Question

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Answer 1

Tgx=-4/3;3π/2<x<2π

tgx/2;cosx/2;sinx/2;ctgx/2=??

1)1+tg²x=1/cos²x

cos²x=1/(1+tg²x)=1/(1+16/9)=9/25

cosx=+√9/25=3/5

2)3π/2<x<2π

3π/4<x/2<π

3)sin²x/2=(1-cosx)/2=(1-3/5)/2=2/5*1/2=1/5

sinx/2=+√(1/5)=√5/5

4)cos²x/2=(1+cosx)/2=(1+3/5)/2=8/5*1/2=4/5

cosx/2=-√(4/5)=-2√5/5

5)tg²x/2=(1-cosx)/(1+cosx)=(1-3/5)/(1+3/5)=

2/5/8/5=2/5*5/8=1/4

tgx/2=-√(1/4)=-1/2

6)ctg²x/2=(1+cosx)/(1-cosx)=

(1+3/5)/(1-3/5)=8/5*5/2=4

ctgx/2=-√4=-2