^2 a = (frac{2sqrt5}{5})^2 - (frac{sqrt5}{5})^2 = \ =frac{4cdot 5}{25} - frac{5}{25} = frac{4}{5}-frac{1}{5}=frac{3}{5}[/latex]     b) [latex]sin a + cos a = frac{2sqrt5}{5} + frac{sqrt5}{5} = frac{3sqrt5}{5}[/latex]   c)   [latex]sin a tg a + cos a ctga=frac{2sqrt5}{5}cdot 2+frac{sqrt5}{5}cdot frac{1}{2}=\ =frac{4sqrt5}{5} + frac{sqrt5}{10} = frac{8sqrt5}{10}+frac{sqrt5}{10} =\ =frac{9sqrt5}{10}[/latex]     d) [latex]2 sin a cos a = 2cdot frac{2sqrt5}{5}cdot frac{sqrt5}{5} = frac{4}{5}[/latex]