Решение во вложенном файле
Будет 8х-2/25х-4 = 1/17х-2
Мы скоротили 8х и 25х, а также 2 и 4
1)находим производную.
2) подставляем точку.
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1)f'=cos(x)-(-2sin(x)cos(x));
подставляем x=0;
1-(0*1)=1;
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2)f'=-(5/2)*(1/корень из(-(x-1)x));
подставляем x=1/2;
−(5 / 2) × (1 / (√(−((0,5) − 1) × (0,5))))=-5;
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3)f'=cos(x)*e^sin(x);
подставляем x=0
(1)*e^0=1;
По условию 3π/2<α<2π
это IV четверть где cosα>0 и sinα<0
значит cosα=¹²/₁₃
найти tg(π/4-α)
решение:
![\displaystyle sin^2a+cos^2a=1\\\\sina= \sqrt{1-cos^2a}= \sqrt{1-( \frac{12}{13})^2}= \sqrt{1- \frac{144}{169}}= \sqrt{ \frac{25}{169}}=| \frac{5}{13}|\\\\sina\ \textless \ 0\\\\sina=- \frac{5}{13}\\\\tga= \frac{sina}{cosa}= \frac{-5}{13}: \frac{12}{13}=- \frac{5}{12}\\\\tg( \frac{ \pi }{4}-a)= \frac{tg( \pi /4)-tga}{1+tg( \pi /4)*tga}= \frac{1-( \frac{-5}{12})}{1+1( \frac{-5}{12})}=\\\\= \frac{ \frac{17}{12}}{ \frac{7}{12}}= \frac{17}{7}](https://tex.z-dn.net/?f=%5Cdisplaystyle+sin%5E2a%2Bcos%5E2a%3D1%5C%5C%5C%5Csina%3D+%5Csqrt%7B1-cos%5E2a%7D%3D+%5Csqrt%7B1-%28+%5Cfrac%7B12%7D%7B13%7D%29%5E2%7D%3D+%5Csqrt%7B1-+%5Cfrac%7B144%7D%7B169%7D%7D%3D+%5Csqrt%7B+%5Cfrac%7B25%7D%7B169%7D%7D%3D%7C+%5Cfrac%7B5%7D%7B13%7D%7C%5C%5C%5C%5Csina%5C+%5Ctextless+%5C+0%5C%5C%5C%5Csina%3D-+%5Cfrac%7B5%7D%7B13%7D%5C%5C%5C%5Ctga%3D+%5Cfrac%7Bsina%7D%7Bcosa%7D%3D+%5Cfrac%7B-5%7D%7B13%7D%3A+%5Cfrac%7B12%7D%7B13%7D%3D-+%5Cfrac%7B5%7D%7B12%7D%5C%5C%5C%5Ctg%28+%5Cfrac%7B+%5Cpi+%7D%7B4%7D-a%29%3D+%5Cfrac%7Btg%28+%5Cpi+%2F4%29-tga%7D%7B1%2Btg%28+%5Cpi+%2F4%29%2Atga%7D%3D+%5Cfrac%7B1-%28+%5Cfrac%7B-5%7D%7B12%7D%29%7D%7B1%2B1%28+%5Cfrac%7B-5%7D%7B12%7D%29%7D%3D%5C%5C%5C%5C%3D+%5Cfrac%7B+%5Cfrac%7B17%7D%7B12%7D%7D%7B+%5Cfrac%7B7%7D%7B12%7D%7D%3D+%5Cfrac%7B17%7D%7B7%7D+++++++++++++++)
Sin 54- sin 18 = 2 sin (54-18 /2) *cos (54+18)/2=2 *sin18*cos72=2 sin^2(18)=1-cos36=
=1-(1+√5)/4
=(3-√5)/4
=3/4-√5/4