Question

Zadanie w załączniku. :)

Dam naj.:)

Answer 1

Wzory:

sin²α+cos²α=1

tgα=sinα/cosα

ctgα=1/tgα

 

a)sinα=1/3

1/9+cos²α=1

cos²α=8/9

cosα=√(8/9)

cosα=(2√2)/3

tgα=1/3:(2√2)/3=1/3*3/(2√2)=1/(2√2)     =2√2/8=√2/4

ctg=1: 1/(2√2)=2√2

 

b)cosα=2/5

4/25+sin²α=1

sin²α=21/25

sinα=√21/5

tgα=(2/5):(√21/5)= (2/5)*(5/√21)=2/√21   =2√21/21

ctgα=1: 2/√21=√21/2

 

c)sin=√2/3

2/9+cos²α=1

cos²α=7/9

cosα=√7/3

tgα=(√2/3) : (√7/3)=(√2/3) * (3/√7)=√2/√7     =√14/7

ctgα=1/ (√2/√7)=√7/√2=√14/2

 

d)cosα=√3/2

3/4+sin²α=1

sin²α=1/4

sinα=1/2

ctgα=(√3/2) : (1/2)=(√3/2) * (2/1)=√3

tgα=1/√3=√3/3