anonymous
  • anonymous
simplify tan x cos x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Remember tangent is just sin(x)/cos(x)
anonymous
  • anonymous
so its cos x (sin x / cos x) ?
anonymous
  • anonymous
Correct, now what happens when you go ahead and multiply the cosine?

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

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