Question

ПОМОГИТЕ, ПОЖАЛУЙСТА!

БУДУ ОЧЕНЬ БЛАГОДАРНА.

Answer 1

1) 1 + sin²12° + cos²12° = 1 + 1 = 2

2)2sin 12° · cos12° = sin24°

3) (tg12° - tg10°)/(1 + tg12° · tg10° = tg(12° - 10°) = tg2°

4) cos50° + cos40° = 2 · cos(50°+40°)/2) · cos(50° - 40°)/2 =

= 2 cos45° · cos5° = 2 · 0.5√2 · cos5° = √2 · cos5°

Answer 2

$1)\; \; 1+\underbrace {sin^212^\circ +cos12^\circ }_{1}=1+1=2\\\\2)\; \; 2\, sin12^\circ \cdot cos12^\circ =sin24^\circ \\\\3)\; \; \frac{tg12^\circ -tg10^\circ }{1+tg12^\circ \cdot tg10^\circ }=tg(12^\circ -10^\circ )=tg2^\circ \\\\4)\; \; cos50^\circ +cos40^\circ =2\, cos\frac{50^\circ +40^\circ }{2}\cdot cos\frac{50^\circ -40^\circ }{2}=2\, cos45^\circ \cdot cos5^\circ =\sqrt2\cdot cos5^\circ$