Question

Daca b=a×tg pi/7 si a^2+b^2 =1, calculati E=a×sin 5pi/14+b×cos 5pi/14
Ajutor va rog cine stie trigonometrie va roooog

Answer 1

$tg\dfrac{\pi}{7}=\dfrac{b}a\ (1),\;sau\ \dfrac{sin\dfrac{\pi}7}{cos\dfrac{\pi}7}=\dfrac{b}a,\ sau\ \dfrac{cos\dfrac{\pi}7}{sin\dfrac{\pi}7}=\dfrac{a}b\Rightarrow\dfrac{cos^2\dfrac{\pi}7}{sin^2\dfrac{\pi}7}=\dfrac{a^2}{b^2}\\\\\dfrac{cos^2\dfrac{\pi}7+sin^2\dfrac{\pi}7}{sin^2\dfrac{\pi}7}=\dfrac{a^2+b^2}{b^2},\ sau\;\dfrac{1}{sin^2\dfrac{\pi}7}=\dfrac{1}{b^2},\ deci\ sin^2\dfrac{\pi}7=b^2\Rightarrow\\\\\Rightarrow\left(sin\dfrac{\pi}7-b\right)\left(sin\dfrac{\pi}{7}+b\right)=0.\\\\Cazul\ 1:\  sin\dfrac{\pi}7=b\ (2).\\\\Din\ rela\c{t}iile\ (1)\ +\ (2)\ avem\ c\breve{a}:\ cos\dfrac{\pi}7=a.\\\\Avem\;deci\ c\breve{a}\ E=cos\dfrac{\pi}7\cdot sin\dfrac{5\pi}{14}+sin\dfrac{\pi}7\cdot cos\dfrac{5\pi}{14}=sin\left(\dfrac{\pi}7+\dfrac{5\pi}{14}\right)=sin\dfrac{\pi}2=1.$

$Cazul\ 2:\  sin\dfrac{\pi}7=-b\ (3).\\\\Din\ rela\c{t}iile\ (1)\ +\ (3)\ avem\ c\breve{a}:\ cos\dfrac{\pi}7=-a.\\\\Avem\;deci\ c\breve{a}\ E=-cos\dfrac{\pi}7\cdot sin\dfrac{5\pi}{14}-sin\dfrac{\pi}7\cdot cos\dfrac{5\pi}{14}=-sin\left(\dfrac{\pi}7+\dfrac{5\pi}{14}\right)=\\\\=-sin\dfrac{\pi}2=-1.\ Deci\ E=\pm 1.$

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