Тригонометрични функции 10 клас

[kristiana13]

някой може ли да ми докаже оградените тъждества

[Добромир Глухаров]

$2sin^2{\alpha}-1=sin^2{\alpha}-(1-sin^2{\alpha})=sin^2{\alpha}-cos^2{\alpha}=(sin{\alpha}+cos{\alpha})(sin{\alpha}-cos{\alpha})=\\=\left(sin{\alpha}+sin{\left(\frac{\pi}{2}-\alpha\right)}\right)\left(sin{\alpha}-sin{\left(\frac{\pi}{2}-\alpha\right)}\right)=\\=2sin{\left(\frac{\alpha+\frac{\pi}{2}-\alpha}{2}\right)}cos{\left(\frac{\alpha-\left(\frac{\pi}{2}-\alpha\right)}{2}\right)}\cdot2sin{\left(\frac{\alpha-\left(\frac{\pi}{2}-\alpha\right)}{2}\right)}cos{\left(\frac{\alpha+\frac{\pi}{2}-\alpha}{2}\right)}=\\=4sin{\frac{\pi}{4}}cos{\frac{\pi}{4}}sin{\left(\alpha-\frac{\pi}{4}\right)}cos{\left(\alpha-\frac{\pi}{4}\right)}=4\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}\cdot sin{\left(\alpha-\frac{\pi}{4}\right)}cos{\left(\frac{\pi}{4}-\alpha\right)}=\\=2sin{\left(\alpha-\frac{\pi}{4}\right)}.sin{\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\alpha\right)\right)}=2sin{\left(\alpha-\frac{\pi}{4}\right)}.sin{\left(\frac{\pi}{4}+\alpha\right)}=2sin{\left(\alpha+\frac{\pi}{4}\right)}.sin{\left(\alpha-\frac{\pi}{4}\right)}$

[Добромир Глухаров]

$\frac{1+cos{\alpha}}{sin{2\alpha}}\cdot\frac{1+cos{2\alpha}}{cos{\alpha}}=\frac{2cos^2{\frac{\alpha}{2}}}{2sin{\alpha}cos{\alpha}}\cdot\frac{2cos^2{\alpha}}{cos{\alpha}}=\\=\frac{2cos^2{\frac{\alpha}{2}}}{sin{\alpha}}=\frac{2cos^2{\frac{\alpha}{2}}}{2sin{\frac{\alpha}{2}}cos{\frac{\alpha}{2}}}=\frac{cos{\frac{\alpha}{2}}}{sin{\frac{\alpha}{2}}}=cotg{\frac{\alpha}{2}}$

[Добромир Глухаров]

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