Question

Срочно, пожалуйста,помогите!((((​прошу!!!!!

Answer 1

$1.\;tg\beta=\frac5{12},\;\beta\in[0;\;\frac\pi2]\\tg\beta=\frac{\sin\beta}{\cos\beta}=\frac{\sin\beta}{\sqrt{1-\sin^2\beta}}\\\\\frac{\sin\beta}{\sqrt{1-\sin^2\beta}}=\frac5{12}\\\\\frac{\sin^2\beta}{1-\sin^2\beta}=\frac{25}{144}\\\\144\sin^2\beta=25-25\sin^2\beta\\169\sin^2\beta=25\\\sin^2\beta=\frac{25}{169}\\\\\sin\beta=\pm\frac5{13}\\\beta\in[0;\;\frac\pi2]\Rightarrow\sin\beta=\frac5{13}$

$\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}|{169}}=\pm\frac{12}{13}\\\beta\in[0;\;\frac\pi2]\Rightarrow\cos\beta=\frac{12}{13}$

$2)\;ctg\alpha=1,\;\alpha\in\left(\frac{3\pi}2;\;2\pi\right)\\ctg\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{\sqrt{1-\sin^2\alpha}}{\sin\alpha}\\\\\frac{\sqrt{1-\sin^2\alpha}}{\sin\alpha}=1\\\\\frac{1-\sin^2\alpha}{\sin^2\alpha}=1\\\sin^2\alpha=1-\sin^2\alpha\\2\sin^2\alpha=1\\\sin^2\alpha=\frac12\\\sin\alpha=\pm\frac1{\sqrt2}\\\[tex]\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac12}=\sqrt{\frac12}=\pm\frac1{\sqrt2}\\\alpha\in\left(\frac{3\pi}2;\;2\pi\right)\Rightarrow\cos\alpha=\frac1{\sqrt2}$[/tex]