Liczby zespolone Dalaya: Obliczyć: ⁵√2+2i lzl = √2² +2² = √8 kąt "fi" = ^π₄ z= √8 ( cos ^π₄ +isin ^π₄ ) Rozwiązanie pierwsze: w₀ = ⁵√√8 (cos ( ^π₄ / 5 ) +isin ( ^π₄ / 5) = p10{8} (cos ^π₂₀ + isin^π₂₀) Jak powinnam obliczyć ^π₂₀ ? w tabeli nie ma takiej wartości natomiast nie wiem jak mogłabym to przekształcić /

Mariusz: cos(5α)=0 cos(2α)=2cos²(α)−1 cos(2(2α))=2(2cos²(α)−1)²−1 cos(4α)=2(4cos⁴(α)−4cos²(α)+1)−1 cos(4α)=8cos⁴(α)−8cos²(α)+1 sin(2α)=2sin(α)cos(α) sin(2*(2α))=2sin(2α)cos(2α) sin(4α)=2(2sin(α)cos(α))(2cos²(α)−1) sin(4α)=8sin(α)cos³(α)−4sin(α)cos(α) sin(4α)=4sin(α)(2cos³(α)−cos(α)) cos(5α)=cos(α+4α) cos(5α)=cos(α)cos(4α)−sin(α)sin(4α) cos(5α)=cos(α)(8cos⁴(α)−8cos²(α)+1)−sin(α)(4sin(α)(2cos³(α)−cos(α))) cos(5α)=cos(α)(8cos⁴(α)−8cos²(α)+1)−4sin²(α)(2cos³(α)−cos(α))) cos(5α)=cos(α)(8cos⁴(α)−8cos²(α)+1)+(4cos²(α)−4)(2cos³(α)−cos(α))) cos(5α)=8cos⁵(α)−8cos³(α)+cos(α)+8cos⁵(α)−4cos³(α)−8cos³(α)+4cos(α) cos(5α)=16cos⁵(α)−20cos³(α)+5cos(α) 16cos⁵(α)−20cos³(α)+5cos(α)=0 cos(α)(16cos⁴(α)−20cos²(α)+5)=0 16cos⁴(α)−20cos²(α)+5=0 cos²(α)=t 16t²−20t+5=0 Δ=400−4*5*16=400−320 Δ=80

	20±4√5
x=
	32

	10±2√5
x=
	16

	10±2√5
cos2(x)=
	16

	√10±2√5
cos(x)=
	4

Ponieważ cosinus jest malejący w pierwszej ćwiartce wybieramy wartość

	√10+2√5
cos(x)=
	4

Teraz korzystamy ze wzoru na cosinusa i sinusa kątów połówkowych Dla kątów w pierwszej ćwiartce

	α		1+cos(α)
cos2(		)=
	2		2

	α		1−cos(α)
sin2(		)=
	2		2

Mariusz:

	π		4+√10+2√5
cos2(		)=
	20		8

	π		4−√10+2√5
sin2(		)=
	20		8

	π		8+2√10+2√5
cos2(		)=
	20		16

	π		8−2√10+2√5
sin2(		)=
	20		16

	π		√8+2√10+2√5
cos(		)=
	20		4

	π		√8−2√10+2√5
sin(		)=
	20		4

(√a+√b)²=a+b+2√ab a+b=8 ab=10+2√5 t²−8t+10+2√5 t²−8t+16−(6−2√5) (t−4)²−(√5−1)² (t−3−√5)(t−5+√5)=0 a=3+√5 b=5−√5

	π		√3+√5		√5−√5
cos(		)=		+
	20		4		4

	π		√3+√5		√5−√5
sin(		)=		−
	20		4		4

	π		√12+2√20		√5−√5
cos(		)=		+
	20		8		4

	π		√12+2√20		√5−√5
sin(		)=		−
	20		8		4

a+b=12 ab=20 t²−12t+20=0 t²−12t+36−16=0 (t−6)²−16=0 (t−6−4)(t−6+4)=0 (t−10)(t−2)=0 a=10 b=2

	π		√10		√2		√5−√5
cos(		)=		+		+
	20		8		8		4

	π		√10		√2		√5−√5
sin(		)=		+		−
	20		8		8		4