devb99
devb99
29-03-2018
Mathematics
contestada
can someone help me out with number 2?
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Banabanana
Banabanana
29-03-2018
[tex]\sin255^o=\sin(300^o-45^o)=\sin300^o\cos45^o-\cos300^o\sin45^o= \\\\=- \frac{ \sqrt{3} }{2}\cdot \frac{ \sqrt{2} }{2}- \frac{1}{2} \cdot \frac{ \sqrt{2} }{2}=- \frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4}= -\frac{ \sqrt{6}+\sqrt{2} }{4}[/tex]
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