Question

Вложение 3,4,5,,,,,,,,,,,,,,,,

Answer 1

$4cos45sin135= 4cdotfrac{ sqrt{2} }{2} cdotfrac{ sqrt{2} }{2} =2 \ sin420cos600=sin60cos120=frac{ sqrt{3} }{2}cdot(- frac{1}{2} )=-frac{ sqrt{3} }{4}$
$frac{ sqrt{3} }{2}-sin frac{pi}{3} =frac{ sqrt{3} }{2}-frac{ sqrt{3} }{2}=0 \ sin frac{6pi}{5}tg frac{7pi}{3} =sin frac{6pi}{5}tg frac{pi}{3} = sqrt{3} sin frac{6pi}{5}$
$frac{3tg^2( alpha +3pi)-1}{3-tg^2( alpha + frac{5pi}{2}) } = frac{3tg^2 alpha-1}{3+ctg^2 alpha }$
$frac{tg2 alpha +tg3 alpha }{1-tg2 alpha tg3 alpha } =tg(2 alpha +3 alpha )=tg5 alpha$
$frac{cos2 alpha }{sin alpha -cos alpha } = frac{cos^2 alpha-sin^2 alpha }{sin alpha -cos alpha } =frac{(cos alpha-sin alpha )(cos alpha+sin alpha ) }{sin alpha -cos alpha } =cos alpha-sin alpha$
$frac{sin3 alpha -sin alpha cos2 alpha }{sin3 alpha +sin alpha } = frac{3sin alpha-4sin^3 alpha -sin alpha( cos^2 alpha-sin^2 alpha ) }{3sin alpha-4sin^3 alpha +sin alpha } = frac{3sin alpha-3sin^3 alpha -sin alpha cos^2 alpha }{4sin alpha-4sin^3 alpha } = \ frac{3-3sin^2 alpha -cos^2 alpha }{4-4sin^2 alpha }$
$cos^4 frac{ alpha }{4} -sin^4 frac{ alpha }{4} -cos alpha = cos^2 frac{ alpha }{2} -cos alpha = frac{1+cos alpha-2cos alpha }{2} = frac{1-cos alpha }{2}$
$cos^4 alpha (1+tg^2 alpha )+sin^2 alpha = frac{cos^4 alpha }{cos^2 alpha } +sin^2 alpha =cos^2 alpha +sin^2 alpha =1$

$sin138>0 \ cos50>0 \ sin138+cos50>0$

$sin frac{2pi}{3} - frac{ sqrt{3} }{2} = frac{ sqrt{3} }{2} - frac{ sqrt{3} }{2} =0$

$sin frac{7pi}{5} \ sin frac{17pi}{10}=sin frac{8.5pi}{5} \ sin frac{7pi}{5}-sin frac{17pi}{10}<0$

$3.14<pi \ tg3.14-tgpi<0$

Answer 2

Использованы тригонометрические формулы и, где возможно, табличные значения триг. функций, а также знаки триг. функций

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