wartosc zul: mamy 3sin x + 4cos x = 5, x∊(0,π/2) oblicz 2sinx + cosx + 4tgx=?

Adamm:

3		4
	sinx+		cosx=1
5		5

cos(arcsin(3/5))sinx+sin(arcsin(3/5))cosx=1 sin(x+arcsin(3/5))=1 x+arcsin(3/5)=π/2+2kπ skoro x∊(0;π/2) to x=π/2−arcsin(3/5) sinx=sin(π/2−arcsin(3/5))=cos(arccos(4/5))=4/5 cosx=cos(π/2−arcsin(3/5))=sin(arcsin(3/5))=3/5 tgx=4/3 2sinx+cosx+4tgx=8/5+3/5+16/3=113/15

Adamm: poprawka cos(arcsin(4/5))sinx+sin(arcsin(4/5))cosx=1 x=π/2−arcsin(4/5) sinx=cos(arccos(3/5))=3/5 cosx=sin(arcsin(4/5))=4/5 tgx=3/4 2sinx+cosx+4tgx=5

Mila:

	π
x∊(0,		)⇒sinx>0 i cosx>0
	2

3sinx+4cosx=5/:5

3		4
	sinx+		cosx=1
5		5

	3		4
cosα=		to sinα=		, α− kąt ostry
	5		5

sinx*cosα+sinα*cosx=1 sin(x+α)=1

	π
x+α=
	2

	π
x=		−α
	2

	π		π		π
2sinx+cosx+4tgx=2sin(		−α)+cos(		−α)+4tg(		−α)=
	2		2		2

	3		4		3		4		3
=2sinα+cosα+4*ctgα=2*		+		+4*		:		=2+4*		=5
	5		5		5		5		4