Question

demonstrati ca sin60*= 2sin30* ori cos30* ori tg30* ori ctg 30*.

Answer 1

sin 60=√3/2
sin 30=1/2
cos 30=√3/2
tg30=√3/3
ctg 30=√3
deci √3/2=1•√3/2•√3/3•√3
√3/2=3•√3/6 <=> √3/2=√3/2 Adevarat

Answer 2

Avem formulele: tgx*ctgx=1, sin2x=2sinx*cosx. Deci: 2sin30°*cos30°*tg30°*ctg30°=(2sin30°*cos30°)*(tg30°*ctg30°)=sin2*30°*1=sin60°. Sau inlocuind valorile functiilor trigonometrice obtiinem: $\frac{ \sqrt{3} }{2}=2* \frac{1}{2}* \frac{ \sqrt{3} }{2}* \sqrt{3}* \frac{ \sqrt{3} }{3}$ simplificand se constata afirmatia adevarata.