Odpowiedź:
Cześć!
Podczas rozwiązywania tych zadań, stale będę korzystać z trzech wzorów :
sin^2\alpha+cos^2\alpha=1sin
2
α+cos
2
α=1
tg\alpha=\frac{sin\alpha}{cos\alpha}tgα=
cosα
sinα
ctg\alpha=\frac{cos\alpha}{sin\alpha}ctgα=
sinα
cosα
a)
cos\alpha=0,3cosα=0,3
sin^2+cos^2\alpha=1sin
2
+cos
2
α=1
sin^2\alpha+(0,3)^2=1sin
2
α+(0,3)
2
=1
sin^2\alpha+0,09=1 \ \ |-0,09sin
2
α+0,09=1 ∣−0,09
sin^2\alpha=\frac{91}{100}sin
2
α=
100
91
sin\alpha=\frac{\sqrt{91}}{10}sinα=
10
91
b)
sin\alpha=0,8sinα=0,8
sin^2\alpha+cos^2\alpha=1sin
2
α+cos
2
α=1
(0,8)^2+cos^2\alpha=1(0,8)
2
+cos
2
α=1
0,64+cos^2\alpha=1 \ \ |-0,640,64+cos
2
α=1 ∣−0,64
cos^2\alpha=0,36cos
2
α=0,36
cos\alpha=0,6cosα=0,6
c)
Tutaj mamy sprawę łatwą, bo wiemy, że cotangens jest odwrotnością tangensa
ctg\alpha=1,2=\frac{12}{10}ctgα=1,2=
10
12
\frac{cos\alpha}{sin\alpha}=\frac{12}{10}
sinα
cosα
=
10
12
tg\alpha=\frac{sin\alpha}{cos\alpha}=\frac{10}{12}tgα=
cosα
sinα
=
12
10
d)
sin\alpha=\frac{1}{4}sinα=
4
1
sin^2\alpha+cos^2\alpha=1sin
2
α+cos
2
α=1
(\frac{1}{4})^2+cos^2\alpha=1(
4
1
)
2
+cos
2
α=1
\frac{1}{16}+cos^2\alpha=1 \ \ |-\frac{1}{16}
16
1
+cos
2
α=1 ∣−
16
1
cos^2\alpha=\frac{15}{16}cos
2
α=
16
15
cos\alpha=\frac{\sqrt{15}}{4}cosα=
4
15
tg\alpha=\frac{1}{4}:\frac{\sqrt{15}}{4}=\frac{1}{4}\cdot\frac{4}{\sqrt{15}}=\frac{1}{\sqrt{15}}=\frac{\sqrt{15}}{15}tgα=
4
1
:
4
15
=
4
1
⋅
15
4
=
15
1
=
15
15
ctg\alpha=\frac{\sqrt{15}}{4}:\frac{1}{4}=\frac{\sqrt{15}}{4}\cdot4=\sqrt{15}ctgα=
4
15
:
4
1
=
4
15
⋅4=
15
e)
cos\alpha=\frac{5}{13}cosα=
13
5
sin^2\alpha+cos^2\alpha=1sin
2
α+cos
2
α=1
sin^2\alpha+(\frac{5}{13})^2=1sin
2
α+(
13
5
)
2
=1
sin^2\alpha+\frac{25}{169}=1 \ \ |-\frac{25}{169}sin
2
α+
169
25
=1 ∣−
169
25
sin^2\alpha=\frac{144}{169}sin
2
α=
169
144
sin\alpha=\frac{12}{13}sinα=
13
12
tg\alpha=\frac{12}{13}:\frac{5}{13}=\frac{12}{13}\cdot\frac{13}{5}=\frac{12}{5}=2,4tgα=
13
12
:
13
5
=
13
12
⋅
5
13
=
5
12
=2,4
ctg\alpha=\frac{5}{13}:\frac{12}{13}=\frac{5}{13}\cdot\frac{13}{12}=\frac{5}{12}ctgα=
13
5
:
13
12
=
13
5
⋅
12
13
=
12
5
Pozdrowienia od zjawy! ;3
