№5 Вычислить a) 005-81 2) 87472-00812-87112.CDS42 COS-SU COS 18. CDS12-85418-81412 6) ty+tg g) 8434-89453 1-200541 1-tyty 8) cos²36 e) VS CDS 105-STU 105 +28TỬ 15 1+ Cos42

a) cos(\(\frac{\pi}{3}\)) \(\cdot\) sin(\(\frac{\pi}{2}\)) / (cos²(\(\frac{\pi}{3}\)) - sin²(\(\frac{\pi}{3}\))) = \(\frac{\frac{1}{2} \cdot 1}{(\frac{1}{2})^2-(\frac{\sqrt{3}}{2})^2}\) = \(\frac{\frac{1}{2}}{\frac{1}{4}-\frac{3}{4}}\) = \(\frac{\frac{1}{2}}{-\frac{2}{4}}\) = \(\frac{\frac{1}{2}}{-\frac{1}{2}}\) = -1

б) sin72 \(\cdot\) cos12 - sin12 \(\cdot\) cos42 / (cos18 \(\cdot\) cos12 - sin18 \(\cdot\) sin12) = \(\frac{sin72 \cdot cos12 - sin12 \cdot cos42}{cos18 \cdot cos12 - sin18 \cdot sin12}\) = \(\frac{sin(72-12)}{cos(18+12)}\) = \(\frac{sin(60)}{cos(30)}\) = \(\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\) = 1

в) (tg(\(\frac{\pi}{6}\)) + tg(\(\frac{\pi}{12}\))) / (1 - tg(\(\frac{\pi}{6}\)) \(\cdot\) tg(\(\frac{\pi}{12}\))) = tg(\(\frac{\pi}{6}\) + \(\frac{\pi}{12}\)) = tg(\(\frac{3\pi}{12}\)) = tg(\(\frac{\pi}{4}\)) = 1

г) (sin34 - sin53) / (1 - 2cos²41) = (sin34 - sin53) / (-cos82) = (sin34 - sin53) / (-cos(90-8)) = (sin34 - sin53) / (-sin8)

д) -2\(\sqrt{3}\)cos105 \(\cdot\) sin105 + 28sin²15 = -\(\sqrt{3}\)sin210 + 28sin²15 = -\(\sqrt{3}\)(-\(\frac{1}{2}\)) + 28sin²15 = \(\frac{\sqrt{3}}{2}\) + 28sin²15 = \(\frac{\sqrt{3}}{2}\) + 1

e) cos²36 / (1 + cos42) = \(\frac{cos^2 36}{1 + cos 42}\)

Ответ: а) -1; б) 1; в) 1; г) ; д) ; e)