Вычислите. (1-sin^2*15)/2cos^2*(pi/8)-1 - вопрос №451641912152281 от сехун2 12.09.2022 10:29

Question

Алгебра: Вычислите. (1-sin^2*15)/2cos^2*(pi/8)-1

сехун2 · Answer

(1-sin²15°)/(2cos²π/8 -1) = cos²15°/cosπ/4 = ((1+cos30°)/2 ) /(1/√2) =√2(1+√3/2)/2 =√2(2+√3)/4 .
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(1-sin²15°)/2cos²π/8 -1 = cos²15° /(1+cosπ/4) - 1 =
(1+cos30°)/2(1 +1/√2) -1=(1+√3/2)/2(1 +1/√2) -1 =(2+√3)/4(1+1/√2) -1 =
(2+√3)/(4+2√2) -1 = (2+√3)*(4-2√2)/(4+2√2)(4-2√2) -1 = (2+√3)*(4-2√2)/8 - 1 =
(8 -4√2 +4√3 -2√6)/8 -1 = (8 -4√2 +4√3 -2√6-8)/8 = (4√3 -4√2 -2√6)/8 = (2√3 -2√2 -√6)/4.