One sided trig identity with conjugates $frac{(cos xcot x)}{cot x-cos x}=frac{cot x+cos x}{(cos xcot x)}$
up vote
-1
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I've been given a trigonometric identity worksheet. We need to use conjugates to solve these identities. One sided proofs. I've spent some time on it but just can't crack it. Here it is, help appreciated. Any tips (such as which side to start on etc) is appreciated.
$$frac{(cos xcot x)}{cot x-cos x}=frac{cot x+cos x}{(cos xcot x)}$$
trigonometry
edited Nov 29 at 4:51
Tianlalu
3,01021038
asked Nov 29 at 4:43
Korvexius
42
add a comment |
up vote
-1
down vote
favorite
I've been given a trigonometric identity worksheet. We need to use conjugates to solve these identities. One sided proofs. I've spent some time on it but just can't crack it. Here it is, help appreciated. Any tips (such as which side to start on etc) is appreciated.
$$frac{(cos xcot x)}{cot x-cos x}=frac{cot x+cos x}{(cos xcot x)}$$
trigonometry
edited Nov 29 at 4:51
Tianlalu
3,01021038
asked Nov 29 at 4:43
Korvexius
42
Hint: use conjugation on either side. Notice symmetry.
– Sean Roberson
Nov 29 at 4:45
I tried applying (cotx+cosx) to the left side and got a few steps in, though I wasn't able to finish it up.
– Korvexius
Nov 29 at 4:46
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I've been given a trigonometric identity worksheet. We need to use conjugates to solve these identities. One sided proofs. I've spent some time on it but just can't crack it. Here it is, help appreciated. Any tips (such as which side to start on etc) is appreciated.
$$frac{(cos xcot x)}{cot x-cos x}=frac{cot x+cos x}{(cos xcot x)}$$
trigonometry
edited Nov 29 at 4:51
Tianlalu
3,01021038
asked Nov 29 at 4:43
Korvexius
42
I've been given a trigonometric identity worksheet. We need to use conjugates to solve these identities. One sided proofs. I've spent some time on it but just can't crack it. Here it is, help appreciated. Any tips (such as which side to start on etc) is appreciated.
$$frac{(cos xcot x)}{cot x-cos x}=frac{cot x+cos x}{(cos xcot x)}$$
trigonometry
trigonometry
edited Nov 29 at 4:51
Tianlalu
3,01021038
asked Nov 29 at 4:43
Korvexius
42
edited Nov 29 at 4:51
Tianlalu
3,01021038
asked Nov 29 at 4:43
Korvexius
42
edited Nov 29 at 4:51
Tianlalu
3,01021038
edited Nov 29 at 4:51
Tianlalu
3,01021038
edited Nov 29 at 4:51
Tianlalu
3,01021038
3,01021038
asked Nov 29 at 4:43
Korvexius
42
asked Nov 29 at 4:43
Korvexius
42
asked Nov 29 at 4:43
Korvexius
42
42
Hint: use conjugation on either side. Notice symmetry.
– Sean Roberson
Nov 29 at 4:45
I tried applying (cotx+cosx) to the left side and got a few steps in, though I wasn't able to finish it up.
– Korvexius
Nov 29 at 4:46
add a comment |
Hint: use conjugation on either side. Notice symmetry.
– Sean Roberson
Nov 29 at 4:45
I tried applying (cotx+cosx) to the left side and got a few steps in, though I wasn't able to finish it up.
– Korvexius
Nov 29 at 4:46
Hint: use conjugation on either side. Notice symmetry.
– Sean Roberson
Nov 29 at 4:45
Hint: use conjugation on either side. Notice symmetry.
– Sean Roberson
Nov 29 at 4:45
I tried applying (cotx+cosx) to the left side and got a few steps in, though I wasn't able to finish it up.
– Korvexius
Nov 29 at 4:46
I tried applying (cotx+cosx) to the left side and got a few steps in, though I wasn't able to finish it up.
– Korvexius
Nov 29 at 4:46
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
As $cos x=cot xcdotsin x$ for $sin xne0,$
$$(cot x+cos x)(cot x-cos x)=cot^2xleft(1-sin^2xright)=?$$
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
add a comment |
up vote
1
down vote
$$dfrac{cos xcot x}{cot x-cos x}=cdots=dfrac{cos x}{1-sin x}$$
Similarly simplify the right hand side
Finally use $$cos^2x=(1-sin x)(1+sin x)$$
$$impliesdfrac{cos x}{1-sin x}=?$$
Alternatively, $$dfrac{cot x-cos x}{cot xcos x}=dfrac1{cos x}-dfrac1{cot x}=sec x-tan x$$
which is $=dfrac1{sec x+tan x}=?$
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
add a comment |
up vote
1
down vote
For $sin xneq0$,
$dfrac{(cos xcot x)}{cot x-cos x}=dfrac{(cos xcot x)}{cos x(csc x-1)}=dfrac{cot x}{csc x-1}cdotdfrac{csc x+1}{csc x+1}=dfrac{csc xcot x+cot x}{cot^2x}=dfrac{dfrac{cos x}{sin^2 x}+dfrac{cos x}{sin x}}{dfrac{cos^2x}{sin^2 x}}=dfrac{dfrac{cos x+sin xcos x}{sin^2 x}}{dfrac{cos^2x}{sin^2 x}}={dfrac{cos x+sin xcos x}{sin^2 x}}cdot{dfrac{sin^2 x}{cos^2 x}}={dfrac{cos x+sin xcos x}{sin x}}cdot{dfrac{sin x}{cos^2 x}}=left(cot x+cos xright)dfrac{1}{cos xcot x}=dfrac{cot x+cos x}{(cos xcot x)}$
answered Nov 29 at 4:53
Yadati Kiran
1,638519
1
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
As $cos x=cot xcdotsin x$ for $sin xne0,$
$$(cot x+cos x)(cot x-cos x)=cot^2xleft(1-sin^2xright)=?$$
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
add a comment |
up vote
3
down vote
As $cos x=cot xcdotsin x$ for $sin xne0,$
$$(cot x+cos x)(cot x-cos x)=cot^2xleft(1-sin^2xright)=?$$
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
add a comment |
up vote
3
down vote
up vote
3
down vote
As $cos x=cot xcdotsin x$ for $sin xne0,$
$$(cot x+cos x)(cot x-cos x)=cot^2xleft(1-sin^2xright)=?$$
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
As $cos x=cot xcdotsin x$ for $sin xne0,$
$$(cot x+cos x)(cot x-cos x)=cot^2xleft(1-sin^2xright)=?$$
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
answered Nov 29 at 4:52
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
1
down vote
$$dfrac{cos xcot x}{cot x-cos x}=cdots=dfrac{cos x}{1-sin x}$$
Similarly simplify the right hand side
Finally use $$cos^2x=(1-sin x)(1+sin x)$$
$$impliesdfrac{cos x}{1-sin x}=?$$
Alternatively, $$dfrac{cot x-cos x}{cot xcos x}=dfrac1{cos x}-dfrac1{cot x}=sec x-tan x$$
which is $=dfrac1{sec x+tan x}=?$
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
add a comment |
up vote
1
down vote
$$dfrac{cos xcot x}{cot x-cos x}=cdots=dfrac{cos x}{1-sin x}$$
Similarly simplify the right hand side
Finally use $$cos^2x=(1-sin x)(1+sin x)$$
$$impliesdfrac{cos x}{1-sin x}=?$$
Alternatively, $$dfrac{cot x-cos x}{cot xcos x}=dfrac1{cos x}-dfrac1{cot x}=sec x-tan x$$
which is $=dfrac1{sec x+tan x}=?$
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
add a comment |
up vote
1
down vote
up vote
1
down vote
$$dfrac{cos xcot x}{cot x-cos x}=cdots=dfrac{cos x}{1-sin x}$$
Similarly simplify the right hand side
Finally use $$cos^2x=(1-sin x)(1+sin x)$$
$$impliesdfrac{cos x}{1-sin x}=?$$
Alternatively, $$dfrac{cot x-cos x}{cot xcos x}=dfrac1{cos x}-dfrac1{cot x}=sec x-tan x$$
which is $=dfrac1{sec x+tan x}=?$
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
$$dfrac{cos xcot x}{cot x-cos x}=cdots=dfrac{cos x}{1-sin x}$$
Similarly simplify the right hand side
Finally use $$cos^2x=(1-sin x)(1+sin x)$$
$$impliesdfrac{cos x}{1-sin x}=?$$
Alternatively, $$dfrac{cot x-cos x}{cot xcos x}=dfrac1{cos x}-dfrac1{cot x}=sec x-tan x$$
which is $=dfrac1{sec x+tan x}=?$
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
edited Nov 29 at 5:17
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
answered Nov 29 at 5:05
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
1
down vote
For $sin xneq0$,
$dfrac{(cos xcot x)}{cot x-cos x}=dfrac{(cos xcot x)}{cos x(csc x-1)}=dfrac{cot x}{csc x-1}cdotdfrac{csc x+1}{csc x+1}=dfrac{csc xcot x+cot x}{cot^2x}=dfrac{dfrac{cos x}{sin^2 x}+dfrac{cos x}{sin x}}{dfrac{cos^2x}{sin^2 x}}=dfrac{dfrac{cos x+sin xcos x}{sin^2 x}}{dfrac{cos^2x}{sin^2 x}}={dfrac{cos x+sin xcos x}{sin^2 x}}cdot{dfrac{sin^2 x}{cos^2 x}}={dfrac{cos x+sin xcos x}{sin x}}cdot{dfrac{sin x}{cos^2 x}}=left(cot x+cos xright)dfrac{1}{cos xcot x}=dfrac{cot x+cos x}{(cos xcot x)}$
answered Nov 29 at 4:53
Yadati Kiran
1,638519
1
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
add a comment |
up vote
1
down vote
For $sin xneq0$,
$dfrac{(cos xcot x)}{cot x-cos x}=dfrac{(cos xcot x)}{cos x(csc x-1)}=dfrac{cot x}{csc x-1}cdotdfrac{csc x+1}{csc x+1}=dfrac{csc xcot x+cot x}{cot^2x}=dfrac{dfrac{cos x}{sin^2 x}+dfrac{cos x}{sin x}}{dfrac{cos^2x}{sin^2 x}}=dfrac{dfrac{cos x+sin xcos x}{sin^2 x}}{dfrac{cos^2x}{sin^2 x}}={dfrac{cos x+sin xcos x}{sin^2 x}}cdot{dfrac{sin^2 x}{cos^2 x}}={dfrac{cos x+sin xcos x}{sin x}}cdot{dfrac{sin x}{cos^2 x}}=left(cot x+cos xright)dfrac{1}{cos xcot x}=dfrac{cot x+cos x}{(cos xcot x)}$
answered Nov 29 at 4:53
Yadati Kiran
1,638519
1
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
add a comment |
up vote
1
down vote
up vote
1
down vote
For $sin xneq0$,
$dfrac{(cos xcot x)}{cot x-cos x}=dfrac{(cos xcot x)}{cos x(csc x-1)}=dfrac{cot x}{csc x-1}cdotdfrac{csc x+1}{csc x+1}=dfrac{csc xcot x+cot x}{cot^2x}=dfrac{dfrac{cos x}{sin^2 x}+dfrac{cos x}{sin x}}{dfrac{cos^2x}{sin^2 x}}=dfrac{dfrac{cos x+sin xcos x}{sin^2 x}}{dfrac{cos^2x}{sin^2 x}}={dfrac{cos x+sin xcos x}{sin^2 x}}cdot{dfrac{sin^2 x}{cos^2 x}}={dfrac{cos x+sin xcos x}{sin x}}cdot{dfrac{sin x}{cos^2 x}}=left(cot x+cos xright)dfrac{1}{cos xcot x}=dfrac{cot x+cos x}{(cos xcot x)}$
answered Nov 29 at 4:53
Yadati Kiran
1,638519
For $sin xneq0$,
$dfrac{(cos xcot x)}{cot x-cos x}=dfrac{(cos xcot x)}{cos x(csc x-1)}=dfrac{cot x}{csc x-1}cdotdfrac{csc x+1}{csc x+1}=dfrac{csc xcot x+cot x}{cot^2x}=dfrac{dfrac{cos x}{sin^2 x}+dfrac{cos x}{sin x}}{dfrac{cos^2x}{sin^2 x}}=dfrac{dfrac{cos x+sin xcos x}{sin^2 x}}{dfrac{cos^2x}{sin^2 x}}={dfrac{cos x+sin xcos x}{sin^2 x}}cdot{dfrac{sin^2 x}{cos^2 x}}={dfrac{cos x+sin xcos x}{sin x}}cdot{dfrac{sin x}{cos^2 x}}=left(cot x+cos xright)dfrac{1}{cos xcot x}=dfrac{cot x+cos x}{(cos xcot x)}$
answered Nov 29 at 4:53
Yadati Kiran
1,638519
edited Nov 29 at 6:03
answered Nov 29 at 4:53
Yadati Kiran
1,638519
answered Nov 29 at 4:53
Yadati Kiran
1,638519
answered Nov 29 at 4:53
Yadati Kiran
1,638519
1,638519
1
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
add a comment |
1
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
1
1
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
It should be $sin xne0iff xne mpi$ for any integer $n$
– lab bhattacharjee
Nov 29 at 5:08
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
@labbhattacharjee: Yeah. Thanks.
– Yadati Kiran
Nov 29 at 5:10
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Thanks. Could you outline the final step a bit more? Are you using that a/b/c = a*c/b rule?
– Korvexius
Nov 29 at 5:11
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Yes but prior to that taking lcm in the numerator. $dfrac{frac{a}{b}}{frac{c}{d}}=dfrac{acdot d}{bcdot c}$.
– Yadati Kiran
Nov 29 at 5:27
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
Sorry, I tried working that out but still couldn't figure out how the full transition of the final step. Could you expand the answer itself some more steps, skipping the lcm part?
– Korvexius
Nov 29 at 5:47
add a comment |
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Hint: use conjugation on either side. Notice symmetry.
– Sean Roberson
Nov 29 at 4:45
I tried applying (cotx+cosx) to the left side and got a few steps in, though I wasn't able to finish it up.
– Korvexius
Nov 29 at 4:46