Question

tgA =4
Найдите. ctgA, sinA, cosA?

Answer 1

$tga=4\\\\ctga=\frac{1}{tga}=\frac{1}{4}\\\\1+tg^2a=\frac{1}{cos^2a}\; \; \to \; \; \; 1+16=\frac{1}{cos^2a}\; \; ,\; \; \frac{1}{cos^2a}=17\; \; \to \; \; cos^2a=\frac{1}{17}\; ,\\\\cosa=\pm \frac{1}{\sqrt{17}}=\pm \frac{\sqrt{17}}{17}\\\\sina=\pm \sqrt{1-cos^2a}=\pm \sqrt{1-\frac{1}{17}}=\pm \sqrt{\frac{16}{17}}=\pm \frac{4}{\sqrt{17}}=\pm \frac{4\sqrt{17}}{17}$