How can I mentally calculate $cos(x), x∈(0.7, 1.2)$
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I'm trying to learn how to calculate trig functions in my head. I'm planning on learning $cos(x), x∈[0,π/2]$ and then using symmetry to calculate the others.
I think the quadratic Maclaurin series at $0$ and the linear at $π/2$ could be calculated in a matter of seconds with some practice. However, I'm struggling to find something that works to 2 D.P. on the interval $(0.7, 1.2)$.
My best idea so far is to use $color{green}{ 1.3-x/1.3}$, but that is neither fast nor accurate to 2 D.P.
Graph of $color{red}{cos(x)}, color{blue}{1-x^2/2}, color{green}{1.3-x/1.3}, color{blue}{π/2-x}$:
Error:
How can I quickly approximate $cos(x)$ for $x∈(0.7, 1.2)$? Or is there a better way to do this?
trigonometry mental-arithmetic
asked Mar 26 '16 at 23:52
Zaz
5441725
add a comment |
up vote
5
down vote
favorite
I'm trying to learn how to calculate trig functions in my head. I'm planning on learning $cos(x), x∈[0,π/2]$ and then using symmetry to calculate the others.
I think the quadratic Maclaurin series at $0$ and the linear at $π/2$ could be calculated in a matter of seconds with some practice. However, I'm struggling to find something that works to 2 D.P. on the interval $(0.7, 1.2)$.
My best idea so far is to use $color{green}{ 1.3-x/1.3}$, but that is neither fast nor accurate to 2 D.P.
Graph of $color{red}{cos(x)}, color{blue}{1-x^2/2}, color{green}{1.3-x/1.3}, color{blue}{π/2-x}$:
Error:
How can I quickly approximate $cos(x)$ for $x∈(0.7, 1.2)$? Or is there a better way to do this?
trigonometry mental-arithmetic
asked Mar 26 '16 at 23:52
Zaz
5441725
A different direction would be to use CORDIC algorithm en.wikipedia.org/wiki/CORDIC
– Jean Marie
Mar 26 '16 at 23:58
Good luck with this. This would be a pretty cool skill to have!
– Alex S
Mar 27 '16 at 0:01
1
If you just want quick off-the-cuff estimates within around 10% error, just memorizing landmarks at every 10 degrees has worked quite well for me, for angles above 20 degrees. I suspect it's probably a much rougher estimate than what you're looking for, but I've found it useful for getting a ballpark and doing quick sanity checks. For angles less than 20 degrees, memorize cos(15-degrees) as well, and for angles less than 10 degrees, go with 0.02 times the angle (twice the angle over 100), or better 0.018 times the angle (subtract one tenth of previous answer from itself).
– brianmearns
May 15 '17 at 12:23
Why do you want to do this?
– JavaMan
Nov 17 at 16:43
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm trying to learn how to calculate trig functions in my head. I'm planning on learning $cos(x), x∈[0,π/2]$ and then using symmetry to calculate the others.
I think the quadratic Maclaurin series at $0$ and the linear at $π/2$ could be calculated in a matter of seconds with some practice. However, I'm struggling to find something that works to 2 D.P. on the interval $(0.7, 1.2)$.
My best idea so far is to use $color{green}{ 1.3-x/1.3}$, but that is neither fast nor accurate to 2 D.P.
Graph of $color{red}{cos(x)}, color{blue}{1-x^2/2}, color{green}{1.3-x/1.3}, color{blue}{π/2-x}$:
Error:
How can I quickly approximate $cos(x)$ for $x∈(0.7, 1.2)$? Or is there a better way to do this?
trigonometry mental-arithmetic
asked Mar 26 '16 at 23:52
Zaz
5441725
I'm trying to learn how to calculate trig functions in my head. I'm planning on learning $cos(x), x∈[0,π/2]$ and then using symmetry to calculate the others.
I think the quadratic Maclaurin series at $0$ and the linear at $π/2$ could be calculated in a matter of seconds with some practice. However, I'm struggling to find something that works to 2 D.P. on the interval $(0.7, 1.2)$.
My best idea so far is to use $color{green}{ 1.3-x/1.3}$, but that is neither fast nor accurate to 2 D.P.
Graph of $color{red}{cos(x)}, color{blue}{1-x^2/2}, color{green}{1.3-x/1.3}, color{blue}{π/2-x}$:
Error:
How can I quickly approximate $cos(x)$ for $x∈(0.7, 1.2)$? Or is there a better way to do this?
trigonometry mental-arithmetic
trigonometry mental-arithmetic
asked Mar 26 '16 at 23:52
Zaz
5441725
asked Mar 26 '16 at 23:52
Zaz
5441725
edited Nov 17 at 16:33
asked Mar 26 '16 at 23:52
Zaz
5441725
asked Mar 26 '16 at 23:52
Zaz
5441725
asked Mar 26 '16 at 23:52
Zaz
5441725
5441725
A different direction would be to use CORDIC algorithm en.wikipedia.org/wiki/CORDIC
– Jean Marie
Mar 26 '16 at 23:58
Good luck with this. This would be a pretty cool skill to have!
– Alex S
Mar 27 '16 at 0:01
1
If you just want quick off-the-cuff estimates within around 10% error, just memorizing landmarks at every 10 degrees has worked quite well for me, for angles above 20 degrees. I suspect it's probably a much rougher estimate than what you're looking for, but I've found it useful for getting a ballpark and doing quick sanity checks. For angles less than 20 degrees, memorize cos(15-degrees) as well, and for angles less than 10 degrees, go with 0.02 times the angle (twice the angle over 100), or better 0.018 times the angle (subtract one tenth of previous answer from itself).
– brianmearns
May 15 '17 at 12:23
Why do you want to do this?
– JavaMan
Nov 17 at 16:43
add a comment |
A different direction would be to use CORDIC algorithm en.wikipedia.org/wiki/CORDIC
– Jean Marie
Mar 26 '16 at 23:58
Good luck with this. This would be a pretty cool skill to have!
– Alex S
Mar 27 '16 at 0:01
1
If you just want quick off-the-cuff estimates within around 10% error, just memorizing landmarks at every 10 degrees has worked quite well for me, for angles above 20 degrees. I suspect it's probably a much rougher estimate than what you're looking for, but I've found it useful for getting a ballpark and doing quick sanity checks. For angles less than 20 degrees, memorize cos(15-degrees) as well, and for angles less than 10 degrees, go with 0.02 times the angle (twice the angle over 100), or better 0.018 times the angle (subtract one tenth of previous answer from itself).
– brianmearns
May 15 '17 at 12:23
Why do you want to do this?
– JavaMan
Nov 17 at 16:43
A different direction would be to use CORDIC algorithm en.wikipedia.org/wiki/CORDIC
– Jean Marie
Mar 26 '16 at 23:58
A different direction would be to use CORDIC algorithm en.wikipedia.org/wiki/CORDIC
– Jean Marie
Mar 26 '16 at 23:58
Good luck with this. This would be a pretty cool skill to have!
– Alex S
Mar 27 '16 at 0:01
Good luck with this. This would be a pretty cool skill to have!
– Alex S
Mar 27 '16 at 0:01
1
1
If you just want quick off-the-cuff estimates within around 10% error, just memorizing landmarks at every 10 degrees has worked quite well for me, for angles above 20 degrees. I suspect it's probably a much rougher estimate than what you're looking for, but I've found it useful for getting a ballpark and doing quick sanity checks. For angles less than 20 degrees, memorize cos(15-degrees) as well, and for angles less than 10 degrees, go with 0.02 times the angle (twice the angle over 100), or better 0.018 times the angle (subtract one tenth of previous answer from itself).
– brianmearns
May 15 '17 at 12:23
If you just want quick off-the-cuff estimates within around 10% error, just memorizing landmarks at every 10 degrees has worked quite well for me, for angles above 20 degrees. I suspect it's probably a much rougher estimate than what you're looking for, but I've found it useful for getting a ballpark and doing quick sanity checks. For angles less than 20 degrees, memorize cos(15-degrees) as well, and for angles less than 10 degrees, go with 0.02 times the angle (twice the angle over 100), or better 0.018 times the angle (subtract one tenth of previous answer from itself).
– brianmearns
May 15 '17 at 12:23
Why do you want to do this?
– JavaMan
Nov 17 at 16:43
Why do you want to do this?
– JavaMan
Nov 17 at 16:43
add a comment |
1 Answer
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Here's a quick approximation. The second-order Taylor series around $x=pi/3$ is
$$T_2(x) = frac12 - frac{sqrt3}2left(x-fracpi3right)-frac14left(x-fracpi3right)^2.$$
Now, $frac{sqrt3}2 = 0.866... approx frac{13}{15}$, and $fracpi3 approx 1.05$, so we have
$$T_2(x) approx frac12 - frac{13}{15}(x-1.05) - frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $frac{sqrt3}2 approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
1
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here's a quick approximation. The second-order Taylor series around $x=pi/3$ is
$$T_2(x) = frac12 - frac{sqrt3}2left(x-fracpi3right)-frac14left(x-fracpi3right)^2.$$
Now, $frac{sqrt3}2 = 0.866... approx frac{13}{15}$, and $fracpi3 approx 1.05$, so we have
$$T_2(x) approx frac12 - frac{13}{15}(x-1.05) - frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $frac{sqrt3}2 approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
1
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
add a comment |
up vote
1
down vote
Here's a quick approximation. The second-order Taylor series around $x=pi/3$ is
$$T_2(x) = frac12 - frac{sqrt3}2left(x-fracpi3right)-frac14left(x-fracpi3right)^2.$$
Now, $frac{sqrt3}2 = 0.866... approx frac{13}{15}$, and $fracpi3 approx 1.05$, so we have
$$T_2(x) approx frac12 - frac{13}{15}(x-1.05) - frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $frac{sqrt3}2 approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
1
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
add a comment |
up vote
1
down vote
up vote
1
down vote
Here's a quick approximation. The second-order Taylor series around $x=pi/3$ is
$$T_2(x) = frac12 - frac{sqrt3}2left(x-fracpi3right)-frac14left(x-fracpi3right)^2.$$
Now, $frac{sqrt3}2 = 0.866... approx frac{13}{15}$, and $fracpi3 approx 1.05$, so we have
$$T_2(x) approx frac12 - frac{13}{15}(x-1.05) - frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $frac{sqrt3}2 approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
Here's a quick approximation. The second-order Taylor series around $x=pi/3$ is
$$T_2(x) = frac12 - frac{sqrt3}2left(x-fracpi3right)-frac14left(x-fracpi3right)^2.$$
Now, $frac{sqrt3}2 = 0.866... approx frac{13}{15}$, and $fracpi3 approx 1.05$, so we have
$$T_2(x) approx frac12 - frac{13}{15}(x-1.05) - frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $frac{sqrt3}2 approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
answered Mar 27 '16 at 2:54
Théophile
19.3k12946
19.3k12946
1
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
add a comment |
1
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
1
1
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
If I were trying to calculate this mentally, I would probably put $z=x-1.05$, and then say $T_2(x)approx (30-z(52+15z))/60$
– G Tony Jacobs
Mar 27 '16 at 15:57
add a comment |
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A different direction would be to use CORDIC algorithm en.wikipedia.org/wiki/CORDIC
– Jean Marie
Mar 26 '16 at 23:58
Good luck with this. This would be a pretty cool skill to have!
– Alex S
Mar 27 '16 at 0:01
1
If you just want quick off-the-cuff estimates within around 10% error, just memorizing landmarks at every 10 degrees has worked quite well for me, for angles above 20 degrees. I suspect it's probably a much rougher estimate than what you're looking for, but I've found it useful for getting a ballpark and doing quick sanity checks. For angles less than 20 degrees, memorize cos(15-degrees) as well, and for angles less than 10 degrees, go with 0.02 times the angle (twice the angle over 100), or better 0.018 times the angle (subtract one tenth of previous answer from itself).
– brianmearns
May 15 '17 at 12:23
Why do you want to do this?
– JavaMan
Nov 17 at 16:43