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?

- anonymous

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- katieb

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- Nnesha

foil! :-)
i guess that will be great

- anonymous

Just with tan
\[ \tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x) \] correct so far?

- Nnesha

yep looks right :-)

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## More answers

- 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

yes
convert 2nd tan to sin /cos
and let me find notebook and a pencil :3 hehe

- Nnesha

make sure question is right :3 :-)

- anonymous

Yes it is right :-)

- Nnesha

is it equal to
tanxsinx or tan^2xsinx ??

- 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

yeah so for the final answer i got sinx tanx
so that's why ..hm are u sure it's sin square ??

- Nnesha

next step is to change tan to sin/cos

- anonymous

Yes, it is sin^2 x tan x

- Nnesha

:( alright

- Nnesha

alright got it!! :-)

- Nnesha

hahah i'm soo happy :P

- 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

\( \huge\rm \frac{sin x}{cos x} -(\sin x)(\cos x) = \) ????
find common denominator :-)

- anonymous

Got ya :-)

- anonymous

one sec

- Nnesha

alright :-)

- anonymous

I am using cos as the common denominator. that is correct?

- Nnesha

yes right

- 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

\[ \frac{(sin x)(\cancel{cos x})}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x \]
multiply cosx/1
so when you multiply first fraction by cos both cos will cancel each other out

- Nnesha

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

Oh right!!! Made a boo boo

- anonymous

One sec, now we should have

- 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

\[ \frac{(sin x)}{\cos x} -\frac{((\sin x (1-cos^2 x))}{cos x} = \sin^2 x \tan x \]

- Nnesha

\[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos \]
multiply |dw:1434231940649:dw|

- 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

Yes that makes sence

- 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

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

we know that common denominator is cos so we multiplied numerator by cos x

- 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

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

first of all we know that cosx is common denominator so that should be at the denominator \[\frac{ ?????? }{ \cos x }\]

- 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

wait a sec that's not cos^2 yet

- Nnesha

let me fix it :-)

- 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

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

Ok, I think we are on the same page now WOW this stuff gets confusing lol
Ok I am where you are at

- Nnesha

what ??

- Nnesha

do you got that denominator thingy ?

- anonymous

So now we have
\[ \large\rm\frac{ sinx }{{ 1} } - \frac{ sinx \cos^2 x }{ 1 } \]

- Nnesha

that's the numerator
what's our common denominator ?

- 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

sinx/1 = sinx
so you can write it as \[\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }\]

- anonymous

Yes correct

- 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

want to make sure!!

- anonymous

Yes I understand
Next ?
\[ \huge\rm \frac{ sinx - sinx(1-sin^2) }{ cosx } \]

- Nnesha

nope next step is to take out common term

- Nnesha

\[ \huge\rm \frac{\color{ReD}{ sinx} -\color{ReD}{ sinx}cos^2x}{ cosx } \]
factor the numerator :-)

- Nnesha

let sinx = a
cosx = b
so you can write it as \[\large\rm (a-ab^2)\]

- anonymous

So this \( (a-ab^2) = (ab^2-a) = (ab+a)(ab-a) \) correct?

- anonymous

I cannot believe this one problem is like this. It has been an hour and 30 minutes lol

- Nnesha

nope what is common in these both terms a - ab^2 ??

- Nnesha

almost done .-.

- anonymous

so a(1-b^2)

- Nnesha

yes right!!!
so now change it back to sin cos

- Nnesha

\[ \huge\rm \frac{\color{ReD}{ sinx}( 1-cos^2x)}{ cosx } \]

- anonymous

and that is the answer :-)

- Nnesha

separate it
\[ \huge\rm \frac{\color{ReD}{ sinx} }{ cosx } \times (1-cos^2x) \]

- anonymous

WOW, I cannot believe we got a problem like this. Some of these things take a long time.

- Nnesha

yay! done!

- Nnesha

practice more and more! then you will be able to finish it just in 5 minutes :3

- 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

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

ohh but still try do 2 question every night :3 = nightmares haha jk

- Nnesha

gO_Od luck!!!

- anonymous

Thank you again!!!

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