1.Wiedzac, ze dla pewnego kata ostrego alfa, sin alfa = 1/3. Obliccz cos alfa, tgalfa i ctg alfa. 2.Wykaz, ze dla kata ostrego alfa rownosc (1+ tgalfa)(1-ctgalfa)=tgalfa - ctgalfa jest tozsamoscia. 3. Z rownosci t= pierwiastek z 2s/g wyznacz s i g 4. z rownosci a-x/b=1/c wyznacz c,x i a
1) sin α=1/3
cosinus liczymy ze wzoru na "jedynke' trygonometryczna
sin²α+cos²α=1
1/9+cos²α=1
cos²α=1-1/9
cos²α=8/9
cosα=√8/9
cosα=(2√2)/3
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tgα=sinα/cosα
tgα=1/3:(2√2)/3
tgα=1/3·3/(2√2)
tgα=1/(2√2)
tgα=√2/4
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ctgα=1/tgα
ctgα=1:√2/4
ctgα=2√2
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2) (1+tg α)(1-ctg α)=tg α-ctg α
1-ctg α+tg α-tgα·ctg α=tg α- ctg α
1-ctg α+tg α-1=tg α-ctg α
tg α-ctg α=tg α-ctg α
L=P
wykorzystano wzor: tg α·ctg α=1
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3) t=√2s/g podnosimy obustronnie do kwadratu
t²=2s/g /·g
gt²=2s
2s=gt² /:2
s=gt²/2
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t=√2s/g /²
t²=2s/g /·g
gt²=2s /:t²
g=2s/t² t≠0
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4) a-x 1
------------=-----------------
b c
wykorzystujemy wlasnosc proporcji
c(a-x)=b /:(a-x)
b
c=------------
a-x
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c(a-x)=b
ac-cx=b
ac=b+cx /:c c≠o
a=b/c +x
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c(a-x)=b
ac-cx=b
ac-b=cx
cx=ac-b /:c c≠0
x=a- b/c
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sprawdz, czy zadanie 4 dobrze przepisalam
jak cos, to pisz na priv
1.
sinα = 1/3
sin²α + cos²α = 1
cos²α = 1 - sin²α = 1 - (1/3)² = 1-1/9 = 9/9 - 1/9 = 8/9
cosα = √(8/9) = 2√2/3
tgα = sinα/cosα = (1/3):(2√2/3) = 1/3 * 3/2√2 = 1/2√2 * √2/√2 = √2/4
tgα * ctgα = 1
ctgα = 1/tgα = 4/√2 * √2/√2 = 4√2/2 = 2√2
2.
(1+tgα)(1-ctgα) = tgα - ctgα
L = (1+tgα)(1-ctgα) = 1-ctgα+tgα-tgα*ctgα = 1-ctgα+tgα-1 = tgα-ctgα
P = tgα-ctgα
L = P
3.
t = √(2s/g) I² (obie strony równania podnosimy do kwadratu)
t² = 2s/g I*g
2s = gt² /:2
s = gt²/2
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2s = gt² /:t²
g = 2s/t²
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4.
(a-x)/2 = 1/c
c(a-x) = b /:(a-x)
c = b/(a-x)
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c(a-x) = b
ac-cx = b
c(a-x) = b /:(a-x)
c = b/(a-x)
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c(a-x) = b
ac-cx = b
cx = ac-b /:c
x = (ac-b)/c
x = a-b/c
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