Question

Вычислить Tg$\alpha/2$ если sin$\alpha$ + cos$\alpha$=$\sqrt{7}/2$

Answer 1

sinx=sinx=2tg(x/2)/(1+tg^2(x/2))

cosx=(1-tg^2(x/2))/(1+tg^2(x/2))

sinx+cosx=sinx=2tg(x/2)/(1+tg^2(x/2))+(1-tg^2(x/2))/(1+tg^2(x/2))=√7/2

обозначу tg(x/2)=y

2y/(1+y^2)+(1-y^2)/(1+y^2)=(2y+1-y^2)/(1+y^2)=√7/2

-y^2+2y+1=√7/2+(√7/2)y^2

y^2(√7/2+1)-2y-1+√7/2=0

D=4-4(√7/2-1)(√7/2+1)=4-4(7/4-1)=4-3=1

y1=tg(x/2)=(2+1)/(√7+2)=3/(√7+2)

y1=tg(x/2)=(211)/(√7+2)=1/(√7+2)

Ответ tg(x/2)={3/(√7+2);1/(√7+2)}

Answer 2

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