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VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0so i know cotx is cos/sin

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0idk if we can change the tan though to sin/cos?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0this one is hard but try this

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0first divide everything by cot x

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0and we know that \[\frac{ \tan x }{ \cot x } = \frac{ \frac{ \sin x }{ \cos x } }{ \frac{ \cos x }{ \sin x } } = \frac{\sin^{2} x }{ \cos^{2} x } = \tan^{2} x\]

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0that would simplify it to: \[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0yeah the second one is: \[\frac{ \tan^{3} x }{ \cot x } = \frac{ \tan x }{ \cot x } * \tan^{2} x = \tan^{2} x * \tan^{2} x = \tan^{4} x\]

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0where did you get the second tan^2x

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0I divided tan^3 x into tan^x * tan^2 x

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1357511455432:dw

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0So this is why we have the following equation so far: \[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0next step is to subtract both sides by tan^4 x

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sec^{4} x  \tan^{4} x = 1 + 2 \tan^{2} x\]

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0and divide both side by2?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0no there is a trick here now. notice the left side is the difference of two squares

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sec^{4} x  \tan^{4} x = (\sec^{2} x)^{2}  (\tan^{2} x)^{2}\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0remember that a^2  b^2 = (a  b)(a + b) so we can now write the left side as: \[\sec^{4} x  \tan^{4} x = (\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x)\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0we kind of did the reverse of FOIL, we factored it

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0So now we have: \[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0but do you remember the identity sec^2 x = tan^2 x + 1

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0yes but there is the 2?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0we'll get to that :) so if sec^2 x = tan^2 x + 1 the if we subtract tan^2 from both sides we get sec^2 x  tan^2 x = 1

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0\[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\] becomes \[(1)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^2{x}\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0No we factored \[\sec^{4} x  \tan^{4} x \] into \[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x)\] because it is a difference of sqaures

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0ok so when you subtract tan^2x?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0There is a trig identity: \[\sec^{2} x = 1 + \tan^{2} x\] i am saying, if we rewrite it as: \[\sec^{2} x  \tan^{2} x = 1\] then the left factor on our left hand side reduces to just 1: \[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x) = (1)(\sec^{2} x + \tan^{2} x) \]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0so we are now left with \[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\]

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0ok you had lost me. now i get it sorry

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0its a complicated problem compared to the last one o_o

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0yes it is. i still have to do 5 more assignments. lol

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0so now we have \[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\] can you see the next step ?

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0substitute the sec^2x with 1+ tan^2x?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0we could, or we could save a step and just subtract tan^2 x from both sides

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0if we subtract tan^2 x from both sides we get \[\sec^{2} x = 1 + \tan^{2} x\]

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0which is an identity

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0but how does this equal to cotxsec^4x?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0the whole idea is to reduce the original equation into a trig identity

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0if you follow the steps then that's precisely what we did

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0to verify that it equals on both sides?

binarymimic
 2 years ago
Best ResponseYou've already chosen the best response.0if you manipulate an equation using algebra then whatever you end up with is equal to what you started with.

VeroZarate
 2 years ago
Best ResponseYou've already chosen the best response.0oh ok thanks so much!
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