Simplify $ frac{4cos(x)}{1-cos(x)} - frac{4sin^2(x)}{(1-cos(x))^2}$ [closed]
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Spent any hour trying, but can't simplify this formula.
$$ frac{4cos(x)}{1-cos(x)} - frac{4sin^2(x)}{(1-cos(x))^2}$$
Calculator CAS tells me the result should be $dfrac{4}{cos(x) - 1}$, but I can't seem to do it myself. Thanks!
algebra-precalculus trigonometry
edited Nov 22 at 20:25
m0nhawk
1,49031128
asked Nov 22 at 19:55
Maxwell Stone
63
closed as off-topic by amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson Nov 22 at 20:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
Spent any hour trying, but can't simplify this formula.
$$ frac{4cos(x)}{1-cos(x)} - frac{4sin^2(x)}{(1-cos(x))^2}$$
Calculator CAS tells me the result should be $dfrac{4}{cos(x) - 1}$, but I can't seem to do it myself. Thanks!
algebra-precalculus trigonometry
edited Nov 22 at 20:25
m0nhawk
1,49031128
asked Nov 22 at 19:55
Maxwell Stone
63
closed as off-topic by amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson Nov 22 at 20:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: start with the common denominator of $(1- cos(x))^2$. Write the first fraction in terms of this denominator. Add the fractions. Simplify.
– amWhy
Nov 22 at 19:58
1
Isn't there a missing bracket in the second denominator?
– user376343
Nov 22 at 20:11
@amWhy Thanks for your help. Forgot how I could split (1-cosx^2) into (1-cosx) and (1+cosx).
– Maxwell Stone
Nov 22 at 20:12
@Mason Yeah that's right what am I thinking. Fixed it.
– Maxwell Stone
Nov 22 at 20:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Spent any hour trying, but can't simplify this formula.
$$ frac{4cos(x)}{1-cos(x)} - frac{4sin^2(x)}{(1-cos(x))^2}$$
Calculator CAS tells me the result should be $dfrac{4}{cos(x) - 1}$, but I can't seem to do it myself. Thanks!
algebra-precalculus trigonometry
edited Nov 22 at 20:25
m0nhawk
1,49031128
asked Nov 22 at 19:55
Maxwell Stone
63
Spent any hour trying, but can't simplify this formula.
$$ frac{4cos(x)}{1-cos(x)} - frac{4sin^2(x)}{(1-cos(x))^2}$$
Calculator CAS tells me the result should be $dfrac{4}{cos(x) - 1}$, but I can't seem to do it myself. Thanks!
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Nov 22 at 20:25
m0nhawk
1,49031128
asked Nov 22 at 19:55
Maxwell Stone
63
edited Nov 22 at 20:25
m0nhawk
1,49031128
asked Nov 22 at 19:55
Maxwell Stone
63
edited Nov 22 at 20:25
m0nhawk
1,49031128
edited Nov 22 at 20:25
m0nhawk
1,49031128
edited Nov 22 at 20:25
m0nhawk
1,49031128
1,49031128
asked Nov 22 at 19:55
Maxwell Stone
63
asked Nov 22 at 19:55
Maxwell Stone
63
asked Nov 22 at 19:55
Maxwell Stone
63
63
closed as off-topic by amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson Nov 22 at 20:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson Nov 22 at 20:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Scientifica, Arnaud D., Nosrati, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: start with the common denominator of $(1- cos(x))^2$. Write the first fraction in terms of this denominator. Add the fractions. Simplify.
– amWhy
Nov 22 at 19:58
1
Isn't there a missing bracket in the second denominator?
– user376343
Nov 22 at 20:11
@amWhy Thanks for your help. Forgot how I could split (1-cosx^2) into (1-cosx) and (1+cosx).
– Maxwell Stone
Nov 22 at 20:12
@Mason Yeah that's right what am I thinking. Fixed it.
– Maxwell Stone
Nov 22 at 20:15
add a comment |
Hint: start with the common denominator of $(1- cos(x))^2$. Write the first fraction in terms of this denominator. Add the fractions. Simplify.
– amWhy
Nov 22 at 19:58
1
Isn't there a missing bracket in the second denominator?
– user376343
Nov 22 at 20:11
@amWhy Thanks for your help. Forgot how I could split (1-cosx^2) into (1-cosx) and (1+cosx).
– Maxwell Stone
Nov 22 at 20:12
@Mason Yeah that's right what am I thinking. Fixed it.
– Maxwell Stone
Nov 22 at 20:15
Hint: start with the common denominator of $(1- cos(x))^2$. Write the first fraction in terms of this denominator. Add the fractions. Simplify.
– amWhy
Nov 22 at 19:58
Hint: start with the common denominator of $(1- cos(x))^2$. Write the first fraction in terms of this denominator. Add the fractions. Simplify.
– amWhy
Nov 22 at 19:58
1
1
Isn't there a missing bracket in the second denominator?
– user376343
Nov 22 at 20:11
Isn't there a missing bracket in the second denominator?
– user376343
Nov 22 at 20:11
@amWhy Thanks for your help. Forgot how I could split (1-cosx^2) into (1-cosx) and (1+cosx).
– Maxwell Stone
Nov 22 at 20:12
@amWhy Thanks for your help. Forgot how I could split (1-cosx^2) into (1-cosx) and (1+cosx).
– Maxwell Stone
Nov 22 at 20:12
@Mason Yeah that's right what am I thinking. Fixed it.
– Maxwell Stone
Nov 22 at 20:15
@Mason Yeah that's right what am I thinking. Fixed it.
– Maxwell Stone
Nov 22 at 20:15
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
HINT
Recall that
$$sin^2 x=1-cos^2 x=(1-cos x)(1+cos x)$$
therefore
$$frac{4cos x}{1-cos x } - frac{4sin^2 x}{(1-cos x)^2}=frac{4cos x}{1-cos x} - frac{4(1-cos x)(1+cos x)}{(1-cos x)^2}$$
answered Nov 22 at 20:03
gimusi
89.1k74495
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
3
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
2
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
|
show 8 more comments
up vote
2
down vote
$dfrac{4cos(x)(1-cos(x))-4sin^2(x)}{(1-cos(x))^2}=dfrac{4(cos x-1)}{(1-cos(x))^2}=dfrac{-4}{1-cos(x)}=dfrac{4}{cos(x) - 1}$.
(Using $sin^2x+cos^2x=1$).
answered Nov 22 at 20:06
Yadati Kiran
1,243417
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
HINT
Recall that
$$sin^2 x=1-cos^2 x=(1-cos x)(1+cos x)$$
therefore
$$frac{4cos x}{1-cos x } - frac{4sin^2 x}{(1-cos x)^2}=frac{4cos x}{1-cos x} - frac{4(1-cos x)(1+cos x)}{(1-cos x)^2}$$
answered Nov 22 at 20:03
gimusi
89.1k74495
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
3
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
2
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
|
show 8 more comments
up vote
0
down vote
accepted
HINT
Recall that
$$sin^2 x=1-cos^2 x=(1-cos x)(1+cos x)$$
therefore
$$frac{4cos x}{1-cos x } - frac{4sin^2 x}{(1-cos x)^2}=frac{4cos x}{1-cos x} - frac{4(1-cos x)(1+cos x)}{(1-cos x)^2}$$
answered Nov 22 at 20:03
gimusi
89.1k74495
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
3
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
2
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
|
show 8 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
HINT
Recall that
$$sin^2 x=1-cos^2 x=(1-cos x)(1+cos x)$$
therefore
$$frac{4cos x}{1-cos x } - frac{4sin^2 x}{(1-cos x)^2}=frac{4cos x}{1-cos x} - frac{4(1-cos x)(1+cos x)}{(1-cos x)^2}$$
answered Nov 22 at 20:03
gimusi
89.1k74495
HINT
Recall that
$$sin^2 x=1-cos^2 x=(1-cos x)(1+cos x)$$
therefore
$$frac{4cos x}{1-cos x } - frac{4sin^2 x}{(1-cos x)^2}=frac{4cos x}{1-cos x} - frac{4(1-cos x)(1+cos x)}{(1-cos x)^2}$$
answered Nov 22 at 20:03
gimusi
89.1k74495
edited Nov 22 at 20:04
answered Nov 22 at 20:03
gimusi
89.1k74495
answered Nov 22 at 20:03
gimusi
89.1k74495
answered Nov 22 at 20:03
gimusi
89.1k74495
89.1k74495
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
3
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
2
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
|
show 8 more comments
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
3
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
2
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
Thanks! I'm new to algebra and completely forgot I could simplify it that way.
– Maxwell Stone
Nov 22 at 20:05
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
@MaxwellStone As a suggestion for clearness don't use $sin (x)$ or $cos (x)$ but simply $sin x$ and $cos x$. Use parenthesis only for $sin (f(x))$ or similar expressions.
– gimusi
Nov 22 at 20:07
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
Thanks, I'll keep that in mind.
– Maxwell Stone
Nov 22 at 20:08
3
3
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
@gimusi You suggested that you omit the parentheses in the interest of clarity. The parentheses are, indeed, often omitted. This does not mean that it is more clear. Personally, I would always advise students to retain the parentheses. This emphasizes that $sin$ is a function which takes an argument. I would also suggest that students use the notation $sin(x)^2$ rather than $sin^2(x)$, as $sin^2(x)$ can be interpreted to mean $sin(sin(x))$, and is ambiguous when the inverse trig functions show up.
– Xander Henderson
Nov 22 at 21:05
2
2
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
If you want to suggest that a student omit the parentheses because it is a common convention (it is) and reduces some visual clutter (this is debatable), great! But justifying it in the name of clarity is, frankly, disingenuous.
– Xander Henderson
Nov 22 at 21:08
|
show 8 more comments
up vote
2
down vote
$dfrac{4cos(x)(1-cos(x))-4sin^2(x)}{(1-cos(x))^2}=dfrac{4(cos x-1)}{(1-cos(x))^2}=dfrac{-4}{1-cos(x)}=dfrac{4}{cos(x) - 1}$.
(Using $sin^2x+cos^2x=1$).
answered Nov 22 at 20:06
Yadati Kiran
1,243417
add a comment |
up vote
2
down vote
$dfrac{4cos(x)(1-cos(x))-4sin^2(x)}{(1-cos(x))^2}=dfrac{4(cos x-1)}{(1-cos(x))^2}=dfrac{-4}{1-cos(x)}=dfrac{4}{cos(x) - 1}$.
(Using $sin^2x+cos^2x=1$).
answered Nov 22 at 20:06
Yadati Kiran
1,243417
add a comment |
up vote
2
down vote
up vote
2
down vote
$dfrac{4cos(x)(1-cos(x))-4sin^2(x)}{(1-cos(x))^2}=dfrac{4(cos x-1)}{(1-cos(x))^2}=dfrac{-4}{1-cos(x)}=dfrac{4}{cos(x) - 1}$.
(Using $sin^2x+cos^2x=1$).
answered Nov 22 at 20:06
Yadati Kiran
1,243417
$dfrac{4cos(x)(1-cos(x))-4sin^2(x)}{(1-cos(x))^2}=dfrac{4(cos x-1)}{(1-cos(x))^2}=dfrac{-4}{1-cos(x)}=dfrac{4}{cos(x) - 1}$.
(Using $sin^2x+cos^2x=1$).
answered Nov 22 at 20:06
Yadati Kiran
1,243417
edited Nov 22 at 21:17
answered Nov 22 at 20:06
Yadati Kiran
1,243417
answered Nov 22 at 20:06
Yadati Kiran
1,243417
answered Nov 22 at 20:06
Yadati Kiran
1,243417
1,243417
add a comment |
add a comment |
Hint: start with the common denominator of $(1- cos(x))^2$. Write the first fraction in terms of this denominator. Add the fractions. Simplify.
– amWhy
Nov 22 at 19:58
1
Isn't there a missing bracket in the second denominator?
– user376343
Nov 22 at 20:11
@amWhy Thanks for your help. Forgot how I could split (1-cosx^2) into (1-cosx) and (1+cosx).
– Maxwell Stone
Nov 22 at 20:12
@Mason Yeah that's right what am I thinking. Fixed it.
– Maxwell Stone
Nov 22 at 20:15