Tổng hợp câu hay và khó chương 6 - Phần 2

Gọi $R$ là bán kính đường tròn ngoại tiếp $\Delta ABC$. Ta có:

$\dfrac{{1 + \cos B}}{{1 - \cos B}} = \dfrac{{2a + c}}{{2a - c}}$$\Leftrightarrow \dfrac{{1 + \cos B}}{{1 - \cos B}} = \dfrac{{2.2R\sin A + 2R\sin C}}{{2.2R\sin A - 2R\sin C}}$$\Leftrightarrow \dfrac{{1 + \cos B}}{{1 - \cos B}} = \dfrac{{2\sin A + \sin C}}{{2\sin A - \sin C}}$$\Leftrightarrow 2\sin A + 2\sin A\cos B - \sin C - \sin C\cos B = 2\sin A - 2\sin A\cos B + \sin C - \sin C\cos B$ $\Leftrightarrow 4\sin A\cos B = 2\sin C$

$\Leftrightarrow 4.\dfrac{a}{{2R}}.\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = 2.\dfrac{c}{{2R}}$

$\Leftrightarrow {a^2} + {c^2} - {b^2} = {c^2}$

$\Leftrightarrow a = b$.

Vậy $\Delta ABC$ cân tại $C$.

Ta có $P = \left( {1 - 2\cos 2\alpha } \right)\left( {2 + 3\cos 2\alpha } \right)$$= \left[ {1 - 2\left( {1 - 2{{\sin }^2}\alpha } \right)} \right]\left[ {2 + 3\left( {1 - 2{{\sin }^2}\alpha } \right)} \right]$.

$= \left[ {1 - 2\left( {1 - 2.\dfrac{4}{9}} \right)} \right]\left[ {2 + 3\left( {1 - 2.\dfrac{4}{9}} \right)} \right]$$= \dfrac{{49}}{{27}}$.

Ta có $\sin a - \cos a = \dfrac{3}{4}$ suy ra ${\left( {\sin a - \cos a} \right)^2} = \dfrac{9}{{16}}$

$\begin{array}{l} \Leftrightarrow {\sin ^2}a + {\cos ^2}a - 2\sin a\cos a = \dfrac{9}{{16}}\\ \Leftrightarrow 1 - \sin 2a = \dfrac{9}{{16}}\\ \Leftrightarrow \sin 2a = 1 - \dfrac{9}{{16}}\\ \Leftrightarrow \sin 2a = \dfrac{7}{{16}} \end{array}$

Vì $- 1 \le \cos x \le 1$, ta có: ${\sin ^4}x + {\cos ^7}x \le {\sin ^4}x + {\cos ^4}x = 1 - \dfrac{1}{2}{\sin ^2}2x \le 1$.

Vậy giá trị lớn nhất của biểu thức ${\sin ^4}x + {\cos ^7}x$ là $1$.

* Ta có $\sin a + \sqrt 3 \cos a = 2\left( {\dfrac{1}{2}\sin a + \dfrac{{\sqrt 3 }}{2}\cos a} \right)$$= 2\left( {\sin a\sin \dfrac{\pi }{6} + \cos a\cos \dfrac{\pi }{6}} \right) = 2\cos \left( {a - \dfrac{\pi }{6}} \right)$.

* Lại có $- 2 \le 2\cos \left( {a - \dfrac{\pi }{6}} \right) \le 2$ suy ra giá trị nhỏ nhất của biểu thức đã cho là $- 2$ khi $\cos \left( {a - \dfrac{\pi }{6}} \right) =  - 1 \Leftrightarrow a = \dfrac{{7\pi }}{6} + k2\pi$, $k \in \mathbb{Z}$.

+ ${\sin ^4}a - {\cos ^4}a = \left( {{{\sin }^2}a - {{\cos }^2}a} \right).\left( {{{\sin }^2}a + {{\cos }^2}a} \right) = \left( {{{\sin }^2}a - {{\cos }^2}a} \right).1 =  - \cos 2a$ nên A sai.

+$2\left( {{{\sin }^4}a + {{\cos }^4}a} \right) = 2\left[ {{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right)}^2} - 2{{\sin }^2}a.{{\cos }^2}a} \right] = 2\left( {1 - 2.\dfrac{1}{4}{{\sin }^2}2a} \right) = 2 - {\sin ^2}2a$ nên B đúng.

+ ${\left( {\sin a - \cos a} \right)^2} = 1 - 2\sin a.\cos a = 1 - \sin 2a$ nên C sai.

+ ${\left( {{{\sin }^2}a + {{\cos }^2}a} \right)^3} = 1$ và $1 + 2{\sin ^4}a.{\cos ^4}a = 1 + 2.{\left( {\dfrac{1}{2}\sin 2a} \right)^4} = 1 + \dfrac{1}{8}{\sin ^4}2a$ nên D sai.

$G = {\cos ^2}\dfrac{\pi }{6} + {\cos ^2}\dfrac{{2\pi }}{6} + ... + {\cos ^2}\dfrac{{5\pi }}{6} + {\cos ^2}\pi$$= \left( {{{\cos }^2}\dfrac{\pi }{6} + {{\cos }^2}\dfrac{{2\pi }}{6}} \right) + \left( {{{\cos }^2}\dfrac{{4\pi }}{6} + {{\cos }^2}\dfrac{{5\pi }}{6}} \right) + \left( {{{\cos }^2}\dfrac{\pi }{2} + {{\cos }^2}\pi } \right)$.

$= \left( {{{\cos }^2}\dfrac{\pi }{6} + {{\cos }^2}\dfrac{\pi }{3}} \right) + \left( {{{\cos }^2}\dfrac{{2\pi }}{6} + {{\cos }^2}\dfrac{\pi }{6}} \right) + 1$.

$= 2\left( {{{\cos }^2}\dfrac{\pi }{6} + {{\cos }^2}\dfrac{\pi }{3}} \right) + 1 = 2\left( {{{\cos }^2}\dfrac{\pi }{6} + {{\sin }^2}\dfrac{\pi }{6}} \right) + 1 = 3$.

$\begin{array}{l}A = \cos 20^\circ  + \cos 40^\circ  + \cos 60^\circ  + ... + \cos 160^\circ  + \cos 180^\circ \\ = \left( {\cos 20^\circ  + \cos 160^\circ } \right) + \left( {\cos 40^\circ  + \cos 140^\circ } \right) + ...\left( {\cos 80^\circ  + \cos 100^\circ } \right) + \cos {180^0}\\ = 0 + 0 + ... + 0 + \left( { - 1} \right)\end{array}$.

$=  - 1$.

${\left( {\dfrac{{\sin \alpha  + \tan \alpha }}{{\cos \alpha  + 1}}} \right)^2} + 1 = {\left( {\dfrac{{\sin \alpha  + \dfrac{{\sin \alpha }}{{\cos \alpha }}}}{{\cos \alpha  + 1}}} \right)^2} + 1 = {\left( {\dfrac{{\sin \alpha \left( {1 + \cos \alpha } \right)}}{{\cos \alpha \left( {\cos \alpha  + 1} \right)}}} \right)^2} + 1 = {\tan ^2}\alpha  + 1 = \dfrac{1}{{{{\cos }^2}\alpha }}$.

$\begin{array}{l}E = \sin \dfrac{\pi }{5} + \sin \dfrac{{2\pi }}{5} + ... + \sin \dfrac{{9\pi }}{5}\\ = \left( {\sin \dfrac{\pi }{5} + \sin \dfrac{{9\pi }}{5}} \right) + \left( {\sin \dfrac{{2\pi }}{5} + \sin \dfrac{{8\pi }}{5}} \right) + ...\\ + \left( {\sin \dfrac{{4\pi }}{5} + \sin \dfrac{{6\pi }}{5}} \right) + \sin \dfrac{{5\pi }}{5}\\ = \left[ {\sin \dfrac{\pi }{5} + \sin \left( {2\pi  - \dfrac{\pi }{5}} \right)} \right] + \left[ {\sin \dfrac{{2\pi }}{5} + \sin \left( {2\pi  - \dfrac{{2\pi }}{5}} \right)} \right]\\ + ... + \left[ {\sin \dfrac{{4\pi }}{5} + \sin \left( {2\pi  - \dfrac{{4\pi }}{5}} \right)} \right] + \sin \pi \\ = \left[ {\sin \dfrac{\pi }{5} + \sin \left( { - \dfrac{\pi }{5}} \right)} \right] + \left[ {\sin \dfrac{{2\pi }}{5} + \sin \left( { - \dfrac{{2\pi }}{5}} \right)} \right]\\+ ... + \left[ {\sin \dfrac{{4\pi }}{5} + \sin \left( { - \dfrac{{4\pi }}{5}} \right)} \right] + 0\\ = \left( {\sin \dfrac{\pi }{5} - \sin \dfrac{\pi }{5}} \right) + \left( {\sin \dfrac{{2\pi }}{5} - \sin \dfrac{{2\pi }}{5}} \right) + ...\\ + \left( {\sin \dfrac{{4\pi }}{5} - \sin \dfrac{{4\pi }}{5}} \right)\\ = 0 + 0 + ... + 0 = 0\end{array}$

${\sin ^4}x = {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2}$$= \dfrac{1}{4}\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)$$= \dfrac{1}{4}\left( {1 - 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right)$$= \dfrac{3}{8} - \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x$$\dfrac{a}{8} - \dfrac{1}{2}\cos 2x + \dfrac{b}{8}\cos 4x$

Vậy $a + b = 3 + 1 = 4$.

${\sin ^8}x + {\cos ^8}x$$= {\left( {{{\sin }^4}x + {{\cos }^4}x} \right)^2} - 2{\sin ^4}x.{\cos ^4}x$$= {\left( {1 - 2{{\sin }^2}x.{{\cos }^2}x} \right)^2} - \dfrac{1}{8}{\sin ^4}2x$

$= {\left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right)^2} - \dfrac{1}{8}{\sin ^4}2x$$= 1 - {\sin ^2}2x + \dfrac{1}{8}{\sin ^4}2x$$= 1 - \dfrac{{1 - \cos 4x}}{2} + \dfrac{1}{8}{\left( {\dfrac{{1 - \cos 4x}}{2}} \right)^2}$

$= 1 - \dfrac{{1 - \cos 4x}}{2} + \dfrac{1}{{32}}\left( {1 - 2\cos 4x + \dfrac{{1 + \cos 8x}}{2}} \right)$$= \dfrac{{35}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{1}{{64}}\cos 8x$

$\Rightarrow a = 35$, $b = 7$, $c = 1$$\Rightarrow a - 5b + c = 1$.

Ta có: $0 < \alpha  < 90^\circ$$\Leftrightarrow 0 < \dfrac{\alpha }{2} < 45^\circ$$\Rightarrow 0 < \sin \dfrac{\alpha }{2} < \dfrac{{\sqrt 2 }}{2}$$\Leftrightarrow 0 < \sqrt {\dfrac{{x - 1}}{{2x}}}  < \dfrac{{\sqrt 2 }}{2}$$\Leftrightarrow x > 0$

${\sin ^2}\dfrac{\alpha }{2} + {\cos ^2}\dfrac{\alpha }{2} = 1$$\Rightarrow \cos \dfrac{\alpha }{2} = \sqrt {1 - {{\sin }^2}\dfrac{\alpha }{2}}$, vì $0 < \dfrac{\alpha }{2} < 45^\circ$

$\Leftrightarrow \cos \dfrac{\alpha }{2} = \sqrt {\dfrac{{x + 1}}{{2x}}}$$\Rightarrow \tan \dfrac{\alpha }{2} = \sqrt {\dfrac{{x - 1}}{{x + 1}}}$

$tan\alpha  = \dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 - {{\tan }^2}\dfrac{\alpha }{2}}} = \dfrac{{2\sqrt {\dfrac{{x - 1}}{{x + 1}}} }}{{1 - \dfrac{{x - 1}}{{x + 1}}}} = \sqrt {{x^2} - 1}$

$\Rightarrow \cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{1}{{\sqrt {{x^2} - 1} }} = \dfrac{{\sqrt {{x^2} - 1} }}{{{x^2} - 1}}$.