anonymous
  • anonymous
sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.
Nnesha
  • Nnesha
\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 -cosine(x) }\] what is the common denominator ?
anonymous
  • anonymous
1-cosine(x)?

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anonymous
  • anonymous
but the second one is sinx/1+cosx
Nnesha
  • Nnesha
well the common denominator is 1-cosine(x) obviously both denominator are same u will get same numerator (sinx)\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } \]
Nnesha
  • Nnesha
huh uh-oh are you sure check your question again
anonymous
  • anonymous
yea i checked it 1+cosx
Nnesha
  • Nnesha
http://prntscr.com/7vfzly
Nnesha
  • Nnesha
\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 +cosine(x) }\] so common denominator is (1-cosinex)(1+cosinex)
anonymous
  • anonymous
but the picture of the equation in the book says 1+cosx
Nnesha
  • Nnesha
alright you typed it wrong
anonymous
  • anonymous
it did that itself it copy pasted that way lol
anonymous
  • anonymous
1 Attachment
Nnesha
  • Nnesha
really ? :o you didin't type it ?
Nnesha
  • Nnesha
shocking how that can translate :o :/
anonymous
  • anonymous
lol no that was weird tho right?
Nnesha
  • Nnesha
\[\huge\rm \frac{ sinx(1+cosx) +sinx(1-cosx)}{ (1-cosx)(1+cosx)}\] common denominator is (1-cosx)(1+cosx) and multiply numerator of first fraction by denominator of 2nd fraction *and multiply numerator of 2nd fraction by denominator of first fraction
Nnesha
  • Nnesha
ye ....
anonymous
  • anonymous
and then you cancel out?
Nnesha
  • Nnesha
no no NO! no then u will get the original question
Nnesha
  • Nnesha
distribute!
anonymous
  • anonymous
ohhh
anonymous
  • anonymous
wait what
Nnesha
  • Nnesha
yes familiar withe the foil method ? apply that at the denominator
anonymous
  • anonymous
oh okay so i leave the numerator alone righ tnow
anonymous
  • anonymous
1-cos2x?
Nnesha
  • Nnesha
yep right now distribute both parentheses by sinx 't the numerator
anonymous
  • anonymous
what? that last sentence you said threw me off i guess the wording did
Nnesha
  • Nnesha
\[\huge\rm \frac{ \color{reD}{sinx}(1+cosx) +\color{red}{sinx}(1-cosx)}{ (1-cosx)^2}\] distribute both parentheses by red sin x
anonymous
  • anonymous
1sinx+sinxcosx+1sinx-sinxcosx
anonymous
  • anonymous
2sinx/1-cos2x
Nnesha
  • Nnesha
and cos^2x equal to what ? do u remember sin/cos identities ?
anonymous
  • anonymous
csc?
Nnesha
  • Nnesha
nope
Nnesha
  • Nnesha
special identity \[\huge\rm sin^2x + \cos^2 x = 1\] solve this for cos^2x
anonymous
  • anonymous
some codes came out for when you said special identity nothing else?
Nnesha
  • Nnesha
sin^2x + cos^2x = 1 solve for cos^2 x
anonymous
  • anonymous
cos^2x=1-sin^2x
Nnesha
  • Nnesha
yes right so replace cos^2x by 1-sin^2x
anonymous
  • anonymous
okay
Nnesha
  • Nnesha
\[\huge\rm \frac{ \color{reD}{2sinx}}{ 1-(1-sin^2x)}\] distribute parentheses by negative one
anonymous
  • anonymous
its coding
Nnesha
  • Nnesha
ahh 2sinx over 1-(1-sin^2x)
anonymous
  • anonymous
sorry lol im being a pain
anonymous
  • anonymous
2sinx/sin^2x
Nnesha
  • Nnesha
yes sin^2x is same as sinx times sin x so 2sin over sinx sin x simplfy
anonymous
  • anonymous
1/sin2x
Nnesha
  • Nnesha
is it sin^2 x ?
anonymous
  • anonymous
yea but with one over it
Nnesha
  • Nnesha
no what about 2 ? it's 2sinx over sinx times sin x
Nnesha
  • Nnesha
|dw:1437522158545:dw|
anonymous
  • anonymous
it all cancels out to 1/sinx
Nnesha
  • Nnesha
no what about 2 ??|dw:1437522218631:dw|
anonymous
  • anonymous
im so lost i dont like this number two lol
Nnesha
  • Nnesha
just look at that drawing who kidnapped that 2 ?
anonymous
  • anonymous
the sinx? i have no idea
Nnesha
  • Nnesha
sin x cancels out right ?? so where is 2 ? look at there
anonymous
  • anonymous
2 over sinx
Nnesha
  • Nnesha
ahh there ugo!
Nnesha
  • Nnesha
i was about ...to .. put u in jail how....darrrrrrrrrrrrrre u kidnapped 2 :/ ;) hey gtg gO_OD luck!
anonymous
  • anonymous
haha okay but what do i do to finish?
Nnesha
  • Nnesha
well 2 over sinx is same as 2 times 1/sinx and 1/sinx = csc 2 times csc

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