Question

transforma in produs tg 15° + tg 75°

Answer 1

$tg \ 15^0= \frac{sin \ 15^0}{cos \ 15^0} \\\\ tg \ 75^0= \frac{sin \ 75^0}{cos \ 75^0} \\\\\\ tg \ 15^0+ tg \ 75^0= \frac{sin \ 15^0}{cos \ 15^0}^{(cos \ 75^0}+ \frac{sin \ 75^0}{cos \ 75^0} ^{(cos \ 15^0} \\\\ \longrightarrow \frac{sin \ 15^0 * cos \ 75^0+ sin \ 75^0 * cos \ 15^0}{cos \ 15^0 * cos \ 75^0} \\\\\\ sin \ 15^0 = sin(45^0-30^0)=sin \ 45^0* cos \ 30^0 - sin \ 30^0* cos \ 45^0 \\\\ sin \ 15^0=\frac{\sqrt{2}}{2}* \frac{\sqrt{3}}{2} - \frac{1}{2}* \frac{\sqrt{2}}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}$


$cos \ 15^0=cos(45^0-30^0)=cos \ 45^0 * cos \ 30^0 +sin \ 45^0* sin \ 30^0 \\\\ cos \ 15^0= \frac{\sqrt{2}}{2}* \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}* \frac{1}{2} \\\\ cos \ 15^0 = \frac{\sqrt{6}+\sqrt{2}}{4} \\\\\\ sin \ 75^0=sin(90^0-15^0)= sin \ 90^0 * cos \ 15^0- sin \ 15^0*cos \ 90^0 \\\\ sin \ 75^0=1* \frac{\sqrt{6}+\sqrt{2}}{4}- \frac{\sqrt{6}-\sqrt{2}}{4}*0 \\\\ sin \ 75^0=\frac{\sqrt{6}+\sqrt{2}}{4} \\\\\\ \Longrightarrow cos \ 75^0=sin \ 15^0= \frac{\sqrt{6}-\sqrt{2}}{4}$


$tg \ 15^0+ tg \ 75^0= \frac{\frac{\sqrt{6}-\sqrt{2}}{4}*\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}*\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}*\frac{\sqrt{6}-\sqrt{2}}{4}}= \\\\\\ = \frac{\frac{(\sqrt{6}-\sqrt{2})^2}{16}+\frac{(\sqrt{6}+\sqrt{2})^2}{16}}{ \frac{6-2}{16}}= \frac{\frac{6-2\sqrt{12}+2+6+2\sqrt{12}+2}{16}}{\frac{4}{16}}= \\\\ = \frac{12+4}{\not16}* \frac{\not16}{4}= \frac{\not16}{\not4} \\\\\\\\ \boxed{tg \ 15^0+ tg \ 75^0=4}$

Answer 2

$\displaystyle Avem: \\ \\ tg15 \textdegree+tg75 \textdegree = \frac{sin(15 \textdegree+ 75 \textdegree)}{cos 15 \textdegree \cdot cos 75 \textdegree }= \frac{sin90 \textdegree}{sin15 \textdegree \cdot cos 15 \textdegree}= \\ \\ = \frac{1}{sin 15 \textdegree \cdot cos 15 \textdegree }= \frac{2}{2 sin 15 \textdegree \cdot cos 15 \textdegree}= \frac{2}{sin30 \textdegree}= \frac{2}{ \frac{1}{2}}=4.$

$\displaystyle \underline{Formule~folosite}: \\ \\ \blacksquare ~tgA+tgB= \frac{sin(A+B)}{cosA \cdot cosB} \\ \\ \blacksquare~cosA=sin(90 \textdegree - A) \\ \\ \blacksquare ~sin2x=2sinx \cdot cosx$