Question

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?

Answer 1

foil! :-)
i guess that will be great

Answer 2

Just with tan


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

Answer 3

yep looks right :-)

Answer 4

Ok, how do I approch it from here? do I turn all the tan into sin / cos? and work it out?
\[ \tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x) \]

Answer 5

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

Answer 6

make sure question is right :3 :-)

Answer 7

Yes it is right :-)

Answer 8

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

Answer 9

I have
\[ \tan x -\sin x + \sin x -(\sin x)(\cos x) = \sin^2 x \tan x \]
\[ \tan x -(\sin x)(\cos x) = \sin^2 x \tan x \]

Answer 10

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

Answer 11

next step is to change tan to sin/cos

Answer 12

Yes, it is sin^2 x tan x

Answer 13

:( alright

Answer 14

alright got it!! :-)

Answer 15

hahah i'm soo happy :P

Answer 16

\( \tan x -(\sin x)(\cos x) = \sin^2 x \tan x \)
\( \frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x \) ????

Answer 17

\( \huge\rm \frac{sin x}{cos x} -(\sin x)(\cos x) = \) ????

find common denominator :-)

Answer 18

Got ya :-)

Answer 19

one sec

Answer 20

alright :-)

Answer 21

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

Answer 22

yes right

Answer 23

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???

Answer 24

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

multiply cosx/1
so when you multiply first fraction by cos both cos will cancel each other out

Answer 25

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

Answer 26

Oh right!!! Made a boo boo

Answer 27

One sec, now we should have

Answer 28

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

Ok I am lost

Answer 29

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

Answer 30

\[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos \]
multiply