Sprawdź, czy:
a)
sin
b)
Mam to zadane na jutro i nie mogę skończyć tych dwóch przykładów :(
sinα/(1 + cos α) + (1 +cosα)/sin α = 2/sin α
Należy sprowadzić do wspólnego mianownika i dodać
L = [sin α * sin α + (1 +cos α)(1 + cos α)]/[( 1+ cos α)*sin α ] =
= [ sin²α + 1 + 2 cosα + cos²α]/[(1 + cos α) * sin α] =
= [ ( sin²α + cos²α ) + 1 + 2 cos α ]/[(1 + cos α) *sin α] =
= [ 1 + 1 + 2 cos α]/[(1 + cos α) * sinα ] =
= [ 2 + 2 cos α]/[(1 + cos α) * sin α] = [ 2*(1 + cos α)]/[(1 + cos α)*sin α ] =
= 2/ sin α = P
=================================================================
( 1 + sin α)( 1/cos α - tg α ) = cos α
Wystraczy pomnożyć po lewej stronie
L = 1*(1/cos α) -1*tg α + sin α *(1/cos α) - sin α* tg α =
= 1/cos α - tg α + sin α/cos α- sin α *(sin α/ cos α) =
= 1/cos α - tg α + tg α - sin²α / cos α =
= 1/cos α - sin²α/ cos α = [ 1 - sin²α]/cos α =
= cos²α / cos α = cos α = P
==========================
Korzystamy z tego,że
sin α / cos α = tg α
oraz sin²α + cos²α = 1 --> 1 - sin²α = cos²α
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a)
sinα/(1 + cos α) + (1 +cosα)/sin α = 2/sin α
Należy sprowadzić do wspólnego mianownika i dodać
L = [sin α * sin α + (1 +cos α)(1 + cos α)]/[( 1+ cos α)*sin α ] =
= [ sin²α + 1 + 2 cosα + cos²α]/[(1 + cos α) * sin α] =
= [ ( sin²α + cos²α ) + 1 + 2 cos α ]/[(1 + cos α) *sin α] =
= [ 1 + 1 + 2 cos α]/[(1 + cos α) * sinα ] =
= [ 2 + 2 cos α]/[(1 + cos α) * sin α] = [ 2*(1 + cos α)]/[(1 + cos α)*sin α ] =
= 2/ sin α = P
=================================================================
b)
( 1 + sin α)( 1/cos α - tg α ) = cos α
Wystraczy pomnożyć po lewej stronie
L = 1*(1/cos α) -1*tg α + sin α *(1/cos α) - sin α* tg α =
= 1/cos α - tg α + sin α/cos α- sin α *(sin α/ cos α) =
= 1/cos α - tg α + tg α - sin²α / cos α =
= 1/cos α - sin²α/ cos α = [ 1 - sin²α]/cos α =
= cos²α / cos α = cos α = P
==========================
Korzystamy z tego,że
sin α / cos α = tg α
oraz sin²α + cos²α = 1 --> 1 - sin²α = cos²α
-----------------------------------------------------------------------------------------