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RD⁴²Best ResponseYou've already chosen the best response.2
Are you sure it's not (cosx)(tanx+sinx cosx) = sinx + sinx cos²x ?
 one year ago

rahrahBest ResponseYou've already chosen the best response.0
oh crap, its actually (cosx)(tanx+sinx cotx)=sinx+cos^2x
 one year ago

RD⁴²Best ResponseYou've already chosen the best response.2
We know that \(\tan x = \dfrac{\sin x}{\cos x}\) and \(\cot x = \dfrac{\cos x}{\sin x}\) So \((\cos x)(\tan x+\sin x \cot x) = (\cos x)\left(\dfrac{\sin x}{\cos x} + \sin x\cdot\dfrac{\cos x}{\sin x}\right)\) \(=(\cos x)\left(\dfrac{\sin x}{\cos x} + \cos x\right)\) \( = \boxed{\sin x + \cos^2 x}\)
 one year ago

rahrahBest ResponseYou've already chosen the best response.0
Q.E.D. ? Thanks a lot for the help otherwise
 one year ago

RD⁴²Best ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Q.E.D.
 one year ago

rahrahBest ResponseYou've already chosen the best response.0
wait, how'd you get sinx + cos^2 at the end? shouldn't sinx/cosx cross out the + cosx?
 one year ago

RD⁴²Best ResponseYou've already chosen the best response.2
No, because it's just addition. you can't cancel cos x out like that.dw:1362452530886:dw
 one year ago

rahrahBest ResponseYou've already chosen the best response.0
shouldnt that make the end result be tanx cosx + (cosx)^2? how did we get sin from the tan?
 one year ago

RD⁴²Best ResponseYou've already chosen the best response.2
\(\tan x = \dfrac{\sin x}{\cos x}\) right? So \(\tan x\cos x\ = \dfrac{\sin x}{\cos x}\cdot \cos x = \sin x\) And cos²x is just another way to write (cos x)²
 one year ago

rahrahBest ResponseYou've already chosen the best response.0
oh, i get it now. Thanks a lot for all your help!
 one year ago
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