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