Question

Nie wiem oco tu chodzi w zad a) b) c)
Proszę o pomoc
Dam naj :)))))

Answer 1

Odpowiedź:

a)

[tex]\frac{cos\alpha - sin\alpha }{cos\alpha *sin\alpha } =\frac{2}{tg 2\alpha *(sin\alpha + cos \alpha) }[/tex]      sin α ≠ 0    i  cos α ≠ 0  i  sin α ≠ - cos α

Mnożymy na krzyż

2 cosα*sin α = ( cos α - sin α)*) cosα + sinα)*tg 2α

2 sinα*cosα = (cos²α - sin²α)*tg 2α

sin 2α = cos 2α* tg 2α

sin 2α = cos 2α*[tex]\frac{sin 2\alpha }{cos 2\alpha }[/tex]

sin 2α = sin 2α

L  = P

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b )  [tex]sin^4\alpha + cos 2\alpha = cos^4\alpha[/tex]

[tex]cos 2\alpha = cos^4\alpha - sin^4\alpha[/tex]

[tex]cos[/tex] [tex]2\alpha = ( cos^2\alpha - sin^2\alpha )*(cos^2\alpha + sin^2\alpha )[/tex]

cos 2α = cos²α - sin²α         bo   cos²α + sin²α = 1

cos 2α = cos 2α

L = P

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c )

[tex]2 sin^4\alpha - sin^2 2\alpha = 2 cos^22\alpha - 2 cos^4\alpha[/tex]

[tex]2 sin^4\alpha - ( 2 sin\alpha *cos\alpha )^2 = 2*( cos^2\alpha - sin^2\alpha )^2 - 2 cos^4\alpha[/tex]

[tex]2 sin^4\alpha - 4 sin^2\alpha *cos^2\alpha =[/tex] [tex]2*cos^4\alpha - 4 cos^2\alpha *sin^2\alpha + 2 sin^4\alpha - 2 cos^4\alpha[/tex]

[tex]2 sin^4\alpha - 4 sin^2\alpha *cos^2\alpha = 2 sin^4\alpha - 4 cos^2\alpha *sin^2\alpha[/tex]

[tex]2 sin^4\alpha - 4 sin^2\alpha *cos^2\alpha = 2 sin^4\alpha - 4 sin^2\alpha *cos^2\alpha[/tex]

L = P

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