Question

464(1,3)
465(1)
заранее спасибо!)))

Answer 1

$1)quad frac{1-2sin^2frac{x}{2}}{2cos^2frac{x}{2}-1}=frac{1-(1-cosx)}{(1+cosx)-1}=frac{cosx}{cosx}=1\\2)quad cos^4a-sin^4a-cos2a=\\=(underbrace {cos^2a-sin^2a}_{cos2a})(underbrace {cos^2a+sin^2a}_{1})-cos2a=cos2a-cos2a=0$

$3)quad frac{1+cos42^circ }{1-cos42^circ } = frac{2cos^284^circ}{2sin^284^circ } =ctg^284^circ=tg^2(90^circ-6^circ )=tg^26^circ\\\Formylu:quad 1-cosa=2sin^2frac{a}{2}; ; ;; ; ; 1+cosa=2cos^2frac{a}{2}\\ili\\ sin^2frac{a}{2}=frac{1-cosa}{2}; ; ; ;; ; ; cos^2frac{a}{2}= frac{1+cosa}{2}$