Question

Trigonometrie. Matematica de liceu

Answer 1

$\displaystyle \alpha \in \left(0; \frac{\pi}{2} \right),~\beta=\left( \frac{3\pi}{2} ;2\pi\right),~sin~\alpha= \frac{3}{5} ,~cos~\beta= \frac{12}{13} \\ \\ \\ ctg(\alpha-\beta)= \frac{ctg~\alpha \cdot ctg~\beta+1}{ctg~\beta-ctg~\alpha} \\ \\ \\ sin^2\alpha+cos^2\alpha=1\Rightarrow cos^2\alpha=1-sin^2\alpha\Rightarrow cos~\alpha= \pm\sqrt{1-sin^2\alpha} \\ \\ \\ \alpha~este~\^in~cadranul~I,~cos~\alpha\ \textgreater \ 0\Rightarrow cos~\alpha= \sqrt{1-sin^2\alpha}}$

$\displaystyle cos~\alpha= \sqrt{1-\left( \frac{3}{5} \right)^2}= \sqrt{1- \frac{9}{25} } = \sqrt{ \frac{25-9}{25} } = \sqrt{ \frac{16}{25} }= \frac{ \sqrt{16} }{ \sqrt{25} } = \frac{4}{5} \\ \\ \\ cos~\alpha= \frac{4}{5} \\ \\ \\ ctg~\alpha= \frac{cos~\alpha}{sin~\alpha}= \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{\not5} \cdot \frac{\not5}{3} = \frac{4}{3}\\ \\ \\ctg~\alpha=\frac{4}{3}$

$\displaystyle sin^2\beta+cos^2\beta=1\Rightarrow sin^2\beta=1-cos^2\beta\Rightarrow sin~\beta=\pm \sqrt{1-cos^2\beta} \\ \\ \\ \beta ~este~\^in~cadranul~IV,~sin~\beta\ \textless \ 0\Rightarrow sin~\beta=- \sqrt{1-cos^2\beta}} \\ \\ \\ sin~\beta=- \sqrt{1-\left( \frac{12}{13} \right)^2}=- \sqrt{1- \frac{144}{169} } =- \sqrt{ \frac{169-144}{169} } =- \sqrt{ \frac{25}{169} } =\\ \\ \\ =- \frac{ \sqrt{25} }{ \sqrt{169} }=- \frac{5}{13} \\ \\ \\ sin~\beta=- \frac{5}{13}$

$\displaystyle ctg~\beta=\frac{cos~\beta}{sin~\beta}=\frac{\frac{12}{13}}{- \frac{5}{13}}=\frac{12}{13}\cdot \left(-\frac{13}{5}\right)=-\frac{12}{5}\\ \\ \\ctg~\beta=-\frac{12}{5}$

$\displaystyle ctg(\alpha-\beta)=\frac{ctg~\alpha\cdot ctg~\beta+1}{ctg~\beta-ctg~\alpha}= \frac{\frac{4}{3}\cdot\left(-\frac{12}{5}\right)+1}{-\frac{12}{5}-\frac{4}{3}}=\frac{ -\frac{48}{15}+1}{\frac{-36-20}{15}}=\\\\\\=\frac{\frac{-48+15}{15}}{-\frac{56}{15} }=\frac{-\frac{33}{15}}{-\frac{56}{15}}=-\frac{33}{15}\cdot\left(-\frac{15}{56}\right)=\frac{33}{56}$

$\displaystyle ctg(\alpha-\beta)=\mathbf{\frac{33}{56}}$