Question

система tgx+tgy=1
x+y=п/4​

Answer 1

Ответ:

(см. объяснение)

Объяснение:

$tg(x)+tg(y)=1\x+y=dfrac{pi}{4}\\dfrac{sin(x+y)}{cos(x)cos(y)}=1\x+y=dfrac{pi}{4}\\dfrac{dfrac{sqrt{2}}{2}}{cos(x)cos(y)}=1\x+y=dfrac{pi}{4}\\\dfrac{cos(x-y)}{2}+dfrac{cos(x+y)}{2}=dfrac{sqrt{2}}{2}\x+y=dfrac{pi}{4}\\\dfrac{cos(x-y)}{2}+dfrac{cos(dfrac{pi}{4})}{2}=dfrac{sqrt{2}}{2}\x+y=dfrac{pi}{4}\\\dfrac{cos(x-y)}{2}=dfrac{sqrt{2}}{4}\x+y=dfrac{pi}{4}\\\cos(x-y)=dfrac{sqrt{2}}{2}\x+y=dfrac{pi}{4}$

Найдем x-y:

$left[begin{array}{c}x-y=dfrac{pi}{4}+2kpi,; kin Z\x-y=dfrac{7pi}{4}+2kpi,; kin Zend{array}right \x+y=dfrac{pi}{4}\\1)\x-y=dfrac{pi}{4}+2kpi,; kin Z\x+y=dfrac{pi}{4}\\\2x=dfrac{pi}{2}+2kpi,; kin Z\x=dfrac{pi}{4}+kpi,; kin Z\y=-kpi,; kin Z\\2)\x-y=dfrac{7pi}{4}+2kpi,; kin Z\x+y=dfrac{pi}{4}\\\x=pi+kpi,; kin Z\y=-dfrac{3pi}{4}-kpi, kin Z$