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sabika13
Group Title
How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 x)
 2 years ago
 2 years ago
sabika13 Group Title
How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 x)
 2 years ago
 2 years ago

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matricked Group TitleBest ResponseYou've already chosen the best response.1
i think ur question is cos2x/(1+sin2x) now cos2x =cos^2x sin^2x=(cosx+sinx)(cosxsinx) and 1+sin2x = cos^2x + sin^2x +2sinxcosx =(cosx+sinx)^2 thus cos2x/(1+sinx)^2= (cosx+sinx)(cosxsinx)/(cosx+sinx)^2 =(cosx sinx)/((cosx+sinx) divide both num and denom by cosx and then we have =(1sinx/cosx)/(1+sinx/cosx) =(1tanx)/(1+tanx) = (tan(pi/4) tanx)/(1+tan(pi/4)tanx)=tan(pie/4 x)
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
for the second last step.. where did the square roots go? isnt it suppose to be: (1sinx/cosx)(1sinx/cos)/(1+sinx/cos)(1+sinx/cosx)
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
actually the factors (cosx+sinx) are there both in num and denom hence they cancel each other
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
but its (cosxsinx) in the num :s
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
(cosx+sinx)(cosxsinx) / (cosx+sinx)^2 =(cosxsinx) / (cosx+sinx)
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
i hope its clear now
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
why 13 u hv included in ur id ? if u don't mind
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
if its (cosx+sinx)^2, it has to equal (cosx+sinx)(cosx+sinx)
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
idk random.
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
(cosx+sinx)^2= (cosx+sinx)(cosx+sinx) (correct ) that is why one of the (cosx+sinx)
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
cancel each other
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
how does cos^2x sin^2x=(cosx+sinx)(cosxsinx)
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
USE ..A*A B*B =(A)^2 (B)^2 = (A+B)(AB)...
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
okay, then why isnt :(cosx+sinx)^2 > (cosxsinx)(cosx+sinx)
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
bcoz a^2 =a*a
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
so (a+b)^2 =(a+b)(a+b)
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
but if u multiply (a+b)(ab) u get (a^2 b^2)
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
so (a^2b^2) does not equal (ab)^2 ?
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
sahi hai..
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
oh i get it..
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
=(1tanx)/(1+tanx) = (tan(pi/4) tanx)/(1+tan(pi/4)tanx)=tan(pie/4 x) can u explain that? .. where did pi/4 come from ?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Tan (pi/4) = 1.
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
is that a trig identity?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
No. If you do that on a calculator you get that.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Tan(pi/4) is always 1. Look on a unit circle.
 2 years ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.1
Okay, thanks!!
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Welcome :) .
 2 years ago
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