Question

functiile trigonometrice ale unghiiurilor optuze CALCULATI. a) sin 150°=. b) cos120°=. c)tg135°=. d)ctg120°=​

Answer 1

Ai câteva identități :

$1. \sin(180 - x) = \sin(x)\\2. \cos(180 - x) = -\cos(x)\\3. \sin(2x) = 2\sin(x)\cos(x)\\4. \cos(2x) = \cos^2{x} - \sin^{2}(x) = 1-2\sin^{2}(x) = 2\cos^{2}(x) - 1\\5. \text{tg}(2x) = \frac{2\text{tg}(x)}{1-\text{tg}^{2}(x)}\\6. \text{ctg}(2x) = \frac{\text{ctg}^2(x) - 1}{2\text{ctg}(x)}$

$a) \sin(150) = \sin(180 - 30) = \sin(30) = \frac{1}{2}\\b) \cos(120) = \cos(2 \cdot 60) = 1 - 2\sin^{2}(60) = 1 - 2(\frac{\sqrt{3}}{2})^2 =1 - 2(\frac{3}{4}) = 1 - \frac{3}{2} = -\frac{1}{2}\\c) \text{tg}(135) = \frac{\sin(135)}{\cos(135)} = \frac{\sin(180 - 45)}{\cos(180 - 45)} = \frac{\sin(45)}{-\cos(45)} = -\text{tg}(45) = -1.$

$d) \text{ctg}(120) = \text{ctg}(2 \cdot 60) = \frac{\text{ctg}^{2}(60) - 1}{2\text{ctg}(60)} = \frac{(\frac{1}{\sqrt{3}})^{2} - 1}{2\frac{1}{\sqrt{3}}} = \frac{\frac{1}{3} - 1}{\frac{2}{\sqrt{3}}} = \\-\frac{\frac{2}{3}}{\frac{2}{\sqrt{3}}} = - \frac{2}{3} \cdot \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}$