TRYGONOMETRIA – rozwiązanie zadań

Zadanie 1

$$\cos\frac{x}{2} = \ ?$$

$$\cos x = - \frac{2}{3}$$

x + x = 1

x = 1 − x

$$\operatorname{}x = 1 - \left( \frac{2}{3} \right)^{2}\ $$

$$\operatorname{s}x = 1 - \frac{4}{9} = \frac{5}{9}$$

$$\sin x = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\ \ \notin IV\ cwiartki\ \ \ \ \ \vee \ \ \ \ \ \sin x = - \frac{\sqrt{5}}{3}\ \ \ \in \ \ IV\ cwiartki$$

$$\cos x = \operatorname{c}\frac{x}{2} - \operatorname{s}\frac{x}{2}$$

$$\cos x = \operatorname{c}\frac{x}{2} - (1 - \operatorname{c}{\frac{x}{2})}$$

$$\cos x = 2\operatorname{c}\frac{x}{2} - 1$$

$$\cos x + 1 = 2\operatorname{}\frac{x}{2}$$

$$\operatorname{}\frac{x}{2} = \frac{\cos x + 1}{2}$$

$$\cos\frac{x}{2} = \sqrt{\frac{\cos x + 1}{2}} = \sqrt{\frac{- \frac{2}{3} + 1}{2}} = \sqrt{\frac{\frac{1}{3}}{2}} = \sqrt{\frac{1}{6}}$$

$$\frac{x}{2} \in \left( \frac{\pi}{2},\ \pi \right)$$

II cwiartka

$$Odpowiedz:\ \ \ \cos\frac{x}{2} = - \sqrt{\frac{1}{6}}$$

Zadanie 2

$$\sqrt{3}\cos x + \sin x = 1$$

sin(α+β)=sinαcosβ + cosαsinβ

$$\frac{\sin{30}}{\cos{30}}\cos x + \sin x = 1\ \ \ |*\ \cos{30}$$

$$\sin{30}\cos x + \sin x\cos{30} = \frac{\sqrt{3}}{2}$$

$$\sin\left( 30 + x \right) = \frac{\sqrt{3}}{2}$$

$$\sin{\left( \frac{\pi}{6} + x \right) = \frac{\sqrt{3}}{2}}$$

$$x_{1} + \frac{\pi}{6} = \frac{\pi}{3} + 2k\pi$$

$$x_{1} = \frac{\pi}{3} - \frac{\pi}{6} + 2k\pi$$

$$x_{1} = \frac{\pi}{6} + 2k\pi$$

$$x_{2} \neq \frac{\pi}{6}$$

$$x_{2} = \pi - \frac{\pi}{3} + 2k\pi$$

$$x_{2} = \frac{3}{6}\pi + 2k\pi$$

$$x_{2} = \frac{\pi}{2} + 2k\pi$$

Zadanie 3

1. $\sin{2x} = \frac{2}{\operatorname{tg}x + \operatorname{ctg}x}$

$$2\sin x\cos x = \frac{2}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}$$

$$P = \frac{2}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}} = \frac{2}{\frac{\operatorname{s}x + \operatorname{c}x}{\sin x\cos x}} = \frac{2\sin x\cos x}{1} = 2\sin x\cos x = L$$

L = P

1. $\frac{\cos x - \cos{3x}}{\sin{3x} - \sin x} = \operatorname{tg}{2x}$

$$\cos{\alpha -}\cos\beta = 2\sin\frac{\alpha + \beta}{2} = \sin\frac{\beta - \alpha}{2}$$

$$\operatorname{tg}{2x} = \frac{\sin{2x}}{\cos{2x}}$$

$$L = \frac{\cos x - \cos{3x}}{\sin{3x} - \sin x} = \frac{2\sin\frac{x + 3x}{2}\sin\frac{3x - x}{2}}{2\cos\frac{x + 3x}{2}\sin\frac{x - 3x}{2}} = \frac{2\sin{2x}\sin{2x}}{2\cos{2x}\sin{- 2x}} = \frac{2\operatorname{}{2x}}{2\cos{2x}\left( - \sin{2x} \right)} = \frac{\sin{2x}}{\cos{2x}} = \operatorname{tg}{2x}$$

tg2x = tg2x

L = P

Zadanie 4

$$y = \operatorname{}x + \operatorname{}x = ({\underset{1}{})}^{2} - 2\operatorname{}x\operatorname{}x = 1 - 2\operatorname{}x\operatorname{}x = 1 - \frac{1}{2}(4\operatorname{}x\operatorname{}{x)} = 1 - \frac{1}{2}{\underset{\operatorname{}x}{}}^{2} = 1 - \frac{1}{2}\left( \sin{2x} \right)^{2} = 1 - \frac{1}{2}\operatorname{s}{2x}$$

$$y = 1 - \frac{1}{2}\operatorname{}x$$

$$y_{\min} = 1 - \frac{1}{2}*0 = 1$$

$$y = 1\frac{1}{2}$$

$$y_{\max} = 1 - \frac{1}{2}*1 = \frac{1}{2}$$

$$y \in \left\langle \frac{1}{2},\ 1 \right\rangle$$

Zadanie 5

$$\frac{\sin x}{\operatorname{}x + 3\operatorname{}x}\ = \ ?$$

$$\operatorname{tg}x = \frac{\sin x}{\cos x} = 2$$

$$1 + \operatorname{}\alpha = \frac{1}{\operatorname{}\alpha}$$

sinx = 2cosx

$$\frac{\sin x}{\operatorname{}x + 3\operatorname{}x} = \frac{2\cos x}{{(2\cos x)}^{3} + 3\operatorname{co}x} = \frac{2\cos x}{8\operatorname{c}x + 3\operatorname{c}x} = \frac{2\cos x}{11\operatorname{co}x} = \frac{2}{11\operatorname{co}x} = \frac{2}{11}*\frac{1}{\operatorname{c}x} = \frac{2}{11}\left( 1 + \operatorname{t}x \right) = \frac{2}{11}\left( 1 + 4 \right) = \frac{2}{11}*5 = \frac{10}{11}$$

$$Odpowiedz:\ \ \ \frac{10}{11}$$

Zadanie 6

$$f\left( x \right) = \frac{\operatorname{s}x - |\sin{x|}}{\sin x}\ \ \ \ \ x \in \left( 0,\ \pi \right) \cup (\pi,\ 2\pi)$$

1)

sinx > 0

$$f\left( x \right) = \frac{\operatorname{s}x - \sin x}{\sin x} = \frac{\sin x(\sin{x - 1})}{\sin x} = \sin x - 1$$

2)

sinx < 0

$$f\left( x \right) = \frac{\operatorname{s}x + \sin x}{\sin x} = \frac{\sin x(\sin x + 1)}{\sin x} = \sin x + 1$$

1. $x_{1} = \frac{\pi}{2},\ \ x_{2} = \frac{3}{2}\pi$

Zadanie 7

y = |sin|x||

f(x) = sinx

f(x) = sin|x|

f(x) = |sin|x||

Zadanie 8

f(x) = |cosx| + cosx

y = cosx + cosx

1)

cosx ≥ 0

f(x) = 2cosx

2)

cosx < 0

f(x) = −cosx + cosx = 0

Zadanie 9

y = sin2x

⟨−2π, 2π⟩

sin(x+2π) = sin2(x+π)

T = π

Zadanie 10

2x = 2π

x = π

sin2x = 2sinxcosx

sinx ≠ 0

1)

$$y = \frac{\sin{2x}}{\sin x} = \frac{2\sin x\cos x}{\sin x} = 2\cos x$$

2)

$$y = \frac{\sin{2x}}{\sin x} = 2\cos x$$

sin2x < 0

spelniona dla ⟨π,0⟩ ∪ ⟨π, 2π⟩

Zadanie 11

$$\frac{\sin{2x}}{1 + \cos{2x}}*\frac{\cos x}{1 + \cos x} = \operatorname{tg}\frac{x}{2}$$

$$L = \frac{2\sin x\cos x}{1 + \operatorname{}x - \operatorname{}x}*\frac{\cos x}{1 + \cos x} = \frac{2\sin x\cos x}{2\operatorname{}x}*\frac{\cos x}{1 + \cos x} = \frac{\sin x\cos x}{\cos x}*\frac{1}{1 + \cos x} = \frac{\sin x}{1 + \cos x} = \frac{2\sin{\frac{x}{2}\cos\frac{x}{2}}}{1 + \operatorname{}\frac{x}{2} - \operatorname{}\frac{x}{2}} = \frac{2\sin{\frac{x}{2}\cos\frac{x}{2}}}{\underset{\cos x = \operatorname{}{\frac{x}{2} - \operatorname{}\frac{x}{2}}}{} + \operatorname{}\frac{x}{2}} = \frac{2\sin{\frac{x}{2}\cos\frac{x}{2}}}{2\operatorname{}\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \operatorname{tg}\frac{x}{2} = P$$