Matematică
frfddf
2015-11-07 04:02:39
sa se arate ca 2*(sin 75-sin15)=radical din 2
Răspunsuri la întrebare
neluprodan565
2015-11-07 05:06:41

2(sin75°-sin15°)=√2 2*2 sin 75°-15°/2*cos75°+15°/2 4*sin60°/2*cos90°/2 4*sin30°*cos45°=4*1/2*√2/2 =√2

SuntGoll
2015-11-07 05:07:56

[latex]2cdot (sin75^{circ}-sin15^{circ}) = 2cdot 2cdot sin dfrac{75^{circ}-15^{circ}}{2}cdot cosdfrac{75^{circ}+15^{circ}}{2} = \ \ = 4cdot sin dfrac{60^{circ}}{2}cdotcos dfrac{90^{circ}}{2} = 4cdot sin 30^{circ}cdot cos 45^{circ} = 4cdot dfrac{1}{2}cdot dfrac{sqrt2}{2} = sqrt2 \ \ \ oxed{$M-am folosit de formula: $ $ $ sin a-sin b = 2cdot sin dfrac{a-b}{2}cdot cos dfrac{a+b}{2} }[/latex]

Adăugați un răspuns