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LaddiusMaximus
 5 years ago
(sinx/(1+cosx)) +((1+cosx)/sinx) totally lost
LaddiusMaximus
 5 years ago
(sinx/(1+cosx)) +((1+cosx)/sinx) totally lost

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you have to do? prove it?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0combine the fractions by finding a common denominator

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i got that, but what do I use, sinx? or 1+cosx?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The Common Denominator will be sinx(1+cosx)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean? how to combine these fractions: 2/3+5/7

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0so the common denominator is (sinx)1+cosx?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that is the common denom

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0okay so now I have (sinx^2x(1+cosx)/sinx(1+cosx)) + (sinx(2+cos^2x)/sinx(1+cosx))?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you will have (sinx)^2+(1+cosx)^2 in the numerator and sinx(1+cosx) is the denominator

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0sin^2x on both sides?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you should have this so far: \[\sin^2x + (1+cosx)^2/sinx(1+cosx)\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok look at your first fraction. what does the bottom lack so that it has the same denominator as the 2nd. sinx, right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0multiply both the top and bottom by sinx on the first fraction

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0hmm i had [\sin^2x(1+cosx)+sinx(2+\cos^2x)/sinx(1+cosx)\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0now look at the 2nd fraction what does the bottom lack so that it has the same denominator as the first fraction, (1+cosx), right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so multiply the top and bottom of the 2nd fraction by (1+cosx). Now the fractions have the same denominator, you can write it as one fraction now

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0I follow you so far.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so we have [sinx*sinx+(1+cosx)(1+cosx)]/[sinx(1+cosx)]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0[(sinx)^2+1+2cosx+(cosx)^2]/[sinx(1+cosx)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I got the complete answer/proof

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0what is (sinx)^2+(cosx)^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0lol gj, but i was laddius

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0do you still follow laddius

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0so the numerator becomes sin^2x + (1+cosx)^2?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0now foil that (1+cosx)^2

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0so now its 1+2cosx+cosx^2?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0now do you see (sinx)^2+(cosx)^2 anywhere in that numerator

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0ahhhhh pythagorean

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0what do you have now?

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0so now I have 2+2cosx/sinx(1+cosx)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok what does the numerator terms have in common

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0HAHA ignore my last 3 posts..

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0does anything cancel?

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0the 2s cancel, right?

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0I have 2+2cosx in the numerator

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0cosx+cosx= 2cosx, right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0didnt you say the numerator terms had a 2 in common so factor that out in you have 2(1+cosx)/[sinx(1+cosx)]

LaddiusMaximus
 5 years ago
Best ResponseYou've already chosen the best response.0crap then the (1+cosx) cancel out and im left with 2cscx
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