Question

la exercitiul 9 please?

Answer 1

$\sin {(90+x)} = \sin 90\cos x + \sin x \cos 90 = \cos{x}\\\\ \sin{(90 - x)} = \sin 90\cos x - \cos 90\sin x = \cos x\\\\ \sin{(90+x)} = \sin{(90-x)}\\\\\implies \sin{20} = \sin{(90 - 70)} = \sin{(90+70)} = \sin{(160)}\\\\\sin{50} = \sin{(90 - 40)} = \sin{(90+40)} = \sin{130}\\\\ \sin{80} = \sin{(90-10)} = \sin{(90+10)} = \sin{100}\\\\\implies \boxed{\sin 20 + \sin 50 + \sin 80 = \sin 160 + \sin 130 + \sin 100 = \sin 100 + \sin 130 + \sin 160}$

Answer 2

Răspuns:

Explicație pas cu pas:

sin a = sin (180° - a)

sin 100 = sin (180 -100) = sin 80

sin 130 = sin(180-130) = sin 50

sin 160 = sin(180-160) = sin 20 si adunandu-le ne da EXACT identitatea care era de demostrat in enunt.