Question

sin a + cos a = $\frac{7}{5}$
tg $\frac{a}{2}$ =?
a ∈ (0, π/4)

Answer 1

$\sin a+\cos a = \dfrac{7}{5} \\ \\ a \in \Big(0,\dfrac{\pi}{4}\Big)\Rightarrow \dfrac{a}{2}\in \Big(0,\dfrac{\pi}{8}\Big) \Rightarrow \tan \dfrac{a}{2} = \Big(\tan 0, \tan \dfrac{\pi}{8}\Big)\\ \\ \tan \dfrac{a}{2} = \dfrac{1-\cos a}{\sin a} \Rightarrow \tan \dfrac{\pi}{8} = \dfrac{1-\cos \frac{\pi}{4}}{\sin\frac{\pi}{4}} = \sqrt 2 - 1\\ \\\\ \Rightarrow\tan \dfrac{a}{2} = \Big(0,\sqrt 2 - 1\Big)$

$t =\tan \dfrac{a}{2}\\ \\ \sin a = \dfrac{2t}{1+t^2} \\ \cos a = \dfrac{1-t^2}{1+t^2}\\ \\ \dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2} = \dfrac{7}{5} \\ \\10t-5t^2+5 = 7+7t^2 \\ \\ 12t^2-10t+2 = 0 \\ \\ 6t^2-5t+1 = 0 \\ \\ \Delta = 25-24 = 1 \Rightarrow t_{1,2}= \dfrac{5\pm 1}{12}\\ \\ \boxed{1}\quad t_1 = \dfrac{1}{3} < \sqrt 2 - 1 \\ \\\boxed{2}\quad t_2 = \dfrac{1}{2} > \sqrt 2 - 1 \\ \\ \Rightarrow\boxed{S = \Big\{\dfrac{1}{3}\Big\}}$