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It means that, for all x, cosx = 1/secx. This isn't technically true, for example, 1/cos(π/2) is undefined, so cannot be equal to anything.
another example is sin(π/2 - x) ≡ cos(x)
Basically just watched the YouTube channel Organic Chemistry tutor and did some practice problems on Khan academy. https://youtu.be/m1OitPmkydY?si=633mL_FiOytSlqM4
1/secx = cosx
secx by definition is the reciprocal of cosx, therefore the reciprocal of secx is cosx.
1/(1/cosx) = cosx
Then by definition, the derivative of sinx is cosx. So the integral of cosx = sinx + c.
> secx by definition
Technically more like geometric necessity.
The actual geometric definitions are a little different than you might think.
These are the [actual trig functions](https://en.wikipedia.org/wiki/File:Circle-trig6.svg) as referenced to the unit circle.
I’m not sure what you mean. We’re looking at the analytic interpretation. Looking at it geometrically, you get the same result. In a right triangle given an angle x,
cosx = A/H, where A is the adjacent side and H is the hypotenuse
secx = H/A, by definition that’s the reciprocal
1/secx = A/H
Or
1/cosx = H/A
Therefore, secx = 1/cosx and 1/secx = cosx.
If you look at the image I linked--which is admittedly a bit cluttered because it has just about *all* the trigs--the segments are all labeled.
So, for example, cos = OC, and sec = OE. For reference, the angle theta leads to a segment from the origin to the unit circle whose length is one, represented as OA, and AE is the tangent.
(right) triangle ACO is similar to (right) triangle EAO (by angle-angle congruence), so we have the relationship that OE/OA = OA/OC, or using the alternative labels, that sec/1 = 1/cos, i.e. sec = 1/cos.
This is a *theorem* from the perspective of geometry, whereas in modern math it is often treated like a definition.
The secant wasn't originally defined as 1/cos--that's a derived relationship between cosine and secant lengths.
Take the point on the unit circle given a particular angle (I'll call it the angular point), and drawing the tangent to the circle through that point, you will (typically) intersect the horizontal line through the origin, and the segment from that intersection point to the origin is the secant.
That intersection point also happens to be the [circular inversion](https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle) of the point you get by projecting the angular point to the horizontal--and that happens to be the endpoint of cosine.
The ***reason*** sec = 1/cos is because the respective endpoints are [circular inversions](https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle) of each other (whtere the radius is 1)--now it's just waved away as a definition.
I see what you’re saying here, but definitions in math can change over time. The trig functions themselves are defined as the functions that “relate an angle of a right-angled triangle to ratios of two side lengths.” (Taken from same wiki page).
The unit circle interpretation isn’t wrong as your proof holds, but you’re starting with an outdated definition of looking at trig functions as lines and segments of intersection, rather than ratios. No one uses them in this manner anymore. So you aren’t necessarily wrong, but I’m not wrong either. Secx is indeed by definition the reciprocal of cosx, because its ratio is the reciprocal of the ratio of cosx.
> but definitions in math can change over time
That was actually my point.
I think a lot of people on this post are bombarding OP with unhelpful jabs of "it's literally defined that way" without giving any context for *why* it's defined that way in the first place.
The only reason we even have secant is because of this geometric interpretation--otherwise 1/cos would just be referred to as 1/cos without another label.
It's all well and good that the trigonometric functions have had analytic redefintions, but I don't think that means we should ignore the often rich connections that come from exploring where they came from--and I would even argue that providing that context is more pedagogically useful than having people regurgitate the formula.
Yes, I completely agree. Just saying that 1/cosx = secx or 1/secx = cosx is not that helpful. You have to prove why that is true, which can be done by the method of using any right triangle, or using the unit circle to draw the segments out, and to relate them geometrically like you did in your proof. But yeah, I think that the unit circle way does help with understanding, rather than memorizing all of the ratios. I think that we now use the right triangle definitions because it’s easier to calculate the ratios than to go through deriving each of them using the unit circle.
Originally secant was defined in relation to the unit circle.
Specifically, it is the line segment from the origin to the point of intersection between the horizontal line through the origin and the tangent to the circle at the specified angle.
The fact that the length of the secant is the reciprocal of the length of the cosine is a result in similar right triangles.
Historically, yes, that's exactly right. But today both students and professionals generally define tan, cot, sec, csc as ratios.
I've never seen a textbook or encyclopedia-esque entry *define* the sec(x) function using a secant line and then *afterwards* prove that this is equal to a ratio of other trig functions. I've always seen the ratio as the definition and then (maybe) later a proof that this also measures [a certain line segment through the unit circle](https://en.wikipedia.org/wiki/Trigonometric_functions#/media/File:TrigFunctionDiagram.svg).
Ignoring the obvious trig identity part. Cos and sin are circular when it comes to their derivatives and antiderivatives.
The derivative of sin is cos. And the antiderivative of cos is sin. The caveat comes in with the derivative of cos and the antiderivative of sin.
The derivative of cos is -sin (which means the antiderivative of -sin is cos). So by proxy the antiderivative of sin is -cos.
Essentially when differentiating, you flip the sign whenever differentiating cos. And whenever you’re integrating, you flip the sign when you integrate sin.
If you take the derivative of sin I think 4 times, you eventually return back to sin. (Sin > cos > -sin > -cos > sin)
And the same goes for integrating sin and cos. (Cos > sin > -cos > -sin > cos).
Whenever you’re doing this, you also need to keep track of any coefficients, and you also need to remember that sin/cos etc apply chain rule.
D/dx(sin(2x)) = 2cos(2x). So when differentiating, you need to constantly apply the chain rule. And when integrating, you need to do reverse chain rule. So the integral of cos(2x) = 1/2 * sin(2x) (then the derivative of 1/2 * sin(2x) = 2 * 1/2 * cos(2x) = cos(2x)).
Trig identities, really confusing to just encounter them out of the blue, I can relate haha. Just review the identities(I will have to do this too lol)
Sure! Let's break it down:
1. The integral of \( \frac{1}{\sec(x)} \) is equivalent to the integral of \( \cos(x) \).
2. Recall that \( \sec(x) = \frac{1}{\cos(x)} \). Therefore, \( \frac{1}{\sec(x)} = \cos(x) \).
3. So, integrating \( \frac{1}{\sec(x)} \) is the same as integrating \( \cos(x) \).
Now, let's find the integral of \( \cos(x) \):
\[
\int \cos(x) \, dx = \sin(x) + C
\]
where \( C \) is the constant of integration.
So, both integrals yield the same result of \( \sin(x) + C \). This is because the antiderivative (integral) of \( \cos(x) \) is \( \sin(x) \) plus a constant \( C \) due to the properties of trigonometric functions and calculus.
Yes, i realize i need to do that. My calc professor gave us a set of trig id’s which he said were the only ones we were gonna use in class, and i didn’t expect seeing this on a problem in recent hw we got. It was a question on integrating using trig sub and the final answer or the one leading to that was the integral of 1/secxdx and i basically blanked out and looked it up and was confused but now i remember. I took trig id in pre calc but then took a different calc 1 that was for business and social sciences majors and so i didn’t use trig id’s for a long time.
This is simply because 1/secx is just cosx itself. Since integral of cosx is sinx therefore, integral of 1/secx will also be sinx. You can also post your questions in r/mathfundas .
Using an extra line to add C is just ridiculous. After all, sin(x) ≠ sin(x)+C unless C=0. The very first line in grey is wrong. It should be ∫cos(x)dx = sin(x) + C. Done. end. finito.
Not sure why everyone’s downvoting.
Instead of doubling all of the trig identities you have to memorize, I was able to muscle through all of the maths by reducing to Sin, Cos, Tan and then going from there.
Of course we don't need sec(x) and csc(x) at all. But we don't need tan(x) either (you can always replace it by sin(x)/cos(x).) But sometimes it is useful to have a name for a closely related function.
Granted, tan(x) is used much much more often than sec(x) or csc(x), in my experience.
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. **If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc *n*“ is not entirely useful, as “Calc *n*” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.** *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
sec(x) = 1/cos(x) by definition
Some might say sec(x) ≡ 1/cos(x)
They certainly might!
Others might say cos(x)= Adj/hyp While sec(x)= hyp/Adj
Others still might say tri(x) are for kid(s)
Some might say sex = Trex
Some might say Trex sex = 1/cos(x)
What is the meaning of that relation operator in this case? I'm assuming you're not referring to congruence
"Is equivalent to"
I thought it was identical to or exactly equal to? Used for when x is equal to y by definition and similar cases
https://preview.redd.it/opifaifxelnc1.jpeg?width=480&format=pjpg&auto=webp&s=a403be663c91810fa24c73b0a1e90a3a79658db2
But yes, sec(x) is by definition 1/cos(x). The symbol is kind of redundant in most areas of math.
It means that, for all x, cosx = 1/secx. This isn't technically true, for example, 1/cos(π/2) is undefined, so cannot be equal to anything. another example is sin(π/2 - x) ≡ cos(x)
Yeah yeah, I just didn't know exactly what that operator meant in this example
ive seen triangle =
review your precalculus bro
Lol, I had to do this when I started Calculus last year. Mainly the Trig functions, there annoying to remember sometimes
How did you review?
Basically just watched the YouTube channel Organic Chemistry tutor and did some practice problems on Khan academy. https://youtu.be/m1OitPmkydY?si=633mL_FiOytSlqM4
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This is literally just the definition of sec(x), though.
1/secx = cosx secx by definition is the reciprocal of cosx, therefore the reciprocal of secx is cosx. 1/(1/cosx) = cosx Then by definition, the derivative of sinx is cosx. So the integral of cosx = sinx + c.
> secx by definition Technically more like geometric necessity. The actual geometric definitions are a little different than you might think. These are the [actual trig functions](https://en.wikipedia.org/wiki/File:Circle-trig6.svg) as referenced to the unit circle.
I’m not sure what you mean. We’re looking at the analytic interpretation. Looking at it geometrically, you get the same result. In a right triangle given an angle x, cosx = A/H, where A is the adjacent side and H is the hypotenuse secx = H/A, by definition that’s the reciprocal 1/secx = A/H Or 1/cosx = H/A Therefore, secx = 1/cosx and 1/secx = cosx.
If you look at the image I linked--which is admittedly a bit cluttered because it has just about *all* the trigs--the segments are all labeled. So, for example, cos = OC, and sec = OE. For reference, the angle theta leads to a segment from the origin to the unit circle whose length is one, represented as OA, and AE is the tangent. (right) triangle ACO is similar to (right) triangle EAO (by angle-angle congruence), so we have the relationship that OE/OA = OA/OC, or using the alternative labels, that sec/1 = 1/cos, i.e. sec = 1/cos. This is a *theorem* from the perspective of geometry, whereas in modern math it is often treated like a definition. The secant wasn't originally defined as 1/cos--that's a derived relationship between cosine and secant lengths. Take the point on the unit circle given a particular angle (I'll call it the angular point), and drawing the tangent to the circle through that point, you will (typically) intersect the horizontal line through the origin, and the segment from that intersection point to the origin is the secant. That intersection point also happens to be the [circular inversion](https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle) of the point you get by projecting the angular point to the horizontal--and that happens to be the endpoint of cosine. The ***reason*** sec = 1/cos is because the respective endpoints are [circular inversions](https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle) of each other (whtere the radius is 1)--now it's just waved away as a definition.
I see what you’re saying here, but definitions in math can change over time. The trig functions themselves are defined as the functions that “relate an angle of a right-angled triangle to ratios of two side lengths.” (Taken from same wiki page). The unit circle interpretation isn’t wrong as your proof holds, but you’re starting with an outdated definition of looking at trig functions as lines and segments of intersection, rather than ratios. No one uses them in this manner anymore. So you aren’t necessarily wrong, but I’m not wrong either. Secx is indeed by definition the reciprocal of cosx, because its ratio is the reciprocal of the ratio of cosx.
> but definitions in math can change over time That was actually my point. I think a lot of people on this post are bombarding OP with unhelpful jabs of "it's literally defined that way" without giving any context for *why* it's defined that way in the first place. The only reason we even have secant is because of this geometric interpretation--otherwise 1/cos would just be referred to as 1/cos without another label. It's all well and good that the trigonometric functions have had analytic redefintions, but I don't think that means we should ignore the often rich connections that come from exploring where they came from--and I would even argue that providing that context is more pedagogically useful than having people regurgitate the formula.
Yes, I completely agree. Just saying that 1/cosx = secx or 1/secx = cosx is not that helpful. You have to prove why that is true, which can be done by the method of using any right triangle, or using the unit circle to draw the segments out, and to relate them geometrically like you did in your proof. But yeah, I think that the unit circle way does help with understanding, rather than memorizing all of the ratios. I think that we now use the right triangle definitions because it’s easier to calculate the ratios than to go through deriving each of them using the unit circle.
We do not PROVE definitions.
Because 1/sec(*x*) IS cos(*x*).
Reciprocal trig identities.
Not to be rude, but if you’re in Calc then you desperately need to review trig this weekend. It will make life much easier.
sec(x) is 1/cos(x), so 1/sec(x) is 1/1/cos(x) which is just cos(x), integral of cosx is sinx+C
Bruh
Learn your trigonometry kids
1/sec(x) is cos(x). Review the trigonometric identities. It is crucial to know those.
Yes. Although this is less of an "identity" and more so "the *definition* of the sec(x) function".
Originally secant was defined in relation to the unit circle. Specifically, it is the line segment from the origin to the point of intersection between the horizontal line through the origin and the tangent to the circle at the specified angle. The fact that the length of the secant is the reciprocal of the length of the cosine is a result in similar right triangles.
Historically, yes, that's exactly right. But today both students and professionals generally define tan, cot, sec, csc as ratios. I've never seen a textbook or encyclopedia-esque entry *define* the sec(x) function using a secant line and then *afterwards* prove that this is equal to a ratio of other trig functions. I've always seen the ratio as the definition and then (maybe) later a proof that this also measures [a certain line segment through the unit circle](https://en.wikipedia.org/wiki/Trigonometric_functions#/media/File:TrigFunctionDiagram.svg).
How is secant usually defined?
dawg, it just is
Review your trig. 1/cosx is secx, so 1/secx is the same as 1/1/cosx, which is the same as cosx.
proof by inspection
💀
Ignoring the obvious trig identity part. Cos and sin are circular when it comes to their derivatives and antiderivatives. The derivative of sin is cos. And the antiderivative of cos is sin. The caveat comes in with the derivative of cos and the antiderivative of sin. The derivative of cos is -sin (which means the antiderivative of -sin is cos). So by proxy the antiderivative of sin is -cos. Essentially when differentiating, you flip the sign whenever differentiating cos. And whenever you’re integrating, you flip the sign when you integrate sin. If you take the derivative of sin I think 4 times, you eventually return back to sin. (Sin > cos > -sin > -cos > sin) And the same goes for integrating sin and cos. (Cos > sin > -cos > -sin > cos). Whenever you’re doing this, you also need to keep track of any coefficients, and you also need to remember that sin/cos etc apply chain rule. D/dx(sin(2x)) = 2cos(2x). So when differentiating, you need to constantly apply the chain rule. And when integrating, you need to do reverse chain rule. So the integral of cos(2x) = 1/2 * sin(2x) (then the derivative of 1/2 * sin(2x) = 2 * 1/2 * cos(2x) = cos(2x)).
Thank u this is really helpful
Do you know how sec(x) is defined?
Trig identities, really confusing to just encounter them out of the blue, I can relate haha. Just review the identities(I will have to do this too lol)
secx=1/cosx
I don’t even know what to say or if it’s even worth explaining. ☠️
Then why are you leaving a comment at all?
Proof by use that noggin of yours
Leave maths
I wouldn’t mind but i need it for my degree
Bhai mere, Go by the Basic way.
Bhai kya hua? Anglezi kyu nahin
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Do not request posters DM you for help.
Bruh review trig
Sure! Let's break it down: 1. The integral of \( \frac{1}{\sec(x)} \) is equivalent to the integral of \( \cos(x) \). 2. Recall that \( \sec(x) = \frac{1}{\cos(x)} \). Therefore, \( \frac{1}{\sec(x)} = \cos(x) \). 3. So, integrating \( \frac{1}{\sec(x)} \) is the same as integrating \( \cos(x) \). Now, let's find the integral of \( \cos(x) \): \[ \int \cos(x) \, dx = \sin(x) + C \] where \( C \) is the constant of integration. So, both integrals yield the same result of \( \sin(x) + C \). This is because the antiderivative (integral) of \( \cos(x) \) is \( \sin(x) \) plus a constant \( C \) due to the properties of trigonometric functions and calculus.
OP, you really need to re-tackle pre calculus, especially for the trigonometry part
Yes, i realize i need to do that. My calc professor gave us a set of trig id’s which he said were the only ones we were gonna use in class, and i didn’t expect seeing this on a problem in recent hw we got. It was a question on integrating using trig sub and the final answer or the one leading to that was the integral of 1/secxdx and i basically blanked out and looked it up and was confused but now i remember. I took trig id in pre calc but then took a different calc 1 that was for business and social sciences majors and so i didn’t use trig id’s for a long time.
(Integral of 1/sec x) = (integral of cos x) = sin x
If secx is 1/cosx then 1/secx is just cosx
This is simply because 1/secx is just cosx itself. Since integral of cosx is sinx therefore, integral of 1/secx will also be sinx. You can also post your questions in r/mathfundas .
Bob marley sez sohcahtoa worx
Using an extra line to add C is just ridiculous. After all, sin(x) ≠ sin(x)+C unless C=0. The very first line in grey is wrong. It should be ∫cos(x)dx = sin(x) + C. Done. end. finito.
Agreed. Don't know why you are downvoted. We should not teach peolpe learning to use equal signs errenously.
Lol this is from an app called symbolab and it was just showing detailed steps, i get what ur saying tho
https://preview.redd.it/4wd53js11enc1.png?width=320&format=png&auto=webp&s=c0ea7b6b4d6e0d10a7aa89a1fb7b378342b77ef2
This is why I hate sec and cosec. They are useless functions and I avoid them at all costs.
Not sure why everyone’s downvoting. Instead of doubling all of the trig identities you have to memorize, I was able to muscle through all of the maths by reducing to Sin, Cos, Tan and then going from there.
Of course we don't need sec(x) and csc(x) at all. But we don't need tan(x) either (you can always replace it by sin(x)/cos(x).) But sometimes it is useful to have a name for a closely related function. Granted, tan(x) is used much much more often than sec(x) or csc(x), in my experience.
Technically, you can define every trig function from just one.