Oblicz wartość wyrażenia: a) [b) oblicz wiedząc, że.... Question from @ola1996

ola1996 @ola1996

November 2018 1 10 Report

Oblicz wartość wyrażenia:

a) [ $\frac{(1-sin \alpha)(1+sin \alpha ) }{sin ^{2} \alpha } + \frac{(1-cos \alpha)(1+cos \alpha ) }{cos ^{2} \alpha } ]*( \frac{1}{sin ^{2} \alpha cos ^{2} \alpha } -2) ^{-1}$

b) $( \sqrt{ \frac{1-cos \alpha }{1+cos \alpha } } + \sqrt{ \frac{1+cos \alpha }{1-cos \alpha } } ) ^{2} - \frac{4}{1-cos ^{2} \alpha }$

oblicz $sin^4 \alpha +cos^4 \alpha$ wiedząc, że $sin \alpha +cos \alpha =0,5$

Zgłoś nadużycie! A)
$\left[ \frac{(1-sin \alpha )(1+sin \alpha )}{ sin^{2} \alpha } + \frac{(1-cos \alpha)(1+cos \alpha ) }{ cos^{2} \alpha }\right]\cdot\left( \frac{1}{ sin^{2} \alpha cos^{2} \alpha } -2\right) ^{-1}=$

$\left[ \frac{1-sin^2\alpha}{ sin^{2} \alpha } + \frac{1-cos^2 \alpha}{ cos^{2} \alpha }\right]\cdot\left( \frac{1-2sin^{2} \alpha cos^{2} \alpha}{ sin^{2} \alpha cos^{2} \alpha }\right) ^{-1}=$

$\left[ \frac{cos^2\alpha}{ sin^{2} \alpha } + \frac{sin^2 \alpha}{ cos^{2} \alpha }\right]\cdot\left( \frac{sin^{2} \alpha cos^{2} \alpha}{1-2sin^{2} \alpha cos^{2} \alpha }\right)=$

$\left[ \frac{cos^4\alpha+sin^4\alpha}{sin^2\alpha cos^{2} \alpha }\right]\cdot\left( \frac{sin^{2} \alpha cos^{2} \alpha}{1-2sin^{2} \alpha cos^{2} \alpha }\right)=$

$\frac{cos^4\alpha+sin^4\alpha}{1-2sin^{2} \alpha cos^{2} \alpha }=$

$\frac{(cos^2\alpha+sin^2\alpha)^2-2cos^2\alpha sin^2\alpha}{1-2sin^{2} \alpha cos^{2} \alpha }=$

$\frac{1-2cos^2\alpha sin^2\alpha}{1-2sin^{2} \alpha cos^{2} \alpha }=1$
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b)
$\left(\sqrt{\frac{1-cos \alpha}{1+cos \alpha}} + \sqrt{\frac{1+cos \alpha}{1-cos \alpha}} \right) ^{2} - \frac{4}{1-cos ^{2} \alpha}=$

$\frac{1-cos \alpha}{1+cos \alpha}} +2 \sqrt{\left( \frac{1-cos \alpha}{1+cos \alpha} \right)\left(\frac{1+cos \alpha}{1-cos \alpha}} \right)}+ \frac{1+cos \alpha}{1-cos \alpha}}- \frac{4}{1-cos ^{2} \alpha}=$

$\frac{1-cos \alpha}{1+cos \alpha}} +2 \sqrt{1}+ \frac{1+cos \alpha}{1-cos \alpha}}- \frac{4}{1-cos ^{2} \alpha}=$

$\frac{1-cos \alpha}{1+cos \alpha}} +2+ \frac{1+cos \alpha}{1-cos \alpha}}- \frac{4}{1-cos ^{2} \alpha}=$

$\frac{(1-cos \alpha)^2+(1+cos \alpha)^2}{(1+cos \alpha)(1-cos \alpha)}+ \frac{2(1-cos^{2} \alpha)}{1-cos ^{2} \alpha}- \frac{4}{1-cos ^{2} \alpha}=$

$\frac{1-2cos \alpha+cos^2\alpha+1+2cos \alpha+cos^2\alpha}{(1+cos \alpha)(1-cos \alpha)}+ \frac{2(1-cos ^{2} \alpha)}{1-cos ^{2} \alpha}- \frac{4}{1-cos ^{2} \alpha}=$

$\frac{2+2cos^2\alpha}{1-cos^2\alpha}+ \frac{2(1-cos ^{2} \alpha)}{1-cos ^{2} \alpha}- \frac{4}{1-cos ^{2} \alpha}=$

$\frac{2+2cos^2\alpha+2(1-cos ^{2})-4}{1-cos ^{2} \alpha}=$

$\frac{2+2cos^2\alpha+2-2cos ^{2}\alpha-4}{1-cos ^{2} \alpha}=0$
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$sin \alpha +cos \alpha =0,5\ / ()^2$

$sin^2 \alpha +2sin\alpha cos\alpha+cos^2 \alpha =0,25$

$1+2sin\alpha cos\alpha=0,25$

$2sin\alpha cos\alpha=0,25-1$

$2sin\alpha cos\alpha=-0,75\ /:2$

$sin\alpha cos\alpha=-\frac{3}{8}$
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$sin^{4} \alpha +cos^{4} \alpha=(sin^2\alpha+cos^2\alpha)^2-2sin^2\alpha cos^2\alpha=$

$1-2(sin\alpha cos\alpha)^2=1-2\cdot\left(-\frac{3}{8} \right)^2=$

$1-2\cdot\frac{9}{64}=1-\frac{9}{32}=\frac{23}{32}$

2 votes Thanks 1

ola1996 a co do nawiasów to chodzi o pierwszy fragment, w którym ich nie ma

ola1996 linijka 5

ola1996 aa, faktycznie. to teraz jeszcze drugi przykład druga linijka

ola1996 aa juz mam

ola1996 tak, już widzę, dziękuję bardzo za pomoc