Đáp án:

Câu 18: $\text{A. P = 1}$

Câu 19: $\text{C. } \dfrac{17}{2}$

Câu 20: $\text{D. 0}$

Giải thích:

Câu 18: $P = \dfrac{\cot^2 x - \cos^2 x}{\cot^2 x} + \dfrac{\sin x \cos x}{\cot x}$

$= 1 - \dfrac{\cos^2 x}{\cot^2 x} + \dfrac{\sin x \cos x}{\dfrac{\cos x}{\sin x}}$

$= 1 - \dfrac{\cos^2 x}{\dfrac{\cos^2 x}{\sin^2 x}} + \sin^2 x$

$= 1 - \sin^2 x + \sin^2 x$

$= 1$

Câu 19: $S = \sin^2 5^0 + \sin^2 10^0 + \sin^2 15^0 + \dots + \sin^2 80^0 + \sin^2 85^0$

$= (\sin^2 5^0 + \sin^2 85^0) + (\sin^2 10^0 + \sin^2 80^0) + \dots + (\sin^2 40^0 + \sin^2 50^0) + \sin^2 45^0$

$= (\sin^2 5^0 + \cos^2 5^0) + (\sin^2 10^0 + \cos^2 10^0) + \dots + (\sin^2 40^0 + \cos^2 40^0) + \sin^2 45^0$

$= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + \left(\dfrac{\sqrt{2}}{2}\right)^2$

$= 8 + \dfrac{1}{2}$

$= \dfrac{17}{2}$

Câu 20: $\cos \alpha + \cos\left(\alpha + \dfrac{\pi}{5}\right) + \cos\left(\alpha + \dfrac{2\pi}{5}\right) + \dots + \cos\left(\alpha + \dfrac{9\pi}{5}\right)$

$= \left[\cos \alpha + \cos\left(\alpha + \dfrac{9\pi}{5}\right)\right] + \left[\cos\left(\alpha + \dfrac{\pi}{5}\right) + \cos\left(\alpha + \dfrac{8\pi}{5}\right)\right] + \dots + \left[\cos\left(\alpha + \dfrac{4\pi}{5}\right) + \cos\left(\alpha + \dfrac{5\pi}{5}\right)\right]$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right)\cos\dfrac{9\pi}{10} + 2\cos\left(\alpha + \dfrac{9\pi}{10}\right)\cos\dfrac{7\pi}{10} + \dots + 2\cos\left(\alpha + \dfrac{9\pi}{10}\right)\cos\dfrac{\pi}{10}$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right) \cdot \left(\cos\dfrac{9\pi}{10} + \cos\dfrac{7\pi}{10} + \cos\dfrac{5\pi}{10} + \cos\dfrac{3\pi}{10} + \cos\dfrac{\pi}{10}\right)$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right) \cdot \left(\cos\dfrac{9\pi}{10} + \cos\dfrac{7\pi}{10} + 0 + \cos\dfrac{3\pi}{10} + \cos\dfrac{\pi}{10}\right)$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right) \cdot \left[\left(\cos\dfrac{9\pi}{10} + \cos\dfrac{\pi}{10}\right) + \left(\cos\dfrac{7\pi}{10} + \cos\dfrac{3\pi}{10}\right)\right]$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right) \cdot \left(2\cos\dfrac{\pi}{2}\cos\dfrac{2\pi}{5} + 2\cos\dfrac{\pi}{2}\cos\dfrac{\pi}{5}\right)$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right) \cdot \left(2 \cdot 0 \cdot \cos\dfrac{2\pi}{5} + 2 \cdot 0 \cdot \cos\dfrac{\pi}{5}\right)$

$= 2\cos\left(\alpha + \dfrac{9\pi}{10}\right) \cdot 0$

$= 0$

Vậy $\cos \alpha + \cos\left(\alpha + \dfrac{\pi}{5}\right) + \cos\left(\alpha + \dfrac{2\pi}{5}\right) + \dots + \cos\left(\alpha + \dfrac{9\pi}{5}\right) = 0$