Question

математика.. пожалуйста помогите срочно надо ....

Answer 1

$1)\; \; -3sin^2x+1-3cos^2x=-3(sin^2x+cos^2x)+1=-3+1=-2\\\\2)\; \; tg100+\frac{sin530}{1+sin640}=tg(90+10)+\frac{sin(360+170)}{1+sin(720-80)}=\\\\=-ctg10+\frac{sin(180-10)}{1-sin80}=-ctg10+\frac{sin10}{1-sin(90-10)}=\\\\=-ctg10+\frac{sin10}{1-cos10}=\frac{sin10}{2sin^220}-ctg10$

$3)\; \; tga=2\\\\ \frac{3sin^2a+12sina\cdot cosa+cos^2a}{sin^2a+sina\cdot cosa-2cos^2a}=\frac{cos^2a\cdot (3tg^2a+12tga+1)}{cos^2a\cdot (tg^2a+tga-2)}=\\\\=\frac{3\cdot 2^2+12\cdot 2+1}{2^2+2-2}=\frac{37}{4}=9,25\\\\4)\; \; sin(\frac{\pi}{2}-\alpha )=cos\alpha \\\\cos72^\circ = \frac{\sqrt5-1}{4}$

$\star \; \; cos72^\circ =cos(90^\circ -18^\circ )=sin18^\circ \\\\4sin^218=\frac{2\cdot sin18\cdot 2sin18\cdot cos18}{cos18}=\frac{2\cdot sin18\cdot sin36}{cos18} =\frac{2\cdot \frac{1}{2}(cos18-cos54)}{cos18}=\\\\=1-\frac{cos54}{cos18}=1-\frac{cos(90-36)}{cos18}=1-\frac{sin36}{cos18}=1-\frac{2sin18\cdot cos18}{cos18}=\\\\=1-2sin18\; \; \to\; \; 4sin^218=1-2sin18\\\\4sin^218+2sin18-1=0\; ,\\\\t=sin18^\circ >0\; ,\; \; 4t^2+2t-1=0\; ,\\\\D/4=1+4=5\; ,\; \; \; t=\frac{-1\pm \sqrt5}{4}\; ,\; \; t>0\; \; \to \\\\sin18^\circ = \frac{-1+\sqrt5}{4}=cos72^\circ$