Zadanie1
wiedząc,że kąt α jest kątem ostrym,oblicz wartość wyrażenia:
a)tgα+cosα/tgα-cosα gdy tgα=20/21
b)√2/2 cos²α+√2sin²α / 4cos²α-3sin²α gdy tgα=√2/3
c)4cosα-tgα / sinα+tgα gdy cosα=15/17
d)2sinα+3cosα / cosα-3sinα gdy tgα=2

zadanie2
oblicz sin²17+cos²17

1

Odpowiedzi

Najlepsza Odpowiedź!
2010-01-16T23:11:08+01:00
A)(tgα+cosα)/(tgα-cosα)
gdy tgα=20/21
sinα/cosα=20/21
21sinα=20cosα /:21
sinα=20/21 cosα

sin²α+cos²α=1
(20/21 cosα)²+cos²α=1
400/441 cos²α +441/441 cos²α=1
841/441 cos²α=1 /:(841/441)
cos²α=441/841
cosα=21/29 bo I ćw

(tgα+cosα)/(tgα-cosα)=(20/21 + 21/29)/ (20/21 -21/29)=
=(580/609 + 441/609)/(580/609 - 441/609)=
=(1021/609) / (139/609)=
=(1021/609) * (609/139)=1021/139

b)(√2/2 cos²α+√2sin²α) / (4cos²α-3sin²α)
gdy tgα=√2/3
sinα/cosα=√2/3
3sinα=√2cosα /:3
sinα=√2/3 cosα

sin²α+cos²α=1
(√2/3 cosα)²+cos²α=1
2/9 cos²α + cos²α=1
2/9 cos²α +9/9 cos²α=1
11/9 cos²α=1 / *9/11
cos²α=9/11

sin²α=1-cos²α
sin²α=1- 9/11
sin²α=2/11

(√2/2 cos²α+√2sin²α) / (4cos²α-3sin²α)=
=(√2/2 *9/11+√2*2/11) / (4*9/11-3*2/11)=
=(9√2/22 + 2√2/11) / (36/11 - 6/11)=
=(9√2/22 + 4√2/22) / (30/11)=
=(13√2/22)/(30/11)=(13√2/22)*(11/30)=13√2/60

c)(4cosα-tgα) / (sinα+tgα)
gdy cosα=15/17

sin²α+cos²α=1
sin²α+(15/17)²=1
sin²α+(225/289)=1
sin²α=1- 225/289
sin²α=289/289 - 225/289
sin²α=64/289
sinα=8/17

tgα=sinα/cosα=(8/17)/(15/17)=8/17 * 17/15=8/15

(4cosα-tgα) / (sinα+tgα)=(4*15/17 - 8/15) / (8/17 + 8/15)=
=(60/17 - 8/15)/(8/17 + 8/15)=
=(900/255 - 136/255)/(120/255 + 136/255)=
=(764/255)/(256/255)=(764/255)*(255/256)=764/256=191/64


d)(2sinα+3cosα) / (cosα-3sinα)
gdy tgα=2

tgα=sinα/cosα=2 /*cosα
sinα=2cosα

sin²α+cos²α=1
(2cosα)²+cos²α=1
4cos²α+cos²α=1
5cos²α=1 /:5
cos²α=1/5
cosα=1/√5 *√5/√5=√5/5

sinα=2cosα=2√5/5

(2sinα+3cosα) / (cosα-3sinα)=
=(2*2√5/5 + 3*√5/5) / (√5/5 - 3*2√5/5)=
=(4√5/5 + 3√5/5) / (√5/5 - 6√5/5)=
=(7√5/5) / (-5√5/5)=(7√5/5)*(5/(-5√5))=-7/5