anonymous
  • anonymous
I bet you can't simplify this:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AlexandervonHumboldt2
  • AlexandervonHumboldt2
simplifying nothing is impossible xD. what is your question?
anonymous
  • anonymous
there is no such mathematical operation as "simplify" so i agree, you cannot do it
anonymous
  • anonymous
lol what is the q

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anonymous
  • anonymous
\[Cos/\sin + 1/\sin \times \sin\]
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
not really clear what that is the equation tool or a screen shot might help
anonymous
  • anonymous
hello misty
anonymous
  • anonymous
um, how do I screen shot ^_^?
AlexandervonHumboldt2
  • AlexandervonHumboldt2
is it \[\frac{ Cos }{ \frac{ \sin+1 }{ \sin*\sin } }\]?
AlexandervonHumboldt2
  • AlexandervonHumboldt2
to make a screen shot press PrtSc on your keyboard, then open paint, paste your picture and then save file and upload it using Attach file button.
anonymous
  • anonymous
no it's cos over sin + 1 over sin times sin if you can read that
Nnesha
  • Nnesha
looks like \[\frac{\color{ReD}{\frac{ \cos }{ \sin +1}} }{ \sin \times \sin }\]
anonymous
  • anonymous
|dw:1446606823293:dw|
anonymous
  • anonymous
try not to be marveled by my picassoesque drawing
misty1212
  • misty1212
try the button below with the little \(\Sigma\)
anonymous
  • anonymous
\[\frac{ \cos }{ \sin }\]
anonymous
  • anonymous
it didnt work
Nnesha
  • Nnesha
\[\large\rm \frac{ \cos^2 \theta }{ \sin^2 \theta } *\csc \theta *\sin \theta \] like this ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
except its plus csc
Nnesha
  • Nnesha
\[\large\rm \frac{ \cos^2 \theta }{ \sin^2 \theta } + \csc \theta *\sin \theta \] like this ?
anonymous
  • anonymous
yes
Nnesha
  • Nnesha
what's the reciprocal of csc ??
anonymous
  • anonymous
1/sin
Nnesha
  • Nnesha
right we can replace cscx with 1/sinx \[\large\rm \frac{ \cos^2 \theta }{ \sin^2 \theta } + \frac{1}{sin \theta} *\sin \theta \] 1/sin times sin = sin/sin \[\large\rm \frac{ \cos^2 \theta }{ \sin^2 \theta } + \frac{sin \theta}{sin \theta} \]
Nnesha
  • Nnesha
pretty sure you can take t from there
anonymous
  • anonymous
Cant I jsut cancel out the 2 sines and leave the 1?
Nnesha
  • Nnesha
that's correct!
anonymous
  • anonymous
then what is my next step when i have cos/sin+1? is there a trig identity?
Nnesha
  • Nnesha
\[\large\rm \frac{ \cos^2 \theta }{ \sin^2 \theta } + 1 \] cos/sin = tan cos^2/sin^2= ?
anonymous
  • anonymous
cot^2?
Nnesha
  • Nnesha
|dw:1446607700306:dw| tan = sin/cos
Nnesha
  • Nnesha
ohh yea that's correct i was thinking about sin/cos which is equal to tan cot^2x is correct
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\large\rm \frac{ \cos^2 \theta }{ \sin^2 \theta } + 1 \] cos/sin = tan cos^2/sin^2= ? \(\color{blue}{\text{End of Quote}}\) i meant to say cos/sin = cot ._. sorry about that ,-,
anonymous
  • anonymous
thats okay ^_^
Nnesha
  • Nnesha
there are two ways to simplify this furthermore if you're familiar with pythagorean Identities \[\cot^2x +1 = \csc^2x \] or keep the cos^/sin^2 fraction and find the common denominator \[\frac{ \cos^2x }{ \sin^2x }+1\]
anonymous
  • anonymous
ohhh, thank you so much. I should probably master these pyth identities.
Nnesha
  • Nnesha
bookmark this website http://www.analyzemath.com/trigonometry/trigonometric_formulas.html
Nnesha
  • Nnesha
\[\frac{ \cos^2 x}{ \sin^2 x}+1\] common denomiantor is sin^2 so \[\frac{ \cos^2 x + \sin^2x }{ \sin^2x }\] use the fact sin^2x+cos^2x= 1 \[\frac{1 }{ \sin^2x }\] and then replace 1/sin2 with its reciprocal

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