anonymous
  • anonymous
Verify the identity. cotangent of x to the second power divided by quantity cosecant of x plus one equals quantity one minus sine of x divided by sine of x
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!
triciaal
  • triciaal
|dw:1443371271009:dw|
triciaal
  • triciaal
the original as written is not clear.
anonymous
  • anonymous
1-2sin^2 x/ sin^2 x / 1+sin x/ sin x = 1- sin^2 x/ sinx * 1+ sinx

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Nnesha
  • Nnesha
\[\huge\rm \frac{ \cot^2 (x) }{ \csc(x) +1 } =\frac{ 1-\sin(x)}{ \sin(x)}\] like this ?
anonymous
  • anonymous
I know if I factor 1-(sinx)^2 I'll get 1+sinx and 1-sinx
anonymous
  • anonymous
yes like that^
Nnesha
  • Nnesha
alright gimme a sec
Nnesha
  • Nnesha
are you working on left or right side ?
anonymous
  • anonymous
left
Nnesha
  • Nnesha
alright first we can rewrite cot and csc in terms of sin and cos
Nnesha
  • Nnesha
csc = 1/sin right so cot = ??
anonymous
  • anonymous
1/tan?
anonymous
  • anonymous
cos/sin
Nnesha
  • Nnesha
yeah but we need to write in terms of sin cos
Nnesha
  • Nnesha
yes right \[\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1 }{ \sin(x) }+1} }\] now first deal with the denominator (red part)
Nnesha
  • Nnesha
\[\frac{ 1 }{ \sin(x) }+1\] find the common denominator
anonymous
  • anonymous
well 1/ sin = csc and 1+ csc = cot^2?
anonymous
  • anonymous
I'm confused.
Nnesha
  • Nnesha
yes right you can use the identity but i found the other one easy
Nnesha
  • Nnesha
but if you use the identity you still have to write csc in terms of sin and cos
anonymous
  • anonymous
so what do I do, I know csc is 1/sin
Nnesha
  • Nnesha
\[\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1 }{ \sin(x) }+1} }\] now first deal with the denominator (red part) we need to find common denominator 1 is same as 1/1 so \[\frac{ 1}{ \sin(x)} +\frac{ 1 }{ 1 }\] common denomiantor is sin x right
Nnesha
  • Nnesha
\[\large\rm \frac{ 1 }{ \color{blue}{\sin(x)} }+\frac{ 1 }{ \color{red}{1}}\] \[\huge\rm \frac{ \color{red}{1}(1) +\color{blue}{sinx }(1)}{ \sin(x) }\] when we find common denominator we should multiply first fraction with the denominator of 2nd one and multiply the numerator of 2nd fraction with the denominator of first fraction
Nnesha
  • Nnesha
let me know if you have a question about that part ??^^^
anonymous
  • anonymous
No it makes sense.
Nnesha
  • Nnesha
alright \[\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1+sin(x) }{ \sin(x) }} }\] now change division to multiplication multiply first fraction with the `reciprocal` of the 2nd fraction
Nnesha
  • Nnesha
here is an example \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } } =\frac{ a }{ b} \times \frac{ d }{ c }\]
Nnesha
  • Nnesha
no multiply top fraction with the reciprocal of bottom one what's reciprocal of 1+sinx/sin ???
anonymous
  • anonymous
sin/1+sinx
Nnesha
  • Nnesha
right so multiply \[\huge\rm \frac{ \cos^2(x) }{\sin^2(x) } \times \frac{ \sin(x) }{1+\sin(x)}\]
Nnesha
  • Nnesha
sin^2 is same as sin times sin so you cancel it out
Nnesha
  • Nnesha
\[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sin^2(x)} } \times \frac{ \sin(x) }{1+\sin(x)}\] sin^2 is same as sin times sin so you cancel it out \[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times sinx} } \times \frac{ \sin(x) }{1+\sin(x)}\]
anonymous
  • anonymous
so does that mean sinx cancels out?
Nnesha
  • Nnesha
yes right
Nnesha
  • Nnesha
\[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times \cancel{sinx}} } \times \frac{\cancel{ \sin(x)} }{1+\sin(x)}\]
Nnesha
  • Nnesha
now use the identity cos^2(x) = ???
anonymous
  • anonymous
cos^2 = cosx * cosx?
Nnesha
  • Nnesha
yes but we need identity
Nnesha
  • Nnesha
sin^2(x) +cos^2(x) =1 solve for cos^2(x)
anonymous
  • anonymous
oh wait its a path. identity.
Nnesha
  • Nnesha
that's the identity
Nnesha
  • Nnesha
ye
anonymous
  • anonymous
That's it?
Nnesha
  • Nnesha
well ye just replace cos^2 with the identity then done
Nnesha
  • Nnesha
cos^2x = ???
anonymous
  • anonymous
sin^2 + cos^2
anonymous
  • anonymous
cos^2=sin^2?
Nnesha
  • Nnesha
how would you cancel out sin^2 from left side \[\huge\rm sin^2x+\cos^2=1\] ?
Nnesha
  • Nnesha
subtract/add/divide /mutliply both sides by sin^2 x ?
anonymous
  • anonymous
I don't understand, how? ^
Nnesha
  • Nnesha
yes i'm asking how would you cancel out sin^2x from left its just like simple equation
Nnesha
  • Nnesha
lets say we have x +y = 1 how would you solve for y ?
anonymous
  • anonymous
you move x to the other side right?
Nnesha
  • Nnesha
yes right what would you get ?
anonymous
  • anonymous
not enough info to solve, because x=?
Nnesha
  • Nnesha
well lets say 2+x =1 how would you solve for x ?
anonymous
  • anonymous
-2 both sides x = -1
Nnesha
  • Nnesha
yes right subtract 2 both sides so for this equation subtract sin2x both sides \[\huge\rm sin^2x+\cos^2=1\]
Nnesha
  • Nnesha
|dw:1443375444006:dw| what would you get at right side ?
anonymous
  • anonymous
cos^2= 1-sin^2
Nnesha
  • Nnesha
yes right now u can replace cos^2x with that \[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times \cancel{sinx}} } \times \frac{\cancel{ \sin(x)} }{1+\sin(x)}\] \[\huge\rm \frac{ 1-sin^2(x)}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}\]
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @clairvoyant1 I know if I factor 1-(sinx)^2 I'll get 1+sinx and 1-sinx \(\color{blue}{\text{End of Quote}}\) now as u said you can write 1-sin^2x as (1+sinx)(1-sinx)
Nnesha
  • Nnesha
\[\huge\rm \frac{ (1+sin(x))(1-sin(x))}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}\]
Nnesha
  • Nnesha
you can cancel out something right after that done!
Nnesha
  • Nnesha
got it ?? :=)
anonymous
  • anonymous
Computer froze logged in on my kindle Lol. Not quite
anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
\[\huge\rm \frac{ (1+sin(x))(1-sin(x))}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}\] what can you cancel out ?
anonymous
  • anonymous
1+sin(x)
Nnesha
  • Nnesha
\[\large\rm \frac{ (1-sin(x))\cancel{(1+sin(x))} }{\color{reD}{sinx } } \times \frac{{ 1}}{\cancel{1+\sin(x)}}\]
Nnesha
  • Nnesha
that's it
Nnesha
  • Nnesha
both sides are the same right ?
anonymous
  • anonymous
Ok could you do a quick recap of what we just went over please.
anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
first rewrite cot and csc in terms of sin and cos then we found the common denominator change division to multiplication use the identity cos^2=1-sin^2 that's it
Nnesha
  • Nnesha
if you need the work you can reread the post :=) feel free to ask q
anonymous
  • anonymous
Thank You so much for your patience and help :)
Nnesha
  • Nnesha
np :=)
anonymous
  • anonymous
Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x
anonymous
  • anonymous
@Nnesha
anonymous
  • anonymous
Can you help me with this one.
Nnesha
  • Nnesha
i'm not sure about that one sorry i just know the basic :(
Nnesha
  • Nnesha
make new post so other people will b able to help chu
anonymous
  • anonymous
Find the domain of the given function. f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two. a) x > 0 b) All real numbers c) x ≥ -3, x ≠ 2 d) x ≠ -8, x ≠ -3, x ≠ 2
Nnesha
  • Nnesha
can you post the pic ?
anonymous
  • anonymous
Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x @campbell_st
campbell_st
  • campbell_st
well cos(4x )= 1 - 2sin^2(2x) and cos(2x) = 1- 2sin^2(x) now make the substitution and simplify
anonymous
  • anonymous
@campbell_st how would I do that?
campbell_st
  • campbell_st
exactly what it says... replace cos(4x) with the right hand side of the expression then the same with cos(2x) and collect like terms. what would probably be best, is to post it as a new question... as there is an lot of information that precedes the question you asked.
anonymous
  • anonymous
alright I'll post it as a new question

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