SugarRainbow
simplify tan x cos x
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qpHalcy0n
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Remember tangent is just sin(x)/cos(x)
SugarRainbow
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so its cos x (sin x / cos x) ?
qpHalcy0n
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Correct, now what happens when you go ahead and multiply the cosine?
SugarRainbow
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sin x
qpHalcy0n
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That's it. The cosines cancel.
SugarRainbow
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okay cool
what bout sec y sin (pi/2-y)? IDK how to do this at all
qpHalcy0n
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\[\sec(y)\sin(\frac{\pi}{2-y})\]
??
SugarRainbow
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but isnt it like pi/2 - y not pi/ 2-y
hartnn
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sin(pi/2-y)=cos y
sec y= 1/cos y
qpHalcy0n
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Ok, so rewrite secant as 1/cos
qpHalcy0n
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Then you can use a fancy pants trig identity.
SugarRainbow
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cos y 1/cos y
qpHalcy0n
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|dw:1347861835303:dw|
Right?
SugarRainbow
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i think so i was kinda looking at what hartnns post too so couldn't you do it like|dw:1347861948541:dw|
hartnn
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sec y =1/ cos y is in numerator, so it would be :|dw:1347862109049:dw|
qpHalcy0n
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No, well what you would have is a multiplication, the cosines would cancel.
SugarRainbow
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oh hey you know how tan x= sinx/ cos x
well what are the rest of those cuz it might help me understand better like how we getting cos y when its sin y
hartnn
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ok, so u want all trignometric formulas ? for converting one ratio to other ?
SugarRainbow
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yes please
hartnn
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i will list some :
\(\huge sin x =1/cosec x \implies cosec x=1/sin x \\\huge cos x = 1/sec x \implies secx=1/cos x \\\huge tan x=1/cot x \implies cot x =1/tanx \\\huge tan x=sin x/cos x \\\huge cot x=cos x/sinx \\
\huge sin^2x+cos^2x=1 \\\huge sec^2x=1+tan^2x\\\huge cosec^2x=1+cot^2x\)
hartnn
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if u want sin x when cos x is given or if u want cos x when sin x is given, u use:
\(sin^2x+cos^2x=1\)
SugarRainbow
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okay
SugarRainbow
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so if its 1+tan^2x/csc^2x that would simplify to sec^2x/csc^2
hartnn
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yes.
or is it ?
\(\large1+\frac{tan^2x}{cosec^2x}\quad?\)
hartnn
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if it is
\(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)
SugarRainbow
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but you put that cos^2x=1+cot^2x
hartnn
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no the last one is cosec^2 x =1+cot^2 x
not cos^2x=1+cot^2x
hartnn
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csc=cosec
SugarRainbow
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oh yeah huh cosec is csc not cos i forgot for a minute lol
SugarRainbow
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so is it 1+ tan^2x/csc^2x or sec^2x/csc^2 and why
hartnn
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what exactly is your question ?
is this : (1+ tan^2x)/csc^2x
hartnn
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(1+ tan^2x)=sec^2 x from the formula i gave u.
SugarRainbow
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no i saying is it my answer or yours and why (idk how to put it on my computer like you did so that was the best i could do)
hartnn
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to tell u what is your answer, i need to know your exact question, that is why i asked whether it is: (1+ tan^2x)/csc^2x
SugarRainbow
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oh yeah it's that and i need to simplify it
hartnn
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ok,for that i have shown u the steps, tell me which step, u didn't understand:
\(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)
SugarRainbow
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oh i get it now earlier you put it wierd so i didn't get it
SugarRainbow
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okay now this one is cot x tan x so it would be (cosx/sinx)(sinx/cosx) correct?
hartnn
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yes, or u could use cot x = 1/ tanx and u get tan x / tan x =1
SugarRainbow
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wait because i have two of each would thay cancel out?
hartnn
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yes, they would.
sin x cos x / sin x cos x =1
SugarRainbow
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oh okay just checking
hartnn
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|dw:1347864987907:dw|
hartnn
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okk.
SugarRainbow
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how bout cot u sin u would that simplify to
(cosu/sinu)sin u= cos u?
hartnn
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thats absolutely correct!
u are improving :)
SugarRainbow
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hooray improvement!
hartnn
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what next ?
SugarRainbow
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1-cos^2 beta/sin beta and that would simplify to ooh okay i got nothin for this one :(
hartnn
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is the numerator (1-cos^2 beta ) ??
SugarRainbow
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yes
hartnn
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from
sin^2 beta +cos^2 beta =1
u have
1-cos^2 beta = sin^2 beta
so u have :
\(\large \frac{sin^2 \beta}{sin \beta}=\frac{sin \beta.sin\beta}{sin \beta}=?\)
SugarRainbow
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sin beta?
hartnn
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yup, but did u get all steps ?
SugarRainbow
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yeah
so is that another function 1-cos^2 beta = sin^2 beta
hartnn
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see the list i have given.
it has
sin^2 x + cos^2 x =1
isn't it ?
from here i got
1-cos^2 x = sin^2 x
clear ?
SugarRainbow
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oh so you just went backwards?
hartnn
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whenever u see sin^2 x or cos^2 x
try to use that formula for simplification.
SugarRainbow
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okay now i have cos x- cos^3 x
SugarRainbow
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like would that be cosx- (cosx)(cosx)/cosx?
hartnn
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nopes.
take cos x common from both the terms, what do u get ?
hartnn
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cosx- (cosx)(cosx)(cosx)
SugarRainbow
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so you get 1
hartnn
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no!
after taking cos x common, u get
cos x (1-(cos x)(cos x) )
=cos x (1- cos^2 x)
and u already know what is (1- cos^2 x)
SugarRainbow
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sin^2x
hartnn
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yes!
so your final answer will be
cos x sin^2x
but did u get all the steps ?
SugarRainbow
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yeah because you have coxx-(cosx)(cosx)(cosx) =cosx-1(cosx)=1-cos^2x=sin^2x
hartnn
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i will draw it :|dw:1347866550463:dw|
SugarRainbow
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oh okay then now my last question is sin^2 u + tan^2 u +cos^2 u/sec u
hartnn
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is this entire thing (sin^2 u + tan^2 u +cos^2 u) in the numerator ?
SugarRainbow
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yes
hartnn
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again using sin^2 x +cos^2 x =1, u have numerator as (1+tan^2 x)
got this ?
and (1+tan^2 x) = sec^2 x.....see formula list.
so u have now
sec^2x/sec x = ??
hartnn
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replace all of my 'x' with 'u'
SugarRainbow
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oh yeah so now you have sec x!
hartnn
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yup :)
hartnn
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sec u ;)
SugarRainbow
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oh yeah lol thanks for your help i going to sleep now so goodnight^-^
hartnn
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a very good night to u and have sweet dreams :)