23 uždavinys
1. Apskaičiuokite f(x) reikšmę, kai $$x = \frac{\pi}{2}$$
Sprendimas.
sin(x)-cos( 2* x) =
|
sin(x)-cos( 2* x) = $$sin(x)-cos(2\cdot x)$$ =
Paaiškinimas: Keitimas $${\normalsize x}$$ = $${\normalsize \frac{\pi}{2}}$$.
sin(
)-cos(
) = $$sin(\frac{\pi}{2})-cos(\frac{2\cdot \pi}{2})$$ =
| π |
| / 2 |
| 2* π |
| / 2 |
$${\normalsize \frac{2\cdot \pi}{2}}$$ = $${\normalsize \pi}$$
sin(
)-cos(π) = $$sin(\frac{\pi}{2})-cos(\pi)$$ =
| π |
| / 2 |
) = $$sin(\frac{\pi}{2})-cos(\pi)$$ = $${\normalsize sin(\frac{\pi}{2})}$$ = $${\normalsize 1}$$
1-cos(π) = $$1-cos(\pi)$$ =
) = $$1-cos(\pi)$$ = $${\normalsize cos(\pi)}$$ = $${\normalsize -1}$$
1+1 = $$1+1$$ =
$${\normalsize 1+1}$$ = $${\normalsize 2}$$
2$$2$$
$$sin(x)-cos(2\cdot x)$$ = $$$$
$$sin(\frac{\pi}{2})-cos(\frac{2\cdot \pi}{2})$$ = $$$$
$$sin(\frac{\pi}{2})-cos(\pi)$$ = $$$$
$$1+1$$ = $$$$
$$2$$ $$$$
Atsakymas: 2
2.Parodykite, kad f(x) = (sin(x)+1)(2sin(x)-1)
Sprendimas.
sin(x)-cos( 2* x) =
|
|
sin(x)-cos( 2* x) = $$sin(x)-cos(2\cdot x)$$ =
$${\normalsize cos(2\cdot x)}$$ = $${\normalsize (cos(x)^{2}-sin(x)^{2})}$$ Paaiškinimas:
Paaiškinimas: Kosinuso argumento žeminimo formulė $${\normalsize cos(2\cdot a) = cos(a)^{2}-sin(a)^{2}}$$
sin(x)-( cos(x)^2- sin(x)^2) = $$sin(x)-(cos(x)^{2}-sin(x)^{2})$$ =
$${\normalsize -(cos(x)^{2}-sin(x)^{2})}$$ = $${\normalsize cos(x)^{2}+sin(x)^{2}}$$
sin(x)- cos(x)^2+ sin(x)^2 = $$sin(x)-cos(x)^{2}+sin(x)^{2}$$ =
$${\normalsize cos(x)^{2} = 1-sin(x)^{2}}$$
sin(x)-(1- sin(x)^2)+ sin(x)^2 = $$sin(x)-(1-sin(x)^{2})+sin(x)^{2}$$ =
$${\normalsize -(1-sin(x)^{2})}$$ = $${\normalsize -1+sin(x)^{2}}$$
sin(x)-1+ sin(x)^2+ sin(x)^2 = $$sin(x)-1+sin(x)^{2}+sin(x)^{2}$$ =
$${\normalsize sin(x)^{2}+sin(x)^{2}}$$ = $${\normalsize 2\cdot sin(x)^{2}}$$
sin(x)-1+ 2* sin(x)^2 = $$sin(x)-1+2\cdot sin(x)^{2}$$ =
2* sin(x)^2+sin(x)-1 = $$2\cdot sin(x)^{2}+sin(x)-1$$ =
$${\normalsize 2\cdot sin(x)^{2}+sin(x)-1}$$ = $${\normalsize 2\cdot (sin(x)-\frac{1}{2})\cdot (sin(x)+1)}$$ Paaiškinimas:
Paaiškinimas: Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 2, b = 1, c = -1.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 1-(-8)}$$ = 9.
x1 = $${\normalsize \frac{-1+\sqrt {9}}{2\cdot 2} = \frac{-1+3}{4} = \frac{2}{4}}$$ = 1/2
x2 = $${\normalsize \frac{-1-\sqrt {9}}{2\cdot 2} = \frac{-1-3}{4} = \frac{-4}{4}}$$ = -1
a = 2, b = 1, c = -1.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 1-(-8)}$$ = 9.
x1 = $${\normalsize \frac{-1+\sqrt {9}}{2\cdot 2} = \frac{-1+3}{4} = \frac{2}{4}}$$ = 1/2
x2 = $${\normalsize \frac{-1-\sqrt {9}}{2\cdot 2} = \frac{-1-3}{4} = \frac{-4}{4}}$$ = -1
2* (sin(x)-
)* (sin(x)+1) = $$2\cdot (sin(x)-\frac{1}{2})\cdot (sin(x)+1)$$ =
| 1 |
| / 2 |
$${\normalsize 2\cdot (sin(x)-\frac{1}{2})}$$ = $${\normalsize (2\cdot sin(x)-\frac{2\cdot 1}{2})}$$
( 2* sin(x)-
)* (sin(x)+1) = $$(2\cdot sin(x)-\frac{2\cdot 1}{2})\cdot (sin(x)+1)$$ =
| 2* 1 |
| / 2 |
$${\normalsize \frac{2\cdot 1}{2}}$$ = $${\normalsize 1}$$
( 2* sin(x)-1)* (sin(x)+1)$$(2\cdot sin(x)-1)\cdot (sin(x)+1)$$
3.Išspręskite lygtį f(x) = 0
Sprendimas.
2* sin(x)-1 =
0
0
|
|
2* sin(x)-1 = 0$$2\cdot sin(x)-1$$ = $$0$$
2* sin(x) = 0+1$$2\cdot sin(x)$$ = $$0+1$$
2* sin(x) = 1$$2\cdot sin(x)$$ = $$1$$
sin(x) =
$$sin(x)$$ = $$\frac{1}{2}$$
| 1 |
| / 2 |
arcsin(sin(x)) = arcsin(
)$$arcsin(sin(x))$$ = $$arcsin(\frac{1}{2})$$
| 1 |
| / 2 |
$${\normalsize arcsin(sin(x))}$$ = $${\normalsize x}$$
x = arcsin(
)$$x$$ = $$arcsin(\frac{1}{2})$$
| 1 |
| / 2 |
$${\normalsize arcsin(\frac{1}{2})}$$ = $${\normalsize \frac{\pi}{6}}$$
x =
$$x$$ = $$\frac{\pi}{6}$$
| π |
| / 6 |
x = ( (-1)^k* (
)+ π* k)$$x$$ = $$((-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k)$$
| π |
| / 6 |
* k)$$x$$ = $$((-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k)$$$${\normalsize ((-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k)}$$ = $${\normalsize (-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k}$$
x = (-1)^k* (
)+ π* k$$x$$ = $$(-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k$$
| π |
| / 6 |
* k$$x$$ = $$(-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k$$$$2\cdot sin(x)-1$$ = $$0$$
$$2\cdot sin(x)$$ = $$1$$
$$sin(x)$$ = $$\frac{1}{2}$$
$$x$$ = $$arcsin(\frac{1}{2})$$
$$x$$ = $$(-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k$$
sin(x)+1 =
0
0
|
|
sin(x)+1 = 0$$sin(x)+1$$ = $$0$$
sin(x) = -1+0$$sin(x)$$ = $$-1+0$$
sin(x) = -1$$sin(x)$$ = $$-1$$
arcsin(sin(x)) = arcsin(-1)$$arcsin(sin(x))$$ = $$arcsin(-1)$$
$${\normalsize arcsin(sin(x))}$$ = $${\normalsize x}$$
x = arcsin(-1)$$x$$ = $$arcsin(-1)$$
$${\normalsize arcsin(-1)}$$ = $${\normalsize -\frac{\pi}{2}}$$
x = -
$$x$$ = $$-\frac{\pi}{2}$$
| π |
| / 2 |
x = -
+ 2* π* k$$x$$ = $$-\frac{\pi}{2}+2\cdot \pi\cdot k$$
| π |
| / 2 |
* k$$x$$ = $$-\frac{\pi}{2}+2\cdot \pi\cdot k$$$$sin(x)+1$$ = $$0$$
$$sin(x)$$ = $$-1$$
$$x$$ = $$arcsin(-1)$$
$$x$$ = $$-\frac{\pi}{2}+2\cdot \pi\cdot k$$
Atsakymas: $$x = (-1)^{k}\cdot (\frac{\pi}{6})+\pi\cdot k$$ ir $$x = -\frac{\pi}{2}+2\cdot \pi\cdot k$$