bardzo proszę o pomoc i wyrażne, zrozumiałe rozwiazania
6.
tgα/cosα=sinα/cos^2α
cos^2α=√(1-sin^2α)
sin^2α= 4*5/25=4/5
cos^2α =1-4/5=1/5
(2√5/5)/(1/5)=2√5
albo
sinα=b/c=2√5/5
b=2√5
c=5
a=√(c^2-b^2)=√(25-20)
a=√5
tgα=b/a= 2√5/√5=2
cosα=a/c=√5/5
tgα/cosα=2*5/√5=2√5
7.
cos α<0,5
α>60
dla 60<α<90
0<cosα<0,5
8.
cosα=3/5=b/c
a=√(5^2-3^2)=√16=4
sinα*tgα= 4/5*4/3=16/15
9.
tgα=2=2/1=a/b
c=√(1+2^2)=√5
a=2
b=1
(cosα+sinα)/cosα=(a/c+b/c)/(a/c)
(cosα+sinα)/cosα=2/√5+1√5/(1/√5)=3
(cosα+sinα)/cosα=1+tgα=1+2=3
10.
tgα=0,4=4/10
tgβ=1/tgα= 10/4=2,5
11.
wspólny mianownik
sin^2α+1+2cosα+cos^2α/((1+cosα)sinα)=
=2(1+cosα)/((1+cosα)sinα)=2/sinα
12.
cos(37)*(1+tan(37)^2)^0,5=1
cos(37)*√(1+sin^2(37)/cos^2(37))
√(1+sin^2(37)/cos^2(37))=√1/cos^2(37)=1/cos(37)
cos(37)*1/cos(37)=1
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
6.
tgα/cosα=sinα/cos^2α
cos^2α=√(1-sin^2α)
sin^2α= 4*5/25=4/5
cos^2α =1-4/5=1/5
(2√5/5)/(1/5)=2√5
albo
sinα=b/c=2√5/5
b=2√5
c=5
a=√(c^2-b^2)=√(25-20)
a=√5
tgα=b/a= 2√5/√5=2
cosα=a/c=√5/5
tgα/cosα=2*5/√5=2√5
7.
cos α<0,5
α>60
dla 60<α<90
0<cosα<0,5
8.
cosα=3/5=b/c
a=√(5^2-3^2)=√16=4
sinα*tgα= 4/5*4/3=16/15
9.
tgα=2=2/1=a/b
c=√(1+2^2)=√5
a=2
b=1
(cosα+sinα)/cosα=(a/c+b/c)/(a/c)
(cosα+sinα)/cosα=2/√5+1√5/(1/√5)=3
albo
(cosα+sinα)/cosα=1+tgα=1+2=3
10.
tgα=0,4=4/10
tgβ=1/tgα= 10/4=2,5
11.
wspólny mianownik
sin^2α+1+2cosα+cos^2α/((1+cosα)sinα)=
=2(1+cosα)/((1+cosα)sinα)=2/sinα
12.
cos(37)*(1+tan(37)^2)^0,5=1
cos(37)*√(1+sin^2(37)/cos^2(37))
√(1+sin^2(37)/cos^2(37))=√1/cos^2(37)=1/cos(37)
cos(37)*1/cos(37)=1