Question

Нужна
И без приколов
нужно решить номер 26.15 на фото
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Answer 1

а) $1-8sin2\beta*cos2\beta = 1-4*2sin2\beta*cos2\beta = 1-4sin4\beta$

б) $tg\beta*(1+cos2\beta)-sin2\beta=\frac{sin\beta}{cos\beta}*(1+2cos^{2}\beta-1)-sin2\beta=\frac{sin\beta}{cos\beta}*2cos^{2}-sin2\beta=sin\beta*2cos\beta-sin2\beta=2sin\beta*cos\beta-sin2\beta=sin2\beta-sin2\beta=0$

в) $\frac{2sin\beta-sin2\beta}{2sin\beta+sin2\beta}=\frac{2sin\beta-2sin\beta*cos\beta}{2sin\beta+2sin\beta*cos\beta}=\frac{2sin\beta(1-cos\beta)}{2sin\beta(1+cos\beta)}=\frac{1-cos\beta}{1+cos\beta}=tg^{2}\frac{\beta}{2}$

г) ctg(45°-β)/1 - ctg²(45°-β) = ctg 45° × ctgβ +1/ctgβ - ctg 45° / 1 - (ctg 45° × ctgβ +1/ctgβ - ctg 45°)² = cotβ+1/ctgβ-1 / 1-(ctgβ+1/ctgβ-1)² = cotβ+1/ctgβ-1 / 1- (ctgβ+1)²/(ctgβ-1)² = cotβ+1/ctgβ-1 / (ctgβ-1)²-ctg(β+1)²/(ctgβ-1)² = cotβ+1/ctgβ-1 / -2×2ctgβ/(ctgβ-1)² = cotβ+1/ctgβ-1 / -4ctgβ/(ctgβ-1)² = - cotβ+1/ctgβ-1 / 4ctgβ/(ctgβ-1)² = - cotβ+1/ctgβ-1 × (ctgβ-1)²/4ctgβ = - (cotβ+1)×(ctgβ-1)/4ctgβ = - ctg²β-1/4ctgβ = 1-ctg²β/4ctgβ