Confused about cosec(x).












1












$begingroup$


The title says it all, let's take this simple example:



Figure 1



I want to find $c$.



Using sine I get:





  1. $sin(30)$ = $4 / c$

  2. $c = 4 / sin(30) = 8$


Now, using $cosec$:




  1. $cosec(30) = c / 4$

  2. $c = cosec(30) * 4 = 8$


But also:




  1. $cosec(30) = c / 4$

  2. $tan(30) = 4 / a$

  3. $4 = tan(30) * a$

  4. $c = cosec(30)cdot tan(30)cdot c = ccdot (c / 4)cdot (4 / a) = c$

  5. OR: $cosec(30) cdot tan(30)cdot c = sec(30)cdot c approx c$


What's gone wrong here, I assumed $c$ would cancel out?



EDIT 1: Why isn't $c$ cancelling out in the last example, even after the fix?



EDIT 2: Fixed steps 2 and 3 (changed $c$ to $a$). Thank you for your help, silly mistakes.






geometry trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Item $2$ is the error. $tan 30=frac4a$ is what it should be.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @MattSamuel thank you, been a long night :)!
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    My advice is too never use $sec$ or $operatorname{cosec}$. And if you are imposed to work with them, translate a soon as possible to $sin,cos$.
    $endgroup$
    – zwim
    5 hours ago












  • $begingroup$
    @zwim, thanks for the advice, I just got a bit OCD over this as it didn't add up (well still doesn't).
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    @zwim I think this advice is a bit extreme. While I see where you're coming from, they have a name for a reason. For instance, $1+tan^2 x=sec^2 x$.
    $endgroup$
    – Matt Samuel
    4 hours ago
















1












$begingroup$


The title says it all, let's take this simple example:



Figure 1



I want to find $c$.



Using sine I get:





  1. $sin(30)$ = $4 / c$

  2. $c = 4 / sin(30) = 8$


Now, using $cosec$:




  1. $cosec(30) = c / 4$

  2. $c = cosec(30) * 4 = 8$


But also:




  1. $cosec(30) = c / 4$

  2. $tan(30) = 4 / a$

  3. $4 = tan(30) * a$

  4. $c = cosec(30)cdot tan(30)cdot c = ccdot (c / 4)cdot (4 / a) = c$

  5. OR: $cosec(30) cdot tan(30)cdot c = sec(30)cdot c approx c$


What's gone wrong here, I assumed $c$ would cancel out?



EDIT 1: Why isn't $c$ cancelling out in the last example, even after the fix?



EDIT 2: Fixed steps 2 and 3 (changed $c$ to $a$). Thank you for your help, silly mistakes.






geometry trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Item $2$ is the error. $tan 30=frac4a$ is what it should be.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @MattSamuel thank you, been a long night :)!
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    My advice is too never use $sec$ or $operatorname{cosec}$. And if you are imposed to work with them, translate a soon as possible to $sin,cos$.
    $endgroup$
    – zwim
    5 hours ago












  • $begingroup$
    @zwim, thanks for the advice, I just got a bit OCD over this as it didn't add up (well still doesn't).
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    @zwim I think this advice is a bit extreme. While I see where you're coming from, they have a name for a reason. For instance, $1+tan^2 x=sec^2 x$.
    $endgroup$
    – Matt Samuel
    4 hours ago














1












1








1





$begingroup$


The title says it all, let's take this simple example:



Figure 1



I want to find $c$.



Using sine I get:





  1. $sin(30)$ = $4 / c$

  2. $c = 4 / sin(30) = 8$


Now, using $cosec$:




  1. $cosec(30) = c / 4$

  2. $c = cosec(30) * 4 = 8$


But also:




  1. $cosec(30) = c / 4$

  2. $tan(30) = 4 / a$

  3. $4 = tan(30) * a$

  4. $c = cosec(30)cdot tan(30)cdot c = ccdot (c / 4)cdot (4 / a) = c$

  5. OR: $cosec(30) cdot tan(30)cdot c = sec(30)cdot c approx c$


What's gone wrong here, I assumed $c$ would cancel out?



EDIT 1: Why isn't $c$ cancelling out in the last example, even after the fix?



EDIT 2: Fixed steps 2 and 3 (changed $c$ to $a$). Thank you for your help, silly mistakes.






geometry trigonometry







share|cite|improve this question











$endgroup$




The title says it all, let's take this simple example:



Figure 1



I want to find $c$.



Using sine I get:





  1. $sin(30)$ = $4 / c$

  2. $c = 4 / sin(30) = 8$


Now, using $cosec$:




  1. $cosec(30) = c / 4$

  2. $c = cosec(30) * 4 = 8$


But also:




  1. $cosec(30) = c / 4$

  2. $tan(30) = 4 / a$

  3. $4 = tan(30) * a$

  4. $c = cosec(30)cdot tan(30)cdot c = ccdot (c / 4)cdot (4 / a) = c$

  5. OR: $cosec(30) cdot tan(30)cdot c = sec(30)cdot c approx c$


What's gone wrong here, I assumed $c$ would cancel out?



EDIT 1: Why isn't $c$ cancelling out in the last example, even after the fix?



EDIT 2: Fixed steps 2 and 3 (changed $c$ to $a$). Thank you for your help, silly mistakes.






geometry trigonometry




geometry trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







Shibalicious

















asked 5 hours ago









ShibaliciousShibalicious

1245




1245








  • 1




    $begingroup$
    Item $2$ is the error. $tan 30=frac4a$ is what it should be.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @MattSamuel thank you, been a long night :)!
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    My advice is too never use $sec$ or $operatorname{cosec}$. And if you are imposed to work with them, translate a soon as possible to $sin,cos$.
    $endgroup$
    – zwim
    5 hours ago












  • $begingroup$
    @zwim, thanks for the advice, I just got a bit OCD over this as it didn't add up (well still doesn't).
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    @zwim I think this advice is a bit extreme. While I see where you're coming from, they have a name for a reason. For instance, $1+tan^2 x=sec^2 x$.
    $endgroup$
    – Matt Samuel
    4 hours ago














  • 1




    $begingroup$
    Item $2$ is the error. $tan 30=frac4a$ is what it should be.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @MattSamuel thank you, been a long night :)!
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    My advice is too never use $sec$ or $operatorname{cosec}$. And if you are imposed to work with them, translate a soon as possible to $sin,cos$.
    $endgroup$
    – zwim
    5 hours ago












  • $begingroup$
    @zwim, thanks for the advice, I just got a bit OCD over this as it didn't add up (well still doesn't).
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    @zwim I think this advice is a bit extreme. While I see where you're coming from, they have a name for a reason. For instance, $1+tan^2 x=sec^2 x$.
    $endgroup$
    – Matt Samuel
    4 hours ago








1




1




$begingroup$
Item $2$ is the error. $tan 30=frac4a$ is what it should be.
$endgroup$
– Matt Samuel
5 hours ago




$begingroup$
Item $2$ is the error. $tan 30=frac4a$ is what it should be.
$endgroup$
– Matt Samuel
5 hours ago












$begingroup$
@MattSamuel thank you, been a long night :)!
$endgroup$
– Shibalicious
5 hours ago




$begingroup$
@MattSamuel thank you, been a long night :)!
$endgroup$
– Shibalicious
5 hours ago




1




1




$begingroup$
My advice is too never use $sec$ or $operatorname{cosec}$. And if you are imposed to work with them, translate a soon as possible to $sin,cos$.
$endgroup$
– zwim
5 hours ago






$begingroup$
My advice is too never use $sec$ or $operatorname{cosec}$. And if you are imposed to work with them, translate a soon as possible to $sin,cos$.
$endgroup$
– zwim
5 hours ago














$begingroup$
@zwim, thanks for the advice, I just got a bit OCD over this as it didn't add up (well still doesn't).
$endgroup$
– Shibalicious
5 hours ago




$begingroup$
@zwim, thanks for the advice, I just got a bit OCD over this as it didn't add up (well still doesn't).
$endgroup$
– Shibalicious
5 hours ago




1




1




$begingroup$
@zwim I think this advice is a bit extreme. While I see where you're coming from, they have a name for a reason. For instance, $1+tan^2 x=sec^2 x$.
$endgroup$
– Matt Samuel
4 hours ago




$begingroup$
@zwim I think this advice is a bit extreme. While I see where you're coming from, they have a name for a reason. For instance, $1+tan^2 x=sec^2 x$.
$endgroup$
– Matt Samuel
4 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$


$$tan(30°)=frac{4}{a}text{ which isn't } frac{4}{c}$$




Now we have that $$c=text{cosec}(30°)·4;text{ and } ;4=tan(30)°·a$$
Thus $$c=text{cosec}(30°)·tan(30°)·a=frac{c}{4}·frac{4}{a}·anot= frac{c}{4}·frac{4}{a}·color{red}{c}$$ as pointed out in the comments






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for spotting this, got it!
    $endgroup$
    – Shibalicious
    5 hours ago










  • $begingroup$
    No problem! @Shibalicious
    $endgroup$
    – Dr. Mathva
    5 hours ago










  • $begingroup$
    So now I got sec(30) * c, which is c ~= 9.23, why?
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    I've edited the answer
    $endgroup$
    – Dr. Mathva
    4 hours ago






  • 1




    $begingroup$
    Spot on, I got it now!
    $endgroup$
    – Shibalicious
    4 hours ago



















1












$begingroup$

$$sec30^{circ}=frac{1}{cos30^{circ}}=frac{2}{sqrt3}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a non sequitur.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
    $endgroup$
    – Michael Rozenberg
    5 hours ago






  • 1




    $begingroup$
    The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
    $endgroup$
    – Matt Samuel
    5 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$


$$tan(30°)=frac{4}{a}text{ which isn't } frac{4}{c}$$




Now we have that $$c=text{cosec}(30°)·4;text{ and } ;4=tan(30)°·a$$
Thus $$c=text{cosec}(30°)·tan(30°)·a=frac{c}{4}·frac{4}{a}·anot= frac{c}{4}·frac{4}{a}·color{red}{c}$$ as pointed out in the comments






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for spotting this, got it!
    $endgroup$
    – Shibalicious
    5 hours ago










  • $begingroup$
    No problem! @Shibalicious
    $endgroup$
    – Dr. Mathva
    5 hours ago










  • $begingroup$
    So now I got sec(30) * c, which is c ~= 9.23, why?
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    I've edited the answer
    $endgroup$
    – Dr. Mathva
    4 hours ago






  • 1




    $begingroup$
    Spot on, I got it now!
    $endgroup$
    – Shibalicious
    4 hours ago
















3












$begingroup$


$$tan(30°)=frac{4}{a}text{ which isn't } frac{4}{c}$$




Now we have that $$c=text{cosec}(30°)·4;text{ and } ;4=tan(30)°·a$$
Thus $$c=text{cosec}(30°)·tan(30°)·a=frac{c}{4}·frac{4}{a}·anot= frac{c}{4}·frac{4}{a}·color{red}{c}$$ as pointed out in the comments






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for spotting this, got it!
    $endgroup$
    – Shibalicious
    5 hours ago










  • $begingroup$
    No problem! @Shibalicious
    $endgroup$
    – Dr. Mathva
    5 hours ago










  • $begingroup$
    So now I got sec(30) * c, which is c ~= 9.23, why?
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    I've edited the answer
    $endgroup$
    – Dr. Mathva
    4 hours ago






  • 1




    $begingroup$
    Spot on, I got it now!
    $endgroup$
    – Shibalicious
    4 hours ago














3












3








3





$begingroup$


$$tan(30°)=frac{4}{a}text{ which isn't } frac{4}{c}$$




Now we have that $$c=text{cosec}(30°)·4;text{ and } ;4=tan(30)°·a$$
Thus $$c=text{cosec}(30°)·tan(30°)·a=frac{c}{4}·frac{4}{a}·anot= frac{c}{4}·frac{4}{a}·color{red}{c}$$ as pointed out in the comments






share|cite|improve this answer











$endgroup$




$$tan(30°)=frac{4}{a}text{ which isn't } frac{4}{c}$$




Now we have that $$c=text{cosec}(30°)·4;text{ and } ;4=tan(30)°·a$$
Thus $$c=text{cosec}(30°)·tan(30°)·a=frac{c}{4}·frac{4}{a}·anot= frac{c}{4}·frac{4}{a}·color{red}{c}$$ as pointed out in the comments







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 4 hours ago

























answered 5 hours ago









Dr. MathvaDr. Mathva

1,975322




1,975322












  • $begingroup$
    Thank you for spotting this, got it!
    $endgroup$
    – Shibalicious
    5 hours ago










  • $begingroup$
    No problem! @Shibalicious
    $endgroup$
    – Dr. Mathva
    5 hours ago










  • $begingroup$
    So now I got sec(30) * c, which is c ~= 9.23, why?
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    I've edited the answer
    $endgroup$
    – Dr. Mathva
    4 hours ago






  • 1




    $begingroup$
    Spot on, I got it now!
    $endgroup$
    – Shibalicious
    4 hours ago


















  • $begingroup$
    Thank you for spotting this, got it!
    $endgroup$
    – Shibalicious
    5 hours ago










  • $begingroup$
    No problem! @Shibalicious
    $endgroup$
    – Dr. Mathva
    5 hours ago










  • $begingroup$
    So now I got sec(30) * c, which is c ~= 9.23, why?
    $endgroup$
    – Shibalicious
    5 hours ago






  • 1




    $begingroup$
    I've edited the answer
    $endgroup$
    – Dr. Mathva
    4 hours ago






  • 1




    $begingroup$
    Spot on, I got it now!
    $endgroup$
    – Shibalicious
    4 hours ago
















$begingroup$
Thank you for spotting this, got it!
$endgroup$
– Shibalicious
5 hours ago




$begingroup$
Thank you for spotting this, got it!
$endgroup$
– Shibalicious
5 hours ago












$begingroup$
No problem! @Shibalicious
$endgroup$
– Dr. Mathva
5 hours ago




$begingroup$
No problem! @Shibalicious
$endgroup$
– Dr. Mathva
5 hours ago












$begingroup$
So now I got sec(30) * c, which is c ~= 9.23, why?
$endgroup$
– Shibalicious
5 hours ago




$begingroup$
So now I got sec(30) * c, which is c ~= 9.23, why?
$endgroup$
– Shibalicious
5 hours ago




1




1




$begingroup$
I've edited the answer
$endgroup$
– Dr. Mathva
4 hours ago




$begingroup$
I've edited the answer
$endgroup$
– Dr. Mathva
4 hours ago




1




1




$begingroup$
Spot on, I got it now!
$endgroup$
– Shibalicious
4 hours ago




$begingroup$
Spot on, I got it now!
$endgroup$
– Shibalicious
4 hours ago











1












$begingroup$

$$sec30^{circ}=frac{1}{cos30^{circ}}=frac{2}{sqrt3}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a non sequitur.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
    $endgroup$
    – Michael Rozenberg
    5 hours ago






  • 1




    $begingroup$
    The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
    $endgroup$
    – Matt Samuel
    5 hours ago


















1












$begingroup$

$$sec30^{circ}=frac{1}{cos30^{circ}}=frac{2}{sqrt3}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a non sequitur.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
    $endgroup$
    – Michael Rozenberg
    5 hours ago






  • 1




    $begingroup$
    The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
    $endgroup$
    – Matt Samuel
    5 hours ago
















1












1








1





$begingroup$

$$sec30^{circ}=frac{1}{cos30^{circ}}=frac{2}{sqrt3}.$$






share|cite|improve this answer









$endgroup$



$$sec30^{circ}=frac{1}{cos30^{circ}}=frac{2}{sqrt3}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









Michael RozenbergMichael Rozenberg

105k1892197




105k1892197












  • $begingroup$
    This is a non sequitur.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
    $endgroup$
    – Michael Rozenberg
    5 hours ago






  • 1




    $begingroup$
    The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
    $endgroup$
    – Matt Samuel
    5 hours ago




















  • $begingroup$
    This is a non sequitur.
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
    $endgroup$
    – Matt Samuel
    5 hours ago










  • $begingroup$
    @Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
    $endgroup$
    – Michael Rozenberg
    5 hours ago






  • 1




    $begingroup$
    The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
    $endgroup$
    – Matt Samuel
    5 hours ago


















$begingroup$
This is a non sequitur.
$endgroup$
– Matt Samuel
5 hours ago




$begingroup$
This is a non sequitur.
$endgroup$
– Matt Samuel
5 hours ago












$begingroup$
@Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
$endgroup$
– Matt Samuel
5 hours ago




$begingroup$
@Dr.Mathva It's just a post of an arbitrary fact. What does this have to do with the question?
$endgroup$
– Matt Samuel
5 hours ago












$begingroup$
@Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
$endgroup$
– Michael Rozenberg
5 hours ago




$begingroup$
@Matt Samuel I wrote where was a mistake. See please better. It's exactly, that you wrote.
$endgroup$
– Michael Rozenberg
5 hours ago




1




1




$begingroup$
The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
$endgroup$
– Matt Samuel
5 hours ago






$begingroup$
The only place I see $sec 30$ is already well after the first error. It's true that it's an error thinking it's equal to $1$, but that's not the important error.
$endgroup$
– Matt Samuel
5 hours ago




















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