Question

9 класс, 100 баллов за три номера

Answer 1

1) sin⁴α + cos⁴α = sin⁴α - 2cos²αsin²α + cos⁴α + 2cos²αsin²α =  (cos²α - sin²α)² + 2cos²αsin²α = cos²2α + 0,5sin²2α = 0,5(1 + cos4α) + 0,25(1 - cos4α) = 0,5 + 0,5cos4α + 0,25 - 0,25cos4α = 0,75 + 0,25cos4α = 3/4 + (cos4α)/4 = (3 + cos4α)/4.

2) sin2α·sin5α·cos(π/2 - 7α) - cos2α·cos(π/2 + 5α)·cos7α = 1/6;

sin2α·sin5α·sin7α + cos2α·sin5α·cos7α = 1/6;

sin5α(sin2α·sin7α + cos2α·cos7α) = 1/6;

sin5αcos(7α - 2α) = 1/6;

sin5αcos5α = 1/6|·2;

2sin5αcos5α = 1/3;

sin10α = 1/3;

cos20α = 1 - 2sin²10α = 1 - 2·1/9 = (9 - 2)/9 = 7/9

3) $\(\frac{cos20^0+cos40^0+cos100^0}{\sqrt{1-cos260^0}}=\frac{2cos30^0\cdot cos10^0+cos100^0}{\sqrt{1+cos80^0}}=2\frac{\frac{\sqrt{3}}{2}cos10^0-\frac{1}{2}sin10^0}{\sqrt{1+cos80^0}}=2\frac{cos30^0cos10^0-sin30^0sin10^0}{\sqrt{2cos^240^0}}=2\frac{cos(30^0+10^0)}{\sqrt{2}cos40^0}=\frac{2cos40^0}{\sqrt{2}cos40^0}=\frac{2}{\sqrt{2}}=\sqrt{2}\)$