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How do calculators calculate the value of trigonometric functions?
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I have an obsession about trig functions. I find them to be so mysterious, because I really don’t understand calculus (other then trying to make sense of it in a few YouTube videos, which doesn’t work for me).
Anyhow I was wondering if someone can explain how calculators find the ratios for sine and cosine by just typing in the degrees.
This seems impossible to me from a logical standpoint. I don’t see any pattern other then as the angle increase the ratio for cosine gets exponentially smaller. If anyone can tell me how you get an exact number for something that has no pattern, please explain it to me in the simplest possible way.
Pretend you're explaining it to your 10-year-old child. I am very visual so anything with pictures would work. If you start getting into Taylor series with calculus, I won’t understand.
I’m basically looking for a simple explanation of what the calculator is doing without getting deep into the math. Thanks in advance.
Cheers
calculus trigonometry
edited 56 mins ago
Blue
45.8k869145
asked 1 hour ago
Physicsrocks
311
 |Â
show 7 more comments
up vote
4
down vote
favorite
I have an obsession about trig functions. I find them to be so mysterious, because I really don’t understand calculus (other then trying to make sense of it in a few YouTube videos, which doesn’t work for me).
Anyhow I was wondering if someone can explain how calculators find the ratios for sine and cosine by just typing in the degrees.
This seems impossible to me from a logical standpoint. I don’t see any pattern other then as the angle increase the ratio for cosine gets exponentially smaller. If anyone can tell me how you get an exact number for something that has no pattern, please explain it to me in the simplest possible way.
Pretend you're explaining it to your 10-year-old child. I am very visual so anything with pictures would work. If you start getting into Taylor series with calculus, I won’t understand.
I’m basically looking for a simple explanation of what the calculator is doing without getting deep into the math. Thanks in advance.
Cheers
calculus trigonometry
edited 56 mins ago
Blue
45.8k869145
asked 1 hour ago
Physicsrocks
311
Oh wow. I'd suggest taking a deep dive into understanding what $sin$ and $cos$ actually mean.
– Rushabh Mehta
1 hour ago
Calculators usually use Taylor series to compute $sin$ and $cos$ to arbitrary precision, so unfortunately, there's no easy answer for that.
– Rushabh Mehta
1 hour ago
If you search here for, say, "calculator sine", you'll find a number of places where this kind of question has been asked before. Look at the answers to those and see if they make sense to you. If they don't, then try to explain here why they don't, so that people have a better idea of the type of answer that will satisfy you. Otherwise, people are likely to simply repeat the earlier explanations, wasting everyone's time. Help us help you.
– Blue
1 hour ago
2
@Blue Agreed. I'm not sure how to answer OP since OP specified that Taylor series is too advanced, but Taylor series are usually the tool of choice to compute these functions.
– Rushabh Mehta
1 hour ago
2
@Physicsrocks Sadly, there is no answer to this question that doesn't involve calculus. Maybe this'll motivate you to start learning some basic calculus! Take any standard high school level text -- it should be sufficient for the purpose.
– Rushabh Mehta
1 hour ago
 |Â
show 7 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have an obsession about trig functions. I find them to be so mysterious, because I really don’t understand calculus (other then trying to make sense of it in a few YouTube videos, which doesn’t work for me).
Anyhow I was wondering if someone can explain how calculators find the ratios for sine and cosine by just typing in the degrees.
This seems impossible to me from a logical standpoint. I don’t see any pattern other then as the angle increase the ratio for cosine gets exponentially smaller. If anyone can tell me how you get an exact number for something that has no pattern, please explain it to me in the simplest possible way.
Pretend you're explaining it to your 10-year-old child. I am very visual so anything with pictures would work. If you start getting into Taylor series with calculus, I won’t understand.
I’m basically looking for a simple explanation of what the calculator is doing without getting deep into the math. Thanks in advance.
Cheers
calculus trigonometry
edited 56 mins ago
Blue
45.8k869145
asked 1 hour ago
Physicsrocks
311
I have an obsession about trig functions. I find them to be so mysterious, because I really don’t understand calculus (other then trying to make sense of it in a few YouTube videos, which doesn’t work for me).
Anyhow I was wondering if someone can explain how calculators find the ratios for sine and cosine by just typing in the degrees.
This seems impossible to me from a logical standpoint. I don’t see any pattern other then as the angle increase the ratio for cosine gets exponentially smaller. If anyone can tell me how you get an exact number for something that has no pattern, please explain it to me in the simplest possible way.
Pretend you're explaining it to your 10-year-old child. I am very visual so anything with pictures would work. If you start getting into Taylor series with calculus, I won’t understand.
I’m basically looking for a simple explanation of what the calculator is doing without getting deep into the math. Thanks in advance.
Cheers
calculus trigonometry
calculus trigonometry
edited 56 mins ago
Blue
45.8k869145
asked 1 hour ago
Physicsrocks
311
edited 56 mins ago
Blue
45.8k869145
asked 1 hour ago
Physicsrocks
311
edited 56 mins ago
Blue
45.8k869145
edited 56 mins ago
Blue
45.8k869145
edited 56 mins ago
Blue
45.8k869145
45.8k869145
asked 1 hour ago
Physicsrocks
311
asked 1 hour ago
Physicsrocks
311
asked 1 hour ago
Physicsrocks
311
311
Oh wow. I'd suggest taking a deep dive into understanding what $sin$ and $cos$ actually mean.
– Rushabh Mehta
1 hour ago
Calculators usually use Taylor series to compute $sin$ and $cos$ to arbitrary precision, so unfortunately, there's no easy answer for that.
– Rushabh Mehta
1 hour ago
If you search here for, say, "calculator sine", you'll find a number of places where this kind of question has been asked before. Look at the answers to those and see if they make sense to you. If they don't, then try to explain here why they don't, so that people have a better idea of the type of answer that will satisfy you. Otherwise, people are likely to simply repeat the earlier explanations, wasting everyone's time. Help us help you.
– Blue
1 hour ago
2
@Blue Agreed. I'm not sure how to answer OP since OP specified that Taylor series is too advanced, but Taylor series are usually the tool of choice to compute these functions.
– Rushabh Mehta
1 hour ago
2
@Physicsrocks Sadly, there is no answer to this question that doesn't involve calculus. Maybe this'll motivate you to start learning some basic calculus! Take any standard high school level text -- it should be sufficient for the purpose.
– Rushabh Mehta
1 hour ago
 |Â
show 7 more comments
Oh wow. I'd suggest taking a deep dive into understanding what $sin$ and $cos$ actually mean.
– Rushabh Mehta
1 hour ago
Calculators usually use Taylor series to compute $sin$ and $cos$ to arbitrary precision, so unfortunately, there's no easy answer for that.
– Rushabh Mehta
1 hour ago
If you search here for, say, "calculator sine", you'll find a number of places where this kind of question has been asked before. Look at the answers to those and see if they make sense to you. If they don't, then try to explain here why they don't, so that people have a better idea of the type of answer that will satisfy you. Otherwise, people are likely to simply repeat the earlier explanations, wasting everyone's time. Help us help you.
– Blue
1 hour ago
2
@Blue Agreed. I'm not sure how to answer OP since OP specified that Taylor series is too advanced, but Taylor series are usually the tool of choice to compute these functions.
– Rushabh Mehta
1 hour ago
2
@Physicsrocks Sadly, there is no answer to this question that doesn't involve calculus. Maybe this'll motivate you to start learning some basic calculus! Take any standard high school level text -- it should be sufficient for the purpose.
– Rushabh Mehta
1 hour ago
Oh wow. I'd suggest taking a deep dive into understanding what $sin$ and $cos$ actually mean.
– Rushabh Mehta
1 hour ago
Oh wow. I'd suggest taking a deep dive into understanding what $sin$ and $cos$ actually mean.
– Rushabh Mehta
1 hour ago
Calculators usually use Taylor series to compute $sin$ and $cos$ to arbitrary precision, so unfortunately, there's no easy answer for that.
– Rushabh Mehta
1 hour ago
Calculators usually use Taylor series to compute $sin$ and $cos$ to arbitrary precision, so unfortunately, there's no easy answer for that.
– Rushabh Mehta
1 hour ago
If you search here for, say, "calculator sine", you'll find a number of places where this kind of question has been asked before. Look at the answers to those and see if they make sense to you. If they don't, then try to explain here why they don't, so that people have a better idea of the type of answer that will satisfy you. Otherwise, people are likely to simply repeat the earlier explanations, wasting everyone's time. Help us help you.
– Blue
1 hour ago
If you search here for, say, "calculator sine", you'll find a number of places where this kind of question has been asked before. Look at the answers to those and see if they make sense to you. If they don't, then try to explain here why they don't, so that people have a better idea of the type of answer that will satisfy you. Otherwise, people are likely to simply repeat the earlier explanations, wasting everyone's time. Help us help you.
– Blue
1 hour ago
2
2
@Blue Agreed. I'm not sure how to answer OP since OP specified that Taylor series is too advanced, but Taylor series are usually the tool of choice to compute these functions.
– Rushabh Mehta
1 hour ago
@Blue Agreed. I'm not sure how to answer OP since OP specified that Taylor series is too advanced, but Taylor series are usually the tool of choice to compute these functions.
– Rushabh Mehta
1 hour ago
2
2
@Physicsrocks Sadly, there is no answer to this question that doesn't involve calculus. Maybe this'll motivate you to start learning some basic calculus! Take any standard high school level text -- it should be sufficient for the purpose.
– Rushabh Mehta
1 hour ago
@Physicsrocks Sadly, there is no answer to this question that doesn't involve calculus. Maybe this'll motivate you to start learning some basic calculus! Take any standard high school level text -- it should be sufficient for the purpose.
– Rushabh Mehta
1 hour ago
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
Calculators, to my recollection, typically use special summation identities for the sine and cosine functions that take in some angle measure, sum up a bunch of things using that measure, and then pop out a measurement.
It's sort of like how you can approximate the exponential function $e^x$ through its power series, which is the infinite sum below. (I know, you said not to get into Taylor series or anything of the like. I'm not getting into the derivation or whatever - if you simply accept for now and take me at my word that this summation makes sense and is 100% valid, it might be easier for now.)
$$e^x = 1 + fracx1! + fracx^22! + fracx^33! + fracx^44! + fracx^55! + ...$$
Imagine sort of stopping that summation at the 10th, or 20th, or 1000th, or 10,000th term - it won't exactly be $e^x$ but it will be an approximation. Not exact, but "close enough" - fractions of a percent off, depending on how far you go.
Lots of such identities exist for various functions - some that are crazy and wacky to look at when taken at first glance, but ones that converge very quickly to "close enough" to the actual value after only a few terms. For example, we can define these summations called "Taylor series" for functions. Their derivation involves the knowledge of calculus, but they have the same core idea as the summation above for $e^x$: take an infinite sum, truncate it at some point, and you have an approximation for the value, which is more accurate the later you truncate.
The Taylor series for sine/cosine and probably the simplest technique for approximation a calculator would use:
$$cos(x) = 1 - fracx^22! + fracx^44!- fracx^66! + fracx^88! - fracx^1010! + ...$$
$$sin(x) = fracx1! - fracx^33! + fracx^55! - fracx^77! + fracx^99! - ...$$
A sort of "refinement" of this idea - I guess anyhow, it's a bit above my head, but I imagine it has the same core principle - is used in some calculators called the "CORDIC algorithm." It's not quite just some basic summation like above, but it has the same core idea. You can probably read up on it here but it seems pretty high-level so don't be surprised - https://ieeexplore.ieee.org/document/7453811
In short, how calculators typically find cosine and sine of functions are by taking known series or algorithms, derived from calculus or even higher math, and using approximation techniques to try and get just "close enough."
The underlying details of why these methods make sense is probably a bit above your head for the time being - I'm not saying it to be rude, but you need a foundation in calculus and derivatives at least to understand the concept of Taylor series, and they're probably some of the more "basic" approximation techniques for stuff of this sort.
answered 1 hour ago
Eevee Trainer
4358
add a comment |Â
up vote
1
down vote
I think asking this question without really understanding Taylor Polynomials (or approximations in general) is a bit of a difficult task. What I would suggest if you like graphs and pictures for example, without actually computing the Taylor expansions is to go onto Desmos and plot the first few. Use this link
https://www.desmos.com/calculator/m8mw0nayab
By the way, the pattern is $(-1)^nfracx^1+2n(1+2n)!$ for n as integers including 0.
If you use enough terms in the approximation (other, more accurate approximations are usually used and the values are stored in a table), then you can get accurate values for each value from 0 to 360deg, and then you can just shift any other value (since sine & cosine are repeating functions every 360deg) to within that range.
Hope this helps, let me know if you want more detail.
answered 1 hour ago
Milan Leonard
708
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Calculators, to my recollection, typically use special summation identities for the sine and cosine functions that take in some angle measure, sum up a bunch of things using that measure, and then pop out a measurement.
It's sort of like how you can approximate the exponential function $e^x$ through its power series, which is the infinite sum below. (I know, you said not to get into Taylor series or anything of the like. I'm not getting into the derivation or whatever - if you simply accept for now and take me at my word that this summation makes sense and is 100% valid, it might be easier for now.)
$$e^x = 1 + fracx1! + fracx^22! + fracx^33! + fracx^44! + fracx^55! + ...$$
Imagine sort of stopping that summation at the 10th, or 20th, or 1000th, or 10,000th term - it won't exactly be $e^x$ but it will be an approximation. Not exact, but "close enough" - fractions of a percent off, depending on how far you go.
Lots of such identities exist for various functions - some that are crazy and wacky to look at when taken at first glance, but ones that converge very quickly to "close enough" to the actual value after only a few terms. For example, we can define these summations called "Taylor series" for functions. Their derivation involves the knowledge of calculus, but they have the same core idea as the summation above for $e^x$: take an infinite sum, truncate it at some point, and you have an approximation for the value, which is more accurate the later you truncate.
The Taylor series for sine/cosine and probably the simplest technique for approximation a calculator would use:
$$cos(x) = 1 - fracx^22! + fracx^44!- fracx^66! + fracx^88! - fracx^1010! + ...$$
$$sin(x) = fracx1! - fracx^33! + fracx^55! - fracx^77! + fracx^99! - ...$$
A sort of "refinement" of this idea - I guess anyhow, it's a bit above my head, but I imagine it has the same core principle - is used in some calculators called the "CORDIC algorithm." It's not quite just some basic summation like above, but it has the same core idea. You can probably read up on it here but it seems pretty high-level so don't be surprised - https://ieeexplore.ieee.org/document/7453811
In short, how calculators typically find cosine and sine of functions are by taking known series or algorithms, derived from calculus or even higher math, and using approximation techniques to try and get just "close enough."
The underlying details of why these methods make sense is probably a bit above your head for the time being - I'm not saying it to be rude, but you need a foundation in calculus and derivatives at least to understand the concept of Taylor series, and they're probably some of the more "basic" approximation techniques for stuff of this sort.
answered 1 hour ago
Eevee Trainer
4358
add a comment |Â
up vote
2
down vote
Calculators, to my recollection, typically use special summation identities for the sine and cosine functions that take in some angle measure, sum up a bunch of things using that measure, and then pop out a measurement.
It's sort of like how you can approximate the exponential function $e^x$ through its power series, which is the infinite sum below. (I know, you said not to get into Taylor series or anything of the like. I'm not getting into the derivation or whatever - if you simply accept for now and take me at my word that this summation makes sense and is 100% valid, it might be easier for now.)
$$e^x = 1 + fracx1! + fracx^22! + fracx^33! + fracx^44! + fracx^55! + ...$$
Imagine sort of stopping that summation at the 10th, or 20th, or 1000th, or 10,000th term - it won't exactly be $e^x$ but it will be an approximation. Not exact, but "close enough" - fractions of a percent off, depending on how far you go.
Lots of such identities exist for various functions - some that are crazy and wacky to look at when taken at first glance, but ones that converge very quickly to "close enough" to the actual value after only a few terms. For example, we can define these summations called "Taylor series" for functions. Their derivation involves the knowledge of calculus, but they have the same core idea as the summation above for $e^x$: take an infinite sum, truncate it at some point, and you have an approximation for the value, which is more accurate the later you truncate.
The Taylor series for sine/cosine and probably the simplest technique for approximation a calculator would use:
$$cos(x) = 1 - fracx^22! + fracx^44!- fracx^66! + fracx^88! - fracx^1010! + ...$$
$$sin(x) = fracx1! - fracx^33! + fracx^55! - fracx^77! + fracx^99! - ...$$
A sort of "refinement" of this idea - I guess anyhow, it's a bit above my head, but I imagine it has the same core principle - is used in some calculators called the "CORDIC algorithm." It's not quite just some basic summation like above, but it has the same core idea. You can probably read up on it here but it seems pretty high-level so don't be surprised - https://ieeexplore.ieee.org/document/7453811
In short, how calculators typically find cosine and sine of functions are by taking known series or algorithms, derived from calculus or even higher math, and using approximation techniques to try and get just "close enough."
The underlying details of why these methods make sense is probably a bit above your head for the time being - I'm not saying it to be rude, but you need a foundation in calculus and derivatives at least to understand the concept of Taylor series, and they're probably some of the more "basic" approximation techniques for stuff of this sort.
answered 1 hour ago
Eevee Trainer
4358
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Calculators, to my recollection, typically use special summation identities for the sine and cosine functions that take in some angle measure, sum up a bunch of things using that measure, and then pop out a measurement.
It's sort of like how you can approximate the exponential function $e^x$ through its power series, which is the infinite sum below. (I know, you said not to get into Taylor series or anything of the like. I'm not getting into the derivation or whatever - if you simply accept for now and take me at my word that this summation makes sense and is 100% valid, it might be easier for now.)
$$e^x = 1 + fracx1! + fracx^22! + fracx^33! + fracx^44! + fracx^55! + ...$$
Imagine sort of stopping that summation at the 10th, or 20th, or 1000th, or 10,000th term - it won't exactly be $e^x$ but it will be an approximation. Not exact, but "close enough" - fractions of a percent off, depending on how far you go.
Lots of such identities exist for various functions - some that are crazy and wacky to look at when taken at first glance, but ones that converge very quickly to "close enough" to the actual value after only a few terms. For example, we can define these summations called "Taylor series" for functions. Their derivation involves the knowledge of calculus, but they have the same core idea as the summation above for $e^x$: take an infinite sum, truncate it at some point, and you have an approximation for the value, which is more accurate the later you truncate.
The Taylor series for sine/cosine and probably the simplest technique for approximation a calculator would use:
$$cos(x) = 1 - fracx^22! + fracx^44!- fracx^66! + fracx^88! - fracx^1010! + ...$$
$$sin(x) = fracx1! - fracx^33! + fracx^55! - fracx^77! + fracx^99! - ...$$
A sort of "refinement" of this idea - I guess anyhow, it's a bit above my head, but I imagine it has the same core principle - is used in some calculators called the "CORDIC algorithm." It's not quite just some basic summation like above, but it has the same core idea. You can probably read up on it here but it seems pretty high-level so don't be surprised - https://ieeexplore.ieee.org/document/7453811
In short, how calculators typically find cosine and sine of functions are by taking known series or algorithms, derived from calculus or even higher math, and using approximation techniques to try and get just "close enough."
The underlying details of why these methods make sense is probably a bit above your head for the time being - I'm not saying it to be rude, but you need a foundation in calculus and derivatives at least to understand the concept of Taylor series, and they're probably some of the more "basic" approximation techniques for stuff of this sort.
answered 1 hour ago
Eevee Trainer
4358
Calculators, to my recollection, typically use special summation identities for the sine and cosine functions that take in some angle measure, sum up a bunch of things using that measure, and then pop out a measurement.
It's sort of like how you can approximate the exponential function $e^x$ through its power series, which is the infinite sum below. (I know, you said not to get into Taylor series or anything of the like. I'm not getting into the derivation or whatever - if you simply accept for now and take me at my word that this summation makes sense and is 100% valid, it might be easier for now.)
$$e^x = 1 + fracx1! + fracx^22! + fracx^33! + fracx^44! + fracx^55! + ...$$
Imagine sort of stopping that summation at the 10th, or 20th, or 1000th, or 10,000th term - it won't exactly be $e^x$ but it will be an approximation. Not exact, but "close enough" - fractions of a percent off, depending on how far you go.
Lots of such identities exist for various functions - some that are crazy and wacky to look at when taken at first glance, but ones that converge very quickly to "close enough" to the actual value after only a few terms. For example, we can define these summations called "Taylor series" for functions. Their derivation involves the knowledge of calculus, but they have the same core idea as the summation above for $e^x$: take an infinite sum, truncate it at some point, and you have an approximation for the value, which is more accurate the later you truncate.
The Taylor series for sine/cosine and probably the simplest technique for approximation a calculator would use:
$$cos(x) = 1 - fracx^22! + fracx^44!- fracx^66! + fracx^88! - fracx^1010! + ...$$
$$sin(x) = fracx1! - fracx^33! + fracx^55! - fracx^77! + fracx^99! - ...$$
A sort of "refinement" of this idea - I guess anyhow, it's a bit above my head, but I imagine it has the same core principle - is used in some calculators called the "CORDIC algorithm." It's not quite just some basic summation like above, but it has the same core idea. You can probably read up on it here but it seems pretty high-level so don't be surprised - https://ieeexplore.ieee.org/document/7453811
In short, how calculators typically find cosine and sine of functions are by taking known series or algorithms, derived from calculus or even higher math, and using approximation techniques to try and get just "close enough."
The underlying details of why these methods make sense is probably a bit above your head for the time being - I'm not saying it to be rude, but you need a foundation in calculus and derivatives at least to understand the concept of Taylor series, and they're probably some of the more "basic" approximation techniques for stuff of this sort.
answered 1 hour ago
Eevee Trainer
4358
answered 1 hour ago
Eevee Trainer
4358
answered 1 hour ago
Eevee Trainer
4358
answered 1 hour ago
Eevee Trainer
4358
4358
add a comment |Â
add a comment |Â
up vote
1
down vote
I think asking this question without really understanding Taylor Polynomials (or approximations in general) is a bit of a difficult task. What I would suggest if you like graphs and pictures for example, without actually computing the Taylor expansions is to go onto Desmos and plot the first few. Use this link
https://www.desmos.com/calculator/m8mw0nayab
By the way, the pattern is $(-1)^nfracx^1+2n(1+2n)!$ for n as integers including 0.
If you use enough terms in the approximation (other, more accurate approximations are usually used and the values are stored in a table), then you can get accurate values for each value from 0 to 360deg, and then you can just shift any other value (since sine & cosine are repeating functions every 360deg) to within that range.
Hope this helps, let me know if you want more detail.
answered 1 hour ago
Milan Leonard
708
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
add a comment |Â
up vote
1
down vote
I think asking this question without really understanding Taylor Polynomials (or approximations in general) is a bit of a difficult task. What I would suggest if you like graphs and pictures for example, without actually computing the Taylor expansions is to go onto Desmos and plot the first few. Use this link
https://www.desmos.com/calculator/m8mw0nayab
By the way, the pattern is $(-1)^nfracx^1+2n(1+2n)!$ for n as integers including 0.
If you use enough terms in the approximation (other, more accurate approximations are usually used and the values are stored in a table), then you can get accurate values for each value from 0 to 360deg, and then you can just shift any other value (since sine & cosine are repeating functions every 360deg) to within that range.
Hope this helps, let me know if you want more detail.
answered 1 hour ago
Milan Leonard
708
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think asking this question without really understanding Taylor Polynomials (or approximations in general) is a bit of a difficult task. What I would suggest if you like graphs and pictures for example, without actually computing the Taylor expansions is to go onto Desmos and plot the first few. Use this link
https://www.desmos.com/calculator/m8mw0nayab
By the way, the pattern is $(-1)^nfracx^1+2n(1+2n)!$ for n as integers including 0.
If you use enough terms in the approximation (other, more accurate approximations are usually used and the values are stored in a table), then you can get accurate values for each value from 0 to 360deg, and then you can just shift any other value (since sine & cosine are repeating functions every 360deg) to within that range.
Hope this helps, let me know if you want more detail.
answered 1 hour ago
Milan Leonard
708
I think asking this question without really understanding Taylor Polynomials (or approximations in general) is a bit of a difficult task. What I would suggest if you like graphs and pictures for example, without actually computing the Taylor expansions is to go onto Desmos and plot the first few. Use this link
https://www.desmos.com/calculator/m8mw0nayab
By the way, the pattern is $(-1)^nfracx^1+2n(1+2n)!$ for n as integers including 0.
If you use enough terms in the approximation (other, more accurate approximations are usually used and the values are stored in a table), then you can get accurate values for each value from 0 to 360deg, and then you can just shift any other value (since sine & cosine are repeating functions every 360deg) to within that range.
Hope this helps, let me know if you want more detail.
answered 1 hour ago
Milan Leonard
708
answered 1 hour ago
Milan Leonard
708
answered 1 hour ago
Milan Leonard
708
answered 1 hour ago
Milan Leonard
708
708
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
add a comment |Â
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
One specific thing that I should add is that the reason we do this is because computers are specifically machines that can only do addition (where we think about multiplication, division and subtraction in terms of addition), and polynomials with rational number substitutions can be solved to great precision with just addition.
– Milan Leonard
1 hour ago
add a comment |Â
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Oh wow. I'd suggest taking a deep dive into understanding what $sin$ and $cos$ actually mean.
– Rushabh Mehta
1 hour ago
Calculators usually use Taylor series to compute $sin$ and $cos$ to arbitrary precision, so unfortunately, there's no easy answer for that.
– Rushabh Mehta
1 hour ago
If you search here for, say, "calculator sine", you'll find a number of places where this kind of question has been asked before. Look at the answers to those and see if they make sense to you. If they don't, then try to explain here why they don't, so that people have a better idea of the type of answer that will satisfy you. Otherwise, people are likely to simply repeat the earlier explanations, wasting everyone's time. Help us help you.
– Blue
1 hour ago
2
@Blue Agreed. I'm not sure how to answer OP since OP specified that Taylor series is too advanced, but Taylor series are usually the tool of choice to compute these functions.
– Rushabh Mehta
1 hour ago
2
@Physicsrocks Sadly, there is no answer to this question that doesn't involve calculus. Maybe this'll motivate you to start learning some basic calculus! Take any standard high school level text -- it should be sufficient for the purpose.
– Rushabh Mehta
1 hour ago