Question

вычислите: tg(a+ pi/4), если cos2a=1/3 и pi

Answer 1

Ответ:

$tg\Big(a+\dfrac{\pi}{4}\Big)=?\ \ ,\ \ cos2a=\dfrac{1}{3}\ \ ,\ \ \pi <a<\dfrac{5\pi}{4}\\\\\\cos2a=cos^2a-sin^2a=\dfrac{cos^2a-sin^2a}{cos^2a+sin^2a}=\dfrac{cos^2a\cdot \Big(1-\dfrac{sin^2a}{cos^2a}\Big)}{cos^2a\cdot \Big(1+\dfrac{sin^2a}{cos^2a}\Big)}=\dfrac{1-tg^2a}{1+tg^2a}\\\\\\\dfrac{1-tg^2a}{1+tg^2a}=\dfrac{1}{3}\ \ \to \ \ \ 3-3tg^2a=1+tg^2a\ \ .\ \ 4tg^2a=2\ \ ,\ \ tg^2a=\dfrac{1}{2}\\\\tga=\pm \dfrac{1}{\sqrt2}=\pm \dfrac{\sqrt2}{2}$

$\pi <a<\dfrac{5\pi}{4}\ \ \Rightarrow \ \ \ tga=\dfrac{\sqrt2}{2}\\\\\\\\tg\Big(a+\dfrac{\pi}{4}\Big)=\dfrac{tga+tg\dfrac{\pi }{4}}{1-tga\cdot tg\dfrac{\pi }{4}}=\dfrac{\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}}{1-\dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt2}{2}}=\dfrac{\sqrt2}{1-\dfrac{1}{2}}=\dfrac{2\sqrt2}{2-1}=\boxed{2\sqrt2\, }$