anonymous
  • anonymous
Trig / Pre Cal/ identities Am I on the right path? Please guide me and do not give me the answers \[ (\tan x + \sin x)(1-cos x) = \sin^2 x \tan x \] Do I multiply like with foil starting with tan or do I turn tan into \( \frac{sin x}{cox x} \) and then use foil?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Nnesha
  • Nnesha
foil! :-) i guess that will be great
anonymous
  • anonymous
Just with tan \[ \tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x) \] correct so far?
Nnesha
  • Nnesha
yep looks right :-)

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More answers

anonymous
  • anonymous
Ok, how do I approch it from here? do I turn all the tan into sin / cos? and work it out? \[ \tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x) \]
Nnesha
  • Nnesha
yes convert 2nd tan to sin /cos and let me find notebook and a pencil :3 hehe
Nnesha
  • Nnesha
make sure question is right :3 :-)
anonymous
  • anonymous
Yes it is right :-)
Nnesha
  • Nnesha
is it equal to tanxsinx or tan^2xsinx ??
anonymous
  • anonymous
I have \[ \tan x -\sin x + \sin x -(\sin x)(\cos x) = \sin^2 x \tan x \] \[ \tan x -(\sin x)(\cos x) = \sin^2 x \tan x \]
Nnesha
  • Nnesha
yeah so for the final answer i got sinx tanx so that's why ..hm are u sure it's sin square ??
Nnesha
  • Nnesha
next step is to change tan to sin/cos
anonymous
  • anonymous
Yes, it is sin^2 x tan x
Nnesha
  • Nnesha
:( alright
Nnesha
  • Nnesha
alright got it!! :-)
Nnesha
  • Nnesha
hahah i'm soo happy :P
anonymous
  • anonymous
\( \tan x -(\sin x)(\cos x) = \sin^2 x \tan x \) \( \frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x \) ????
Nnesha
  • Nnesha
\( \huge\rm \frac{sin x}{cos x} -(\sin x)(\cos x) = \) ???? find common denominator :-)
anonymous
  • anonymous
Got ya :-)
anonymous
  • anonymous
one sec
Nnesha
  • Nnesha
alright :-)
anonymous
  • anonymous
I am using cos as the common denominator. that is correct?
Nnesha
  • Nnesha
yes right
anonymous
  • anonymous
Ok I got \[ \frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x \] \[ \frac{(sin x)(cos x)}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x \] correct???
Nnesha
  • Nnesha
\[ \frac{(sin x)(\cancel{cos x})}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x \] multiply cosx/1 so when you multiply first fraction by cos both cos will cancel each other out
Nnesha
  • Nnesha
so there isn't supposed to be any cosx \[ \frac{(sin x)}{\cos x} -\frac{((\sin x cos^2x))}{cos x} = \sin^2 x \tan x \] like this
anonymous
  • anonymous
Oh right!!! Made a boo boo
anonymous
  • anonymous
One sec, now we should have
anonymous
  • anonymous
\[ \frac{(sin x)}{\cos x} -\frac{((\sin x cos^2x))}{cos x} = \sin^2 x \tan x \] \[ \frac{(sin x)}{\cos x} -\frac{((\sin x (1-\sin^2 x)2x))}{cos x} = \sin^2 x \tan x \] Ok I am lost
anonymous
  • anonymous
\[ \frac{(sin x)}{\cos x} -\frac{((\sin x (1-cos^2 x))}{cos x} = \sin^2 x \tan x \]
Nnesha
  • Nnesha
\[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos \] multiply |dw:1434231940649:dw|
Nnesha
  • Nnesha
\[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos \] \[\huge\rm \frac{ \sin x - sinx cosx (cosx) }{ cosx }\]you will get this let me if that doesn't make sense
anonymous
  • anonymous
Yes that makes sence
Nnesha
  • Nnesha
alright now multiply cos x times cos x = ?? \[\huge\rm \frac{ \sin x - sinx \color{Red}{cosx (cosx)} }{ cosx }\]you will get this let me if that doesn't make sense
anonymous
  • anonymous
Was this originally \[\frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \] \[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ 1 } \times \cos \] \[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos x }{ 1 } \times \cos \] Is that what we did there?
Nnesha
  • Nnesha
we know that common denominator is cos so we multiplied numerator by cos x
Nnesha
  • Nnesha
if the original fraction already have a same denominator as common denominator so u just have to carrry it down|dw:1434232700973:dw|
anonymous
  • anonymous
If we did that we have \[ \cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos \] \[ \sin x - (\sin x)(\cos^2 x) \] ??
Nnesha
  • Nnesha
first of all we know that cosx is common denominator so that should be at the denominator \[\frac{ ?????? }{ \cos x }\]
Nnesha
  • Nnesha
after that we have to multiply both fraction by cos If we did that we have \[ \cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos \] \[\large\rm \frac{?????????}{cosx }\] ??
Nnesha
  • Nnesha
wait a sec that's not cos^2 yet
Nnesha
  • Nnesha
let me fix it :-)
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nixy If we did that we have \[ \cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos \] \[ \sin x - (\sin x)(\cos^2 x) \] ?? \(\color{blue}{\text{End of Quote}}\) we know common denominator is cosx now we have to find numerator \[\frac{ ?????? }{ \cos x }\] original equation is \[\huge\rm \frac{ sinx }{ cosx } - \frac{ sinx \cos x }{ 1 }\] to find out numerator we should multiply both fraction by cosx(which is common denominator ) \[\large\rm cosx \times \frac{ sinx }{ cosx } - \frac{ sinx \cos x }{ 1 } \times cosx\]
Nnesha
  • Nnesha
and that's how \[\large\rm \cancel{cosx} \times \frac{ sinx }{\cancel{ cosx} } - \frac{ sinx \cos^2 x }{ 1 } \] that's the numerator part
anonymous
  • anonymous
Ok, I think we are on the same page now WOW this stuff gets confusing lol Ok I am where you are at
Nnesha
  • Nnesha
what ??
Nnesha
  • Nnesha
do you got that denominator thingy ?
anonymous
  • anonymous
So now we have \[ \large\rm\frac{ sinx }{{ 1} } - \frac{ sinx \cos^2 x }{ 1 } \]
Nnesha
  • Nnesha
that's the numerator what's our common denominator ?
anonymous
  • anonymous
From what section?? I am going from here \[ \large\rm \cancel{cosx} \times \frac{ sinx }{\cancel{ cosx} } - \frac{ sinx \cos^2 x }{ 1 } \]
Nnesha
  • Nnesha
sinx/1 = sinx so you can write it as \[\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }\]
anonymous
  • anonymous
Yes correct
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha sinx/1 = sinx so you can write it as \[\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }\] \(\color{blue}{\text{End of Quote}}\) do you understand this step ??
Nnesha
  • Nnesha
want to make sure!!
anonymous
  • anonymous
Yes I understand Next ? \[ \huge\rm \frac{ sinx - sinx(1-sin^2) }{ cosx } \]
Nnesha
  • Nnesha
nope next step is to take out common term
Nnesha
  • Nnesha
\[ \huge\rm \frac{\color{ReD}{ sinx} -\color{ReD}{ sinx}cos^2x}{ cosx } \] factor the numerator :-)
Nnesha
  • Nnesha
let sinx = a cosx = b so you can write it as \[\large\rm (a-ab^2)\]
anonymous
  • anonymous
So this \( (a-ab^2) = (ab^2-a) = (ab+a)(ab-a) \) correct?
anonymous
  • anonymous
I cannot believe this one problem is like this. It has been an hour and 30 minutes lol
Nnesha
  • Nnesha
nope what is common in these both terms a - ab^2 ??
Nnesha
  • Nnesha
almost done .-.
anonymous
  • anonymous
so a(1-b^2)
Nnesha
  • Nnesha
yes right!!! so now change it back to sin cos
Nnesha
  • Nnesha
\[ \huge\rm \frac{\color{ReD}{ sinx}( 1-cos^2x)}{ cosx } \]
anonymous
  • anonymous
and that is the answer :-)
Nnesha
  • Nnesha
separate it \[ \huge\rm \frac{\color{ReD}{ sinx} }{ cosx } \times (1-cos^2x) \]
anonymous
  • anonymous
WOW, I cannot believe we got a problem like this. Some of these things take a long time.
Nnesha
  • Nnesha
yay! done!
Nnesha
  • Nnesha
practice more and more! then you will be able to finish it just in 5 minutes :3
anonymous
  • anonymous
You work them one way then come to find out it is wrong then work them another and repeat and finally you get it lol This was the worst one yet!!! thank you for your help!!!!
anonymous
  • anonymous
I do not have time to practice we have a week to fit a lot in and then we move on. Next week we probably will not even mess with these. idk
Nnesha
  • Nnesha
ohh but still try do 2 question every night :3 = nightmares haha jk
Nnesha
  • Nnesha
gO_Od luck!!!
anonymous
  • anonymous
Thank you again!!!

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