| π | ||
mam pewien problem a mianowicie jak z wzoru funkcji tg(x+ | ) dojść do wzoru | |
| 4 |
| 1+sin2x | ||
| cos2x |
| π |
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tg(x + | ) = | = | |||||||||||
| 4 |
|
| sinx + cosx | (sinx + cosx)2 | |||
= | = | = | ||
| cosx − sinx | (cosx − sinx)(cosx + sinx) |
| sin2x + cos2 + 2sinxcosx | 1 + sin2x | |||
= | = | |||
| cos2x − sin2x | cos2x |