oblicz Mateo:

	π		3π		5π		7π		9π
16 sin		* sin		* sin		* sin		* sin
	18		18		18		18		18

M:

M:

Mila:

	1
sin(60−x)*sinx*sin(60+x)=		sin(3*x)
	4

16*sin 10^o*sin30⁰*sin50^o*sin70^o*sin(90^o)=

	1
16*		*1*sin 10o*sin50o*sin70o*)=
	2

	1		1		1
=8* sin(60o−10o)*sin(10o)*sin(60+10o)=8*		sin(3*10o)=8*		*		=1
	4		4		2

Mariusz:

	9π
sin		= 1
	18

	π		3π		5π		7π
16sin		*sin		*sin		*sin		=
	18		18		18		18

	π   sin( )   9		3π   sin( )   9
=		*		*
	π   cos( )   18		3π   cos( )   18

5π   sin( )   9		7π   sin( )   9
	*
5π   cos( )   18		7π   cos( )   18

	π   sin( )   9		3π   sin( )   9
=		*		*
	π   π   sin( − )   2   18		π   3π   sin( − )   2   18

5π   sin( )   9		7π   sin( )   9
	*
π   5π   sin( − )   2   18		π   7π   sin( − )   2   18

	π   sin( )   9		3π   sin( )   9
=		*		*
	4π   sin( )   9		3π   sin( )   9

5π   sin( )   9		7π   sin( )   9
	*
2π   sin( )   9		π   sin( )   9

	5π   sin( )   9		7π   sin( )   9
=		*
	4π   sin( )   9		2π   sin( )   9

	5π   sin( )   9		7π   sin( )   9
=		*
	4π   sin(π− )   9		2π   sin(π− )   9

	5π   sin( )   9		7π   sin( )   9
=		*
	5π   sin( )   9		7π   sin( )   9

=1

Zosia: Inny sposób: korzystając ze wzorów: 2sinαcosα= sin(2α) i sin(90⁰−α)= cosα i cos(90^o−α)=sinα 16sin10^o*sin30^o*sin50^o*sin70^o*sin90^o=

	1		sin10o*cos10o
=16*		*1*		*cos20o*cos40o=
	2		cos10o

	4sin20o*cos20o*cos40o		2sin40o*cos40o		sin80o
=		=		=		=1
	cos10o		sin80o		sin80o

Mila: Można tak, zamiast ostatniej linijki 21:39 8*sin (10)* sin(50)*sin(70)=

4*(2sin10*cos10)*sin50*cos20
	=
cos10

	2*(2*sin20*cos 20)*sin(50)		2sin40*cos40
=		=		=
	cos 10		cos10

	sin(80)
=		=1
	sin(80)

Mila:

Mariusz: Podobne zadanie T_n(cos(θ)) = cos(n*θ) cos(n*θ) = 0

	π
n*θ =		+ (k−1)π , k∊Z
	2

	(2k−1)π
n*θ =
	2

	(2k−1)π
θ =
	2n

x = cos(θ)

	(2k−1)π
xk = cos(		) , k = 1,2,..,n
	2n

	(2k−1)π
Tn(x) = a∏k=1n(x−cos(		))
	2n

T_n(1) = 1

	(2k−1)π
1 = a∏k=1n(1−cos(		))
	2n

	1
a =
	(2k−1)π   ∏k=1n(1−cos( ))   2n

	1
a = ∏k=1n
	(2k−1)π   1−cos( ))   2n

	(2k−1)π   x−cos( )   2n
Tn(x) = ∏k=1n
	(2k−1)π   1−cos( )   2n

	(2k−1)π		(2k−1)π		(2k−1)π
1 − cos(		) = cos2(		) + sin2(		)
	2n		4n		4n

	(2k−1)π		(2k−1)π
−cos2(		) + sin2(		)
	4n		4n

	(2k−1)π		(2k−1)π
1 − cos(		) = 2sin2(		)
	2n		4n

	(2k−1)π		(2k−1)π
∏k=1n(1−cos(		)) = ∏k=1n(2sin2(		))
	2n		4n

	(2k−1)π
∏k=1n(1−cos(		)) =
	2n

	(2k−1)π		(2k−1)π
∏k=1n(2sin(		))∏k=1n(sin(		))
	4n		4n

	(2k−1)π
∏k=1n(1−cos(		)) =
	2n

	(2k−1)π		(2(n−k+1)−1)π
∏k=1n(2sin(		))∏k=1n(sin(		))
	4n		4n

	(2k−1)π
∏k=1n(1−cos(		)) =
	2n

	(2k−1)π		π		(2k−1)π
∏k=1n(2sin(		))∏k=1n(sin(		−		))
	4n		2		4n

	(2k−1)π
∏k=1n(1−cos(		)) =
	2n

	(2k−1)π		(2k−1)π
∏k=1n(2sin(		))∏k=1n(cos(		))
	4n		4n

	(2k−1)π
∏k=1n(1−cos(		)) =
	2n

	(2k−1)π		(2k−1)π
∏k=1n(2sin(		)cos(		))
	4n		4n

	(2k−1)π		(2k−1)π
∏k=1n(1−cos(		)) =∏k=1n(sin(		))
	2n		2n

	(2k−1)π   x−cos( )   2n
Tn(x) = ∏k=1n
	(2k−1)π   sin( )   2n

No właśnie jako zadanie domowe policzcie taki iloczyn

	(2k−1)π
∏k=1n(sin(		))
	2n

Możenie się pochwalić obliczeniami w tym wątku