Question

Помогите сделать срочно по математике

Answer 1

$1.\;\sin^2\frac\pi4-2\cos\frac\pi3+3tg0^o=\left(\sin\frac\pi4\right)^2-2\cdot\frac12+3\cdot0=\left(\frac1{\sqrt2}\right)^2-1=\frac12-1=-\frac12\\\\2.\;\sin\alpha=\frac5{13}\\\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\pm\frac{12}{13}\\\frac\pi2<\alpha<\pi\Rightarrow\cos\alpha<0\\\cos\alpha=-\frac{12}{13}\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac5{13}:\frac{12}{13}=\frac5{13}\cdot\frac{13}{12}=\frac5{12}\\ctg\alpha=\frac1{tg\alpha}=\frac{12}5=2\frac25=2,4$

$3.\;\cos^2x-\cos^4x+\sin^4x=\cos^2x(1-\cos^2x)+\sin^4x=\cos^2x\sin^2x+\sin^4x=\\=\sin^2x(\cos^2x+\sin^2x)=\sin^2x\\\\4.\;\sin^4\alpha-\cos^4\alpha+\cos^2\alpha=\sin^4\alpha+\cos^2\alpha(1-\cos^2\alpha)=\sin^4\alpha+\cos^2\alpha\sin^2\alpha=\\=\sin^2\alpha(\sin^2\alpha+\cos^2\alpha)=\sin^2\alpha$