Chứng minh rằng:

a) sin x – cos x = \(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right)\);

b) \(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{1 - \tan x}}{{1 + \tan x}}\,\,\,\)\(\left( {x \ne \frac{\pi }{2} + k\pi ,\,\,x \ne \frac{{3\pi }}{4} + k\pi ,k \in \mathbb{Z}} \right)\).

Lời giải:

a) Vì \(\frac{\pi }{2} < a < \pi \) nên cos a < 0.

Mặt khác, từ sin² a + cos² a = 1 suy ra

cos a = \( - \sqrt {1 - {{\sin }^2}a} = - \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = - \frac{{2\sqrt 2 }}{3}\).

Ta có: sin 2a = 2sin a cos a = \(2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) = - \frac{{4\sqrt 2 }}{9}\).

\(\cos 2a = 1 - 2{\sin ^2}a = 1 - 2.{\left( {\frac{1}{3}} \right)^2} = \frac{7}{9}\).

\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{{4\sqrt 2 }}{9}}}{{\frac{7}{9}}} = - \frac{{4\sqrt 2 }}{7}\).  

b) Ta có: (sin a + cos a)² = \({\left( {\frac{1}{2}} \right)^2}\)\( \Leftrightarrow {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = \frac{1}{4}\)

\( \Leftrightarrow 1 + \sin 2a = \frac{1}{4} \Leftrightarrow \sin 2a = - \frac{3}{4}\).

Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\pi < 2a < \frac{{3\pi }}{2}\), do đó cos 2a < 0. Mặt khác từ sin² (2a) + cos² (2a) = 1

Suy ra \(\cos 2a = - \sqrt {1 - {{\sin }^2}\left( {2a} \right)} = - \sqrt {1 - {{\left( { - \frac{3}{4}} \right)}^2}} = - \frac{{\sqrt 7 }}{4}\).

Lời giải:

a) Vì \(\frac{\pi }{2} < a < \pi \) nên cos a < 0.

Mặt khác, từ sin² a + cos² a = 1 suy ra

cos a = \( - \sqrt {1 - {{\sin }^2}a} = - \sqrt {1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}} = - \frac{{\sqrt 6 }}{3}\).

Ta có: \(\cos \left( {a + \frac{\pi }{6}} \right)\)\( = \cos a\cos \frac{\pi }{6} - \sin a\sin \frac{\pi }{6}\)

\( = \left( { - \frac{{\sqrt 6 }}{3}} \right).\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 3 }}.\frac{1}{2} = \frac{{ - \sqrt 6 - 1}}{{2\sqrt 3 }} = - \frac{{\sqrt 3 + 3\sqrt 2 }}{6}\).

b) Vì \(\pi < a < \frac{{3\pi }}{2}\) nên sin a < 0, do đó \(\tan a = \frac{{\sin a}}{{\cos a}} > 0\).

Mặt khác từ \(1 + {\tan ^2}a = \frac{1}{{{{\cos }^2}a}}\)

Suy ra \(\tan a = \sqrt {\frac{1}{{{{\cos }^2}a}} - 1} = \sqrt {\frac{1}{{{{\left( { - \frac{1}{3}} \right)}^2}}} - 1} = 2\sqrt 2 \).

Ta có: \(\tan \left( {a - \frac{\pi }{4}} \right)\)\( = \frac{{\tan a - \tan \frac{\pi }{4}}}{{1 + \tan a\tan \frac{\pi }{4}}}\)\( = \frac{{2\sqrt 2 - 1}}{{1 + 2\sqrt 2 .1}} = \frac{{9 - 4\sqrt 2 }}{7}\).