trygonometria ssss:

	1
wyznacz wartość sin4 x + cos4 x wiedząc, że sinx + cosx =
	4

Mariusz:

	1
sin2(x)+cos2(x)+2sinxcosx=
	16

	1
1+2sinxcosx=
	16

	15
sin(2x)=−
	16

	1
(sin2(x)+cos2(x))2=sin4x+cos4(x)+		4sin2xcos2(x)
	2

	225
1=sin4x+cos4(x)+
	512

	512−225		287
sin4x+cos4(x)=		=
	512		512

(sin(x)+cos(x))⁴=sin⁴{x}+4sin³(x)cos(x)+6sin²(x)cos²(x)+4sin(x) cos³(x)+cos⁴(x) (sin(x)+cos(x))⁴=sin⁴(x)+cos⁴(x)

	3
+4sin(x)cos(x)(sin2(x)+cos2(x))+		(4sin2(x)cos2(x))
	2

	3
(sin(x)+cos(x))4=sin4(x)+cos4(x)+4sin(x)cos(x)+		(4sin2(x)cos2(x))
	2

(sin(x)+cos(x))²=sin²(x)+cos²(x)+2sin(x)cos(x) (sin(x)+cos(x))²−1=2sin(x)cos(x)

	3
(sin(x)+cos(x))4=sin4(x)+cos4(x)+2((sin(x)+cos(x))2−1)+		((sin(x)+cos(x))2−1)
	2

(sin(x)+cos(x))⁴=sin⁴(x)+cos⁴(x)+2(sin(x)+cos(x))²−2+

3
	((sin(x)+cos(x))4−2(sin(x)+cos(x))2+1)
2

	1		1
sin4(x)+cos4(x)=−		(sin(x)+cos(x))4+(sin(x)+cos(x))2+
	2		2

	1		32		256
sin4(x)+cos4(x)=−		+		+
	512		512		512

	287
sin4(x)+cos4(x)=
	512

PW:

	1
sinx + cosx =		,
	4

skąd wynika że

	1
(sinx + cosx)2 = (		)2
	4

	1
sin2x + 2sinxcosx + cos2x =		,
	16

po zastosowaniu "jedynki trygonometrycznej"

	1
1 + 2sinxcosx =
	16

	15
(1) 2sinxcosx = −		.
	16

Mamy policzyć (2) sin⁴x + cos⁴x = (sin²x + cos²x)² − 2sin²xcos²x = 1 − 2sin²xcos²x. Wyliczamy z (1) podnosząc stronami do kwadratu:

	225
4sin2xcos2x =
	256

	225
2sin2xcos2x =		,
	512

to podstawione do (2) daje

	225		287
sin4x + cos4x = 1 −		=		.
	512		512