Odpowiedź:
a)
6 cos²α = 5 sin α oraz sin²α + cos²α = 1 ⇒ cos²α = 1 - sin²α
więc mamy
6*(1 - sin²α ) = 5 sin α
6 - 6 sin² α = 5 sin α
6 sin² α + 5 sin α - 6 = 0
sin α = t
6 t² + 5 t - 6 = 0 Δ = 5² - 4*6*(-6) = 25 + 144 = 169 √Δ = 13
t = [tex]\frac{- 5 - 13}{2*6} < 0 - odpada[/tex] lub t = [tex]\frac{- 5 + 13}{12} = \frac{8}{12} = \frac{2}{3}[/tex]
Odp. sin α = [tex]\frac{2}{3}[/tex] α < 90°
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b) 4 sin²α = 2 + 7 cos α
4*( 1 - cos²α ) = 2 + 7 cos α
4 - 4 cos²α = 2 + 7 cos α
4 cos²α + 7 cos α - 2 = 0
cos α = t
4 t² + 7 t - 2 = 0
Δ = 49 - 4*4*(-2) = 49 + 32 = 81 √Δ = 9
t = [tex]= \frac{- 7 - 9}{2*4} = - 2 - odpada[/tex] lub t = [tex]\frac{- 7 + 9}{8} = \frac{1}{4}[/tex]
więc
cos α = [tex]\frac{1}{4}[/tex]
sin²α = 1 - cos²α = 1 - [tex]\frac{1}{16} = \frac{15}{16}[/tex]
Odp. sin α = [tex]\sqrt{\frac{15}{16} } = \frac{\sqrt{15} }{4}[/tex]
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c ) 2 sin α = 3 ctg α
2 sin α = 3*[tex]\frac{cos \alpha }{sin \alpha }[/tex] / * sin α
2 sin²α = 3 cos α
2*(1 - cos²α ) = 3 cos α
2 - 2 cos²α = 3 cos α
2 cos²α + 3 cos α - 2 = 0
t = cos α
2 t² + 3 t - 2 = 0
Δ = 9 - 4*2*(- 2) = 9 + 16 = 25 √Δ = 5
t = [tex]\frac{- 3 - 5}{2*2} = - 2 - odpada[/tex] lub t = [tex]\frac{- 3 + 5}{4} = \frac{2}{4} = \frac{1}{2}[/tex]
cos α = [tex]\frac{1}{2}[/tex]
sin²α = 1 - cos²α = 1 - [tex]\frac{1}{4} = \frac{3}{4}[/tex]
sin α = [tex]\sqrt{\frac{3}{4} } = \frac{\sqrt{3} }{2}[/tex] [tex]\alpha = 60^o[/tex]
d )
[tex]sin^4\alpha - cos^4\alpha = \frac{1}{8}[/tex]
( [tex]sin^2\alpha - cos^2\alpha )*(sin^2\alpha + cos^2\alpha ) = \frac{1}{8}[/tex] Ponieważ sin²α+ cos²α = 1
sin²α - cos²α = [tex]\frac{1}{8}[/tex]
sin²α - ( 1 - sin²α ) = [tex]\frac{1}{8}[/tex]
2 sin²α = [tex]\frac{1}{8} + 1 = \frac{9}{8}[/tex] / : 2
sin²α = [tex]\frac{9}{16}[/tex]
sin α = [tex]\sqrt{\frac{9}{16} } = \frac{3}{4}[/tex]
================
Szczegółowe wyjaśnienie:
Dla α ∈ ( 0° , 90° ) jest;
0 < sin α < 1 oraz 0 < cos α < 1
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Verified answer
Odpowiedź:
a)
6 cos²α = 5 sin α oraz sin²α + cos²α = 1 ⇒ cos²α = 1 - sin²α
więc mamy
6*(1 - sin²α ) = 5 sin α
6 - 6 sin² α = 5 sin α
6 sin² α + 5 sin α - 6 = 0
sin α = t
6 t² + 5 t - 6 = 0 Δ = 5² - 4*6*(-6) = 25 + 144 = 169 √Δ = 13
t = [tex]\frac{- 5 - 13}{2*6} < 0 - odpada[/tex] lub t = [tex]\frac{- 5 + 13}{12} = \frac{8}{12} = \frac{2}{3}[/tex]
Odp. sin α = [tex]\frac{2}{3}[/tex] α < 90°
==============
b) 4 sin²α = 2 + 7 cos α
4*( 1 - cos²α ) = 2 + 7 cos α
4 - 4 cos²α = 2 + 7 cos α
4 cos²α + 7 cos α - 2 = 0
cos α = t
4 t² + 7 t - 2 = 0
Δ = 49 - 4*4*(-2) = 49 + 32 = 81 √Δ = 9
t = [tex]= \frac{- 7 - 9}{2*4} = - 2 - odpada[/tex] lub t = [tex]\frac{- 7 + 9}{8} = \frac{1}{4}[/tex]
więc
cos α = [tex]\frac{1}{4}[/tex]
sin²α = 1 - cos²α = 1 - [tex]\frac{1}{16} = \frac{15}{16}[/tex]
Odp. sin α = [tex]\sqrt{\frac{15}{16} } = \frac{\sqrt{15} }{4}[/tex]
========================
c ) 2 sin α = 3 ctg α
2 sin α = 3*[tex]\frac{cos \alpha }{sin \alpha }[/tex] / * sin α
2 sin²α = 3 cos α
2*(1 - cos²α ) = 3 cos α
2 - 2 cos²α = 3 cos α
2 cos²α + 3 cos α - 2 = 0
t = cos α
2 t² + 3 t - 2 = 0
Δ = 9 - 4*2*(- 2) = 9 + 16 = 25 √Δ = 5
t = [tex]\frac{- 3 - 5}{2*2} = - 2 - odpada[/tex] lub t = [tex]\frac{- 3 + 5}{4} = \frac{2}{4} = \frac{1}{2}[/tex]
cos α = [tex]\frac{1}{2}[/tex]
więc
sin²α = 1 - cos²α = 1 - [tex]\frac{1}{4} = \frac{3}{4}[/tex]
sin α = [tex]\sqrt{\frac{3}{4} } = \frac{\sqrt{3} }{2}[/tex] [tex]\alpha = 60^o[/tex]
==============
d )
[tex]sin^4\alpha - cos^4\alpha = \frac{1}{8}[/tex]
( [tex]sin^2\alpha - cos^2\alpha )*(sin^2\alpha + cos^2\alpha ) = \frac{1}{8}[/tex] Ponieważ sin²α+ cos²α = 1
więc
sin²α - cos²α = [tex]\frac{1}{8}[/tex]
sin²α - ( 1 - sin²α ) = [tex]\frac{1}{8}[/tex]
2 sin²α = [tex]\frac{1}{8} + 1 = \frac{9}{8}[/tex] / : 2
sin²α = [tex]\frac{9}{16}[/tex]
sin α = [tex]\sqrt{\frac{9}{16} } = \frac{3}{4}[/tex]
================
Szczegółowe wyjaśnienie:
Dla α ∈ ( 0° , 90° ) jest;
0 < sin α < 1 oraz 0 < cos α < 1