Question

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Answer 1

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Answer 2

$23) \frac{1-sin( \frac{ \pi }{2}+2 \beta )}{2sin \beta }= \frac{1-cos2 \beta }{2sin \beta }= \frac{1-(1-2sin^2 \beta) }{2sin \beta }= \frac{2sin^2 \beta }{2sin \beta}=sin \beta$
$24) \frac{1-cos( \pi + \beta )}{sin( \pi - \beta )}= \frac{1-(-cos \beta )}{sin \beta }= \frac{1+cos \beta }{sin \beta }=ctg \frac{ \alpha }{2}$
$25) \frac{1+cos4 \beta }{cos^2\beta-sin^2\beta}= \frac{1+2cos^22\beta-1}{cos2 \beta}=2cos2 \beta$
$26) \frac{1+cos2 \beta}{1-cos2\beta}= \frac{1+2cos^2\beta-1}{1-1+2sin2\beta}=\frac{2cos^2\beta}{2sin^2\beta}=tg^2 \beta$
$27) \frac{1-sin2 \beta}{1+sin2\beta}= \frac{(sin \beta-cos\beta)^2}{(sin\beta+cos\beta)^2}=( \frac{tg \beta-1}{tg \beta+1})^2=tg^2( \beta- \frac{\pi}{4}})$
$28)1+cos4x=1+2cos^22x-1=2cos^22x$
$29)1-cos4x=1-(1-2sin^22x)=2sin^22x$
УДАЧИ ВАМ ВО ВСЁМ)))!