Odpowiedzi

2010-03-06T22:13:29+01:00
Udowodnij tożsamości:
a) (1÷sinα-1÷cosα)(tgα+ctgα)=(cosα-sinα)÷(sin²α×cos²α)

L=(1÷sinα-1÷cosα)(sinα/cosα+cosα/sinα)=
L=1÷sinα*sinα/cosα+ 1÷sinα*cosα/sinα- 1÷cosα* sinα/cosα-1÷cosα*cosα/sinα=
1/cosα+ cosα/sin²α- sinα/cos²α-1/sinα=
[cosα*sin²α+cos³α-sin³α-cos²α*sinα]/(sin²α*cos²α)=
[cosα*(1-cos²α)+cos³α-sin³α-(1-sin²α)*sinα]/(sin²α*cos²α)=
[cosα-cos³α+cos³α-sin³α-sinα+sin³α]/(sin²α*cos²α)=
[cosα-sinα]/(sin²α*cos²α)=P

b) (cosα÷sinα)-(sinα÷cosα)=(ctgα-1)(tgα+1)
P=(ctgα-1)(tgα+1)=ctgα*tgα+ctgα-tgα-1=1+ctgα-tgα-1=ctgα-tgα=P

c) sin⁴α-cos⁴α=sin²α-cos²α
L=sin⁴α-cos⁴α=(sin²α-cos²α)(sin²α+cos²α)=sin²α-cos²α=P

d) (ctgα-tgα)÷(sinα+cosα)=1÷sinα-1÷cosα
L=(ctgα-tgα)÷(sinα+cosα)=(cosα/sinα-sinα/cosα)÷(sinα+cosα)=
[(cos²α-sin²α)/sinα*cosα]÷(sinα+cosα)=
[(cosα-sinα)(cosα+sinα)/sinα*cosα]÷(sinα+cosα)=
(cosα-sinα)(cosα+sinα)/sinα*cosα*(sinα+cosα)=
(cosα-sinα)/sinα*cosα=cosα/sinα*cosα-sinα/sinα*cosα=
1/sinα-1/cosα=P
Najlepsza Odpowiedź!
2010-03-06T22:14:00+01:00
A) (1/sinα - 1/cosα)(tgα + ctgα) = (cosα - sinα)/(sin²α * cos²α)

L = ((cosα - sinα)/cosα * sinα)(sinα/cosα + cosα/sinα) = ((cosα - sinα)/cosα * sinα)((sin²α + cos²α)/sinα * cosα) = (cosα - sinα)*(sin²α + cos²α)/cos²α * sin²α = (cosα - sinα)/cos²α * sin²α = P

b) (cosα/sinα) - (sinα/cosα) = (ctgα - 1)(tgα + 1)

P = (ctgα - 1)(tgα + 1) = ctgα * tgα - tgα + ctgα - 1 = ctgα * 1/ctgα - 1 - tgα + ctgα = 0 + (cosα/sinα) - (sinα/cosα) = (cosα/sinα) - (sinα/cosα) = L

c) sin⁴α - cos⁴α = sin²α - cos²α

L = sin⁴α - cos⁴α = (sin²α)² - (cos²α)² = (sin²α - cos²α)(sin²α + cos²α) = sin²α - cos²α = P

d) (ctgα - tgα)/(sinα + cosα) = 1/sinα - 1/cosα

L = ((cosα/sinα) - (sinα/cosα))/(sinα + cosα) = ((cos²α - sin²α)/sinα*cosα)/(sinα + cosα) = (cos²α - sin²α)/[sinα*cosα(sinα + cosα)] = (cosα - sinα)(sinα + cosα)/(sinα*cosα(sinα + cosα)) = (cosα - sinα)/sinα*cosα = 1/sinα - 1/cosα = P

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