1 | ||
1) arccos | =α, α∊<0,π> | |
4 |
1 | ||
1) arccos | =α, α∊<0,π> | |
4 |
1 | 1 | 1 | 1 | |||||
α=2sin(arccos | )*cos(arccos | )= | *sin(arccos | ) | ||||
4 | 4 | 2 | 4 |
π | ||
sin zamienić na cos z jedynki trygonometrycznej, czy arccos x = | − arcsin x? | |
2 |
1 | 1 | 1 | 1 | 1 | 15 | 15 | |||||||
*(1−cos2(arccoss | )) = | *(1−( | )2) = | * | = | ||||||||
2 | 4 | 2 | 4 | 2 | 16 | 32 |
π | ||
arccos x = | − arcsin x? | |
2 |
1 | π | 1 | 1 | 1 | 1 | 3 | 3 | ||||||||
*sin( | +arcsin(− | ) = | *(1− | ) = | *(− | ) = − | |||||||||
2 | 2 | 4 | 2 | 4 | 2 | 4 | 8 |
√15 | ||
Natomiast wynik ma być | ||
8 |
1 | 1 | ||
*sin(arccos | )= | ||
2 | 4 |
1 | |
*√1−cos2(arccos(1/4))= | |
2 |
1 | 1 | √15 | √15 | |||||
= | √1−(1/16)= | * | = | |||||
2 | 2 | √16 | 8 |
1 | π | 1 | |||
sin( | −arcsin | ) tu korzystasz z wzoru sin(α−β) | |||
2 | 2 | 4 |
1 | 1 | |||
sin(2arccos | )=sinα, gdzie 2arccos | =α, α∊<0,π> | ||
4 | 4 |
1 | α | |||
arccos | = | /cos | ||
4 | 2 |
α | 1 | |||
cos | = | |||
2 | 4 |
α | ||
cosα= 2 cos2 | −1 ze wzoru cos2x=cos2x−sin2x=2cos2x−1 | |
2 |
1 | ||
cosα=2*( | )2−1 | |
4 |
−7 | ||
cosα= | (α−kąt rozwarty) | |
8 |
√15 | ||
sinα= | ||
8 |
1 | ||
2) y=ctg(arccos( | ) ? taki przykład? | |
3 |
cos(arccos(1/3)) | ||
y= | i sin(arccos(1/3))≠0 | |
sin(arccos(1/3)) |
1 | ||
arccos | =α | |
3 |
(1/3) | ||
y= | ||
sin(arccos(1/3)) |
1 | 1 | |||
arccos | =α ⇔cosα= | |||
3 | 3 |
2√2 | ||
sinα= | ||
3 |
| 1 | √2 | ||||||||||
y= | = | = | ||||||||||
| 2√2 | 4 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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