Can $cos(36^circ)$ be calculated using derivatives?Compute $cos'(0)$How can I verify the Pythagorean Trig Identity using approximations for $sin x$ and $cos x$ derived from their infinite series representations?Functions and derivatives using limitsHow were the sine, cosine and tangent tables originally calculated?$cos(47^circ)sin(32^circ)$ approximation by differentialsFinding the Equation of a Line Using Horizontal Tangents and Derivativesfind minimum value of $2^sin^2(theta)+2^cos^2(theta)$How can you measure angle degrees in nature?Derivative of $tan x$ without using derivative of $sin x$ and $cos x$Evaluating $fracsin A + sin B + sin Ccos A + cos B + cos C$ for a triangle with sides $2$, $3$, $4$
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Can $cos(36^circ)$ be calculated using derivatives?
Compute $cos'(0)$How can I verify the Pythagorean Trig Identity using approximations for $sin x$ and $cos x$ derived from their infinite series representations?Functions and derivatives using limitsHow were the sine, cosine and tangent tables originally calculated?$cos(47^circ)sin(32^circ)$ approximation by differentialsFinding the Equation of a Line Using Horizontal Tangents and Derivativesfind minimum value of $2^sin^2(theta)+2^cos^2(theta)$How can you measure angle degrees in nature?Derivative of $tan x$ without using derivative of $sin x$ and $cos x$Evaluating $fracsin A + sin B + sin Ccos A + cos B + cos C$ for a triangle with sides $2$, $3$, $4$
$begingroup$
Can $cos(36°)$ be calculated using known cosines like $cos(37°)$ and the derivative formula $$f'(x) approx fracf(x+ Δx) - f(x)Delta x$$
?
(We have to let $f(x):=cos(x)$, so we can let $x=37°$, or in radian form, and let $f(x+Delta x)$ be $cos(x+Delta x)$ and $Delta x=-1°$ or in radian form, so we can say $cos(37°-1°)$. I just do not know the process. I understand it is approximate, but I cannot calculate it approximately.
derivatives trigonometry
edited Mar 29 at 18:08
David G. Stork
12k41735
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
$endgroup$
|
show 4 more comments
$begingroup$
Can $cos(36°)$ be calculated using known cosines like $cos(37°)$ and the derivative formula $$f'(x) approx fracf(x+ Δx) - f(x)Delta x$$
?
(We have to let $f(x):=cos(x)$, so we can let $x=37°$, or in radian form, and let $f(x+Delta x)$ be $cos(x+Delta x)$ and $Delta x=-1°$ or in radian form, so we can say $cos(37°-1°)$. I just do not know the process. I understand it is approximate, but I cannot calculate it approximately.
derivatives trigonometry
edited Mar 29 at 18:08
David G. Stork
12k41735
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
$endgroup$
$begingroup$
Is this in degrees?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 16:53
5
$begingroup$
I'm quite surprise that you know $cos(37)$...
$endgroup$
– user657324
Mar 29 at 16:53
1
$begingroup$
Yes it is in degrees. My deepest apologies for not knowing formula scripts. I just want knowledge in derivatives. Please forgive me. Cos(37deg) is 0.8 and Sin(37deg) is 0.6 we had to learn it at school.
$endgroup$
– Alireza Honarvar
Mar 29 at 16:55
3
$begingroup$
the cosine of $36°$ is really special: $$cos(36°)=fracvarphi2=frac1+sqrt54$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:56
2
$begingroup$
@AlirezaHonarvar But, actually $$cos(37°)not=0.8$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:57
|
show 4 more comments
$begingroup$
Can $cos(36°)$ be calculated using known cosines like $cos(37°)$ and the derivative formula $$f'(x) approx fracf(x+ Δx) - f(x)Delta x$$
?
(We have to let $f(x):=cos(x)$, so we can let $x=37°$, or in radian form, and let $f(x+Delta x)$ be $cos(x+Delta x)$ and $Delta x=-1°$ or in radian form, so we can say $cos(37°-1°)$. I just do not know the process. I understand it is approximate, but I cannot calculate it approximately.
derivatives trigonometry
edited Mar 29 at 18:08
David G. Stork
12k41735
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
$endgroup$
Can $cos(36°)$ be calculated using known cosines like $cos(37°)$ and the derivative formula $$f'(x) approx fracf(x+ Δx) - f(x)Delta x$$
?
(We have to let $f(x):=cos(x)$, so we can let $x=37°$, or in radian form, and let $f(x+Delta x)$ be $cos(x+Delta x)$ and $Delta x=-1°$ or in radian form, so we can say $cos(37°-1°)$. I just do not know the process. I understand it is approximate, but I cannot calculate it approximately.
derivatives trigonometry
derivatives trigonometry
edited Mar 29 at 18:08
David G. Stork
12k41735
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
edited Mar 29 at 18:08
David G. Stork
12k41735
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
edited Mar 29 at 18:08
David G. Stork
12k41735
edited Mar 29 at 18:08
David G. Stork
12k41735
edited Mar 29 at 18:08
David G. Stork
12k41735
12k41735
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
asked Mar 29 at 16:49
Alireza HonarvarAlireza Honarvar
274
274
$begingroup$
Is this in degrees?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 16:53
5
$begingroup$
I'm quite surprise that you know $cos(37)$...
$endgroup$
– user657324
Mar 29 at 16:53
1
$begingroup$
Yes it is in degrees. My deepest apologies for not knowing formula scripts. I just want knowledge in derivatives. Please forgive me. Cos(37deg) is 0.8 and Sin(37deg) is 0.6 we had to learn it at school.
$endgroup$
– Alireza Honarvar
Mar 29 at 16:55
3
$begingroup$
the cosine of $36°$ is really special: $$cos(36°)=fracvarphi2=frac1+sqrt54$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:56
2
$begingroup$
@AlirezaHonarvar But, actually $$cos(37°)not=0.8$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:57
|
show 4 more comments
$begingroup$
Is this in degrees?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 16:53
5
$begingroup$
I'm quite surprise that you know $cos(37)$...
$endgroup$
– user657324
Mar 29 at 16:53
1
$begingroup$
Yes it is in degrees. My deepest apologies for not knowing formula scripts. I just want knowledge in derivatives. Please forgive me. Cos(37deg) is 0.8 and Sin(37deg) is 0.6 we had to learn it at school.
$endgroup$
– Alireza Honarvar
Mar 29 at 16:55
3
$begingroup$
the cosine of $36°$ is really special: $$cos(36°)=fracvarphi2=frac1+sqrt54$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:56
2
$begingroup$
@AlirezaHonarvar But, actually $$cos(37°)not=0.8$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:57
$begingroup$
Is this in degrees?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 16:53
$begingroup$
Is this in degrees?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 16:53
5
5
$begingroup$
I'm quite surprise that you know $cos(37)$...
$endgroup$
– user657324
Mar 29 at 16:53
$begingroup$
I'm quite surprise that you know $cos(37)$...
$endgroup$
– user657324
Mar 29 at 16:53
1
1
$begingroup$
Yes it is in degrees. My deepest apologies for not knowing formula scripts. I just want knowledge in derivatives. Please forgive me. Cos(37deg) is 0.8 and Sin(37deg) is 0.6 we had to learn it at school.
$endgroup$
– Alireza Honarvar
Mar 29 at 16:55
$begingroup$
Yes it is in degrees. My deepest apologies for not knowing formula scripts. I just want knowledge in derivatives. Please forgive me. Cos(37deg) is 0.8 and Sin(37deg) is 0.6 we had to learn it at school.
$endgroup$
– Alireza Honarvar
Mar 29 at 16:55
3
3
$begingroup$
the cosine of $36°$ is really special: $$cos(36°)=fracvarphi2=frac1+sqrt54$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:56
$begingroup$
the cosine of $36°$ is really special: $$cos(36°)=fracvarphi2=frac1+sqrt54$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:56
2
2
$begingroup$
@AlirezaHonarvar But, actually $$cos(37°)not=0.8$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:57
$begingroup$
@AlirezaHonarvar But, actually $$cos(37°)not=0.8$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:57
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If we know that $cos(37^circ)approx 0.8$ and $sin(37^circ)approx 0.6$ then we can use the fact that
$$fracddx cos(x)=-sin(x)$$
to approximate $cos(36^circ)$ as
$$cos(36^circ)approx 0.8+0.6(fracpi180) = 0.810...$$
Where $1^circ = fracpi180$ radians which must be used here.
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
$endgroup$
add a comment |
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$begingroup$
If we know that $cos(37^circ)approx 0.8$ and $sin(37^circ)approx 0.6$ then we can use the fact that
$$fracddx cos(x)=-sin(x)$$
to approximate $cos(36^circ)$ as
$$cos(36^circ)approx 0.8+0.6(fracpi180) = 0.810...$$
Where $1^circ = fracpi180$ radians which must be used here.
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
$endgroup$
add a comment |
$begingroup$
If we know that $cos(37^circ)approx 0.8$ and $sin(37^circ)approx 0.6$ then we can use the fact that
$$fracddx cos(x)=-sin(x)$$
to approximate $cos(36^circ)$ as
$$cos(36^circ)approx 0.8+0.6(fracpi180) = 0.810...$$
Where $1^circ = fracpi180$ radians which must be used here.
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
$endgroup$
add a comment |
$begingroup$
If we know that $cos(37^circ)approx 0.8$ and $sin(37^circ)approx 0.6$ then we can use the fact that
$$fracddx cos(x)=-sin(x)$$
to approximate $cos(36^circ)$ as
$$cos(36^circ)approx 0.8+0.6(fracpi180) = 0.810...$$
Where $1^circ = fracpi180$ radians which must be used here.
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
$endgroup$
If we know that $cos(37^circ)approx 0.8$ and $sin(37^circ)approx 0.6$ then we can use the fact that
$$fracddx cos(x)=-sin(x)$$
to approximate $cos(36^circ)$ as
$$cos(36^circ)approx 0.8+0.6(fracpi180) = 0.810...$$
Where $1^circ = fracpi180$ radians which must be used here.
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
answered Mar 29 at 17:08
Peter ForemanPeter Foreman
5,9091317
5,9091317
add a comment |
add a comment |
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$begingroup$
Is this in degrees?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 16:53
5
$begingroup$
I'm quite surprise that you know $cos(37)$...
$endgroup$
– user657324
Mar 29 at 16:53
1
$begingroup$
Yes it is in degrees. My deepest apologies for not knowing formula scripts. I just want knowledge in derivatives. Please forgive me. Cos(37deg) is 0.8 and Sin(37deg) is 0.6 we had to learn it at school.
$endgroup$
– Alireza Honarvar
Mar 29 at 16:55
3
$begingroup$
the cosine of $36°$ is really special: $$cos(36°)=fracvarphi2=frac1+sqrt54$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:56
2
$begingroup$
@AlirezaHonarvar But, actually $$cos(37°)not=0.8$$
$endgroup$
– Dr. Mathva
Mar 29 at 16:57