Question

Uprość wyrażenie:
a) sinα × tgα + cosα
b) tg²α - 1/cos²α
c) tgα/tgα+ctgα
d) (tgα + ctgα)² - (tgα - ctgα)²
e) (1/cos²α - 1)(1/sin²α - 1)
f) (cosα + 1/vos)² + sin²α - tg²α

Oblicz cosα, jeśli:
a) sin²α + cosα +1 = 0
b) 3sin²α + 7cosα +3 = 0
c) 3ctgα = 2sinα

Answer 1

A) sinα × tgα + cosα =
sinα *sinα/cosα+cosα=
sin²α/cosα+cosα=
(sin²α+cos²α)/cosα=1/cosα
b) tg²α - 1/cos²α=
(sinα/cosα)²-1/cos²α=
( sin²α-1)/cos²α=
(sin²α-sin²α-cos²α)/cos²α= -1
c) tgα/tgα+ctgα= 1+ctgα
d) (tgα + ctgα)² - (tgα - ctgα)² = tg²α+2tgα*ctgα+ctg²α-(tg²α-2tgα*ctgα+ctg²α)=4tgα *ctgα=4tgα*1/tgα=4
e) (1/cos²α - 1)(1/sin²α - 1)= (1-cos²α)/cos²α*(1-sin²α)/sin²α=
sin²α/cos²α*cos²α/sin²α=1
f) (cosα + 1/vos)² + sin²α - tg²α= [(cos²α+1)/cos²α]²+sin²α+sin²α/cos²α=
(cos⁴α+2cos²α+1+sin²α*cos⁴α+sin²α*cos²α)/cos⁴α=
[cos⁴α+3cos²α+sin²α+sin²αcos²α(cos²α+1)]/cos⁴α=

Oblicz cosα, jeśli:
a) sin²α + cosα +1 = 0
(1-cos²α+cosα+1)=0
-cos²α+cosα+2=0
Δ=1+8=9 √Δ=3
cosα=-1-3/-2=2
cosα=-1+3/-2=-1
b) 3sin²α + 7cosα +3 = 0
3(1-cos²α)+7cosα+3=0
-3cos²α+7cosα+6=0
Δ=49+72=121
√Δ=11
cosα=-7-11/-6=3
cosα=-7+11/-6=-4/6=-2/3
c) 3ctgα = 2sinα
3*cosα/sinα=2sinα
3cosα=2sin²α
-2(1-cos²α)+3cosα=0
2cos²α+3cosα-2=0
Δ=9+16=25
√Δ=5
cosα=-3-5/4=-2
cosα=-3+5/4=½