Question

Na jutro.! . ;/

Przedstaw w prostszej postaci:

a)(sin α + cos α)² - 2sin α cos α

b)(tg α + ctg α)² - (th α - ctg α)²

c)(cos α tg α)² - (sin α ctg α)²

d) sin α cos α (tg α + ctg α)

Answer 1

a) (sinL+cosL)^{2}-2sinLcosL=

sin^{2}L+2sinLcosL+cos^{2}L-2sinLcosL=

sin^{2}L+cos^{2}L=1

 

b) (tgL+ctgL)^{2}-(tgL-ctgL)^{2}=

tg^{2}L+2tgLctgL+ctg^{2}L-(tg^{2}-2tgLctgL+ctg^{2}L)=

tg^{2}L+2tgLctgL+ctg^{2}L-tg^{2}L+2tgLctgL-ctg^{2}L=

2tgLctgL+2tgLctgL=4tgLctgL

 

c) (cosLtgL)^{2}-(sinLctgL)^{2}

=(cosL\cdot\frac{sinL}{cosL})^{2}-(sinL\cdot\frac{cosL}{sinL})^{2}=

sin^{2}L-cos^{2}L

 

d)sinLcosL(tgL+ctgL)=sinLcosL(\frac{sinL}{cosL}\cdot\frac{cosL}{sinL})=sinLcosL

 

W razie jakichkolwiek pytań śmiało pisz to odpowiem:)