Odpowiedź:
a ) = sin ( 90° + 30°)* cos (90° + 60°) + tg ( 90° + 45° ) =
= cos 30°*( - sin 60° ) - ctg 45°° = [tex]- \frac{\sqrt{3} }{2} * \frac{\sqrt{3} }{2} - 1 = - \frac{3}{4} - 1 = - 1\frac{3}{4}[/tex]
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b) =( ( sin ( 90° + 45°) - tg ( 90° + 30°))*( cos ( 90° + 45°) - ctg ( 90° + 60°)) =
= ( cos 45° + ctg 30°)*( - sin 45° + tg 60° ) =
= ( [tex]\frac{\sqrt{2} }{2} + \sqrt{3} )*(- \frac{\sqrt{2} }{2}+ \sqrt{3} ) = (\sqrt{3} )^2 - (0,5\sqrt{2} )^2 =[/tex] 3 - 0,5 = 2,5
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Odpowiedź:
a ) = sin ( 90° + 30°)* cos (90° + 60°) + tg ( 90° + 45° ) =
= cos 30°*( - sin 60° ) - ctg 45°° = [tex]- \frac{\sqrt{3} }{2} * \frac{\sqrt{3} }{2} - 1 = - \frac{3}{4} - 1 = - 1\frac{3}{4}[/tex]
--------------------------------------------------------------------------------------------
b) =( ( sin ( 90° + 45°) - tg ( 90° + 30°))*( cos ( 90° + 45°) - ctg ( 90° + 60°)) =
= ( cos 45° + ctg 30°)*( - sin 45° + tg 60° ) =
= ( [tex]\frac{\sqrt{2} }{2} + \sqrt{3} )*(- \frac{\sqrt{2} }{2}+ \sqrt{3} ) = (\sqrt{3} )^2 - (0,5\sqrt{2} )^2 =[/tex] 3 - 0,5 = 2,5
===============================================================
Szczegółowe wyjaśnienie: