Question

TRIGONOMETRIE
Demonstrați egalitatile ​

Answer 1

Răspuns:

Ex. 9)

a) $tg 55-tg35=2tg20$

$\frac{sin55}{cos55} -\frac{sin35}{cos35}=\frac{sin55*cos35-sin35*cos55}{cos55*cos35} <=>$

Voi rezolva separat sin55*cos35-sin35*cos55 si cos55*cos35 si apoi voi inlocui in ecuatie. Fac asta ptr. ca nu se vede absolut nimic daca pun fractie dupa fractie dupa fractie...

$sin55*cos35=\frac{sin(55+35)+sin(55-35)}{2} =\frac{sin(90)+sin(20)}{2} =\frac{1+sin(20)}{2}$

$sin35*cos55=\frac{sin(35+55)+sin(35-55)}{2} = \frac{sin(90)+sin(-20)}{2}=\frac{1-sin(20)}{2}$

$sin55*cos35-sin35*cos55=\frac{1+sin(20)-(1+sin(20))}{2}=\frac{2sin(20)}{2}$

$cos55*cos35=\frac{cos(55+35)+cos(55-35)}{2}=\frac{cos(90)+cos(20)}{2}=\frac{0+cos(20)}{2}$

Inlocuim:

$\frac{\frac{2sin20}{2} }{\frac{cos20}{2} } = 2*\frac{sin20}{cos20}=2tg20$

b)

$tg9-tg27-tg63+tg81=4$

$tg9+tg81-(tg27+tg63)=tg9+tg(90-9)-(tg27+tg(90-27))\\=tg9+ctg9-(tg27+ctg) = \frac{sin9}{cos9}+\frac{cos9}{sin9}-(\frac{sin27}{cos27}+\frac{sin0}{cos0})$

$=\frac{sin^29+cos^29}{sin9*cos9}-(\frac{sin^227+cos^227}{sin27*cos27}) = \frac{1}{\frac{sin(9+9)+sin(9-9)}{2} } -\frac{1}{\frac{sin(27+27)+sin(27-27)}{2} }$

$=\frac{2}{sin(18)}-\frac{2}{sin(54)}=\frac{2sin(54)-2sin(18)}{sin(18)*sin(54)}=\frac{2*2*(sin18*cos54)}{sin18*cos54} =4$