Question

Ma ajuta cineva cu rezolvarea? Va rog frumos

Answer 1

 

$a)\\\sin^2130^o+\cos^250^o=\\=\sin^2(180-50^o)+\cos^250^o=\\=\sin^250^o+\cos^250^o=1\\\\b)\\\sin^250^o+\sin^240^o=\\=\sin^250^o+\sin^2(90^o-50^o)=\\=\sin^250^o+\cos^250^o=1\\\\c)\\\sin^260^o-\cos^230^o=\\\sin^260^o-\cos^2(90^o-60^o)=\\\sin^260^o-\sin^260^o=0\\\\d)\displaystyle\\\cos^230^o-\sin^245^o-\cos^260^o=\\\\=\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{2}}{2}\right)^2-\left(\frac{1}{2}\right)^2=\\\\\\=\frac{3}{4}-\frac{2}{4}-\frac{1}{4}=\frac{3-2-1}{4}=0$

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La punctul e) este o greseala de transcriere.

Este minus intre sin si cos NU plus.

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$e)\\\sin\,75^o-\cos\,15^o=\\=\sin(90^o-15^o)-\cos\,15^o=\\=\cos\,15^o-\cos\,15^o=0\\\\f)\\\cos\,20^o+\cos\,50^o+\cos\,160^o+\cos\,130^o=~~~(\bf~rearanjem)\\=\cos\,20^o+\cos\,160^o+\cos\,50^o+\cos\,130^o=\\=\cos\,20^o+\cos(180^o-20^o)+\cos\,50^o+\cos(180^o-50^o)=\\=\cos\,20^o-\cos(20^o)+\cos\,50^o-\cos(50^o)=0+0=0$