Question

Срочно!!!!!! Помогите пожалуйста

Answer 1

$1)\; \; a=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}=15^\circ \\\\sina=sin(\frac{\pi}{3}-\frac{\pi}{4})=sin\frac{\pi}{3}\cdot cos\frac{\pi}{4}-sin\frac{\pi}{4}\cdot cos\frac{\pi}{3}=\\\\=\frac{\sqrt3}{2}\cdot \frac{\sqrt2}{2}-\frac{\sqrt2}{2}\cdot \frac{1}{2}=\frac{\sqrt3\cdot \sqrt2-\sqrt2}{4}=\frac{\sqrt6-\sqrt2}{4}=\frac{\sqrt3-1}{2\sqrt2}\\\\cosa=cos(\frac{\pi }{3}-\frac{\pi}{4})=cos\frac{\pi}{3}\cdot cos\frac{\pi}{4}+sin\frac{\pi}{3}\cdot sin\frac{\pi}{4}=\\\\=\frac{1}{2}\cdot \frac{\sqrt2}{2}+\frac{\sqrt3}{2}\cdot \frac{\sqrt2}{2}=\frac{\sqrt2+\sqrt6}{4}=\frac{\sqrt3+1}{2\sqrt2}$

$tga=\frac{sina}{cosa}=\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}{(\sqrt3-1)(\sqrt3+1)}=\frac{4-2\sqrt3}{3+1}=\frac{2(2-\sqrt3)}{4}=2-\sqrt3\\\\ctga=\frac{1}{tga}=\frac{1}{2-\sqrt3}=\frac{2+\sqrt3}{(2-\sqrt3)(2+\sqrt3)}=\frac{2+\sqrt3}{4-3}=2+\sqrt3\\\\\\sin\frac{\pi }{12}=sin15^\circ =\frac{\sqrt3-1}{2\sqrt2}=\frac{\sqrt6-\sqrt2}{4}\; ,\; \; cos\frac{\pi }{12}=cos15^\circ =\frac{\sqrt3+1}{2\sqrt2}=\frac{\sqrt6+\sqrt2}{4}\; ,\\\\tg{\frac{\pi}{12}=tg15^\circ =2-\sqrt3\; \; ,\; \; ctg \frac{\pi }{12}=2+\sqrt3$

$2)\; \; cosa=-\frac{5}{13}\\\\\pi <a<\frac{3\pi}{2}\; \; \to \; \; sina<0\; ,\; \; tga>0\; ,\; \; ctga>0\\\\sina=-\sqrt{1-cos^2a}=-\sqrt{1-\frac{25}{169}}=-\frac{12}{13}\\\\sin2a=2\, sina\, cosa=2\cdot (- \frac{12}{13})\cdot (-\frac{5}{13})=\frac{120}{169}\\\\cos2a=2cos^2a-1=2\cdot \frac{25}{169}-1=-\frac{119}{169}\\\\tg2a=\frac{sin2a}{cos2a}=-\frac{120}{119}\; \; ,\; \; ctga=-\frac{119}{120}$

$3)\; \; 2sin(\frac{\pi}{4}+a)\cdot sin(\frac{\pi}{4}-a)+sin^2a=\\\\=2\cdot \frac{1}{2}\cdot \Big (cos(\frac{\pi}{4}+a-\frac{\pi}{4}+a)-cos(\frac{\pi }{4}-a+\frac{\pi }{4}-a)\Big )+sin^2a=\\\\=cos2a-cos\frac{\pi }{2}+sin^2a=(cos^2a-sin^2a)-0+sin^2a=cos^2a\; ;\\\\cos^2a=cos^2a\; .\\\\\star \; \; sina\cdot sin\beta =\frac{1}{2}\cdot (cos(a-\beta )-cos(a+\beta ))\; \; \star$