Odpowiedź:
zad 9.17
sinα = √3/2
sin²α = (√3/2)² = 3/4
1 - cos²α = 3/4
- cos²α = - 1 + 3/4
cos²α = 1 - 3/4 = 1/4
cosα = √(1/4) = 1/2
sin²α - 3cos²α = 3/4 - 3 * 1/4 = 3/4 - 3/4 = 0
zad 9.18
sin²α + sin²α * cos²α + cos⁴α = sin²α(1 + cos²α) + cos⁴α =
= sin²α(sin²α + cos²α + cos²α) + cos⁴α = sin²α(sin²α + 2cos²α) + cos⁴α =
= sin⁴α + 2sin²αcos²α + cos⁴α = (sin²α + cos²α)² = 1² = 1
Odp: C
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Odpowiedź:
zad 9.17
sinα = √3/2
sin²α = (√3/2)² = 3/4
1 - cos²α = 3/4
- cos²α = - 1 + 3/4
cos²α = 1 - 3/4 = 1/4
cosα = √(1/4) = 1/2
sin²α - 3cos²α = 3/4 - 3 * 1/4 = 3/4 - 3/4 = 0
zad 9.18
sin²α + sin²α * cos²α + cos⁴α = sin²α(1 + cos²α) + cos⁴α =
= sin²α(sin²α + cos²α + cos²α) + cos⁴α = sin²α(sin²α + 2cos²α) + cos⁴α =
= sin⁴α + 2sin²αcos²α + cos⁴α = (sin²α + cos²α)² = 1² = 1
Odp: C