Mixing
I don't understand why $cos(a+2 pi)= sin(π/2−a) $ is true [closed]
$begingroup$
When I place them on a unit circle they create a triangle and shouldn't be the same? Could someone explain it please, because my teacher skipped it
trigonometry
edited Jan 2 at 13:08
mrtaurho
4,06121234
asked Jan 2 at 13:07
ythhtrgythhtrg
91
$endgroup$
closed as unclear what you're asking by José Carlos Santos, Martin R, amWhy, Adrian Keister, Andrei Jan 2 at 19:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
When I place them on a unit circle they create a triangle and shouldn't be the same? Could someone explain it please, because my teacher skipped it
trigonometry
edited Jan 2 at 13:08
mrtaurho
4,06121234
asked Jan 2 at 13:07
ythhtrgythhtrg
91
$endgroup$
closed as unclear what you're asking by José Carlos Santos, Martin R, amWhy, Adrian Keister, Andrei Jan 2 at 19:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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Do you see why $cos(a+2pi)=cos a$?
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– Lord Shark the Unknown
Jan 2 at 13:09
$begingroup$
Use the formula for cosine of sum and sine of subtraction
$endgroup$
– Makina
Jan 2 at 13:11
1
$begingroup$
What exactly did you place on the unit circle? The two angles generally don't point to the same place on the circle, but then you take the sine of one and the cosine of the other, so why shouldn't they be the same? Actually proving that they are the same is more work than that, of course.
$endgroup$
– David K
Jan 2 at 13:29
add a comment |
$begingroup$
When I place them on a unit circle they create a triangle and shouldn't be the same? Could someone explain it please, because my teacher skipped it
trigonometry
edited Jan 2 at 13:08
mrtaurho
4,06121234
asked Jan 2 at 13:07
ythhtrgythhtrg
91
$endgroup$
When I place them on a unit circle they create a triangle and shouldn't be the same? Could someone explain it please, because my teacher skipped it
trigonometry
trigonometry
edited Jan 2 at 13:08
mrtaurho
4,06121234
asked Jan 2 at 13:07
ythhtrgythhtrg
91
edited Jan 2 at 13:08
mrtaurho
4,06121234
asked Jan 2 at 13:07
ythhtrgythhtrg
91
edited Jan 2 at 13:08
mrtaurho
4,06121234
edited Jan 2 at 13:08
mrtaurho
4,06121234
edited Jan 2 at 13:08
mrtaurho
4,06121234
4,06121234
asked Jan 2 at 13:07
ythhtrgythhtrg
91
asked Jan 2 at 13:07
ythhtrgythhtrg
91
asked Jan 2 at 13:07
ythhtrgythhtrg
91
91
closed as unclear what you're asking by José Carlos Santos, Martin R, amWhy, Adrian Keister, Andrei Jan 2 at 19:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by José Carlos Santos, Martin R, amWhy, Adrian Keister, Andrei Jan 2 at 19:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Do you see why $cos(a+2pi)=cos a$?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 13:09
$begingroup$
Use the formula for cosine of sum and sine of subtraction
$endgroup$
– Makina
Jan 2 at 13:11
1
$begingroup$
What exactly did you place on the unit circle? The two angles generally don't point to the same place on the circle, but then you take the sine of one and the cosine of the other, so why shouldn't they be the same? Actually proving that they are the same is more work than that, of course.
$endgroup$
– David K
Jan 2 at 13:29
add a comment |
2
$begingroup$
Do you see why $cos(a+2pi)=cos a$?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 13:09
$begingroup$
Use the formula for cosine of sum and sine of subtraction
$endgroup$
– Makina
Jan 2 at 13:11
1
$begingroup$
What exactly did you place on the unit circle? The two angles generally don't point to the same place on the circle, but then you take the sine of one and the cosine of the other, so why shouldn't they be the same? Actually proving that they are the same is more work than that, of course.
$endgroup$
– David K
Jan 2 at 13:29
2
2
$begingroup$
Do you see why $cos(a+2pi)=cos a$?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 13:09
$begingroup$
Do you see why $cos(a+2pi)=cos a$?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 13:09
$begingroup$
Use the formula for cosine of sum and sine of subtraction
$endgroup$
– Makina
Jan 2 at 13:11
$begingroup$
Use the formula for cosine of sum and sine of subtraction
$endgroup$
– Makina
Jan 2 at 13:11
1
1
$begingroup$
What exactly did you place on the unit circle? The two angles generally don't point to the same place on the circle, but then you take the sine of one and the cosine of the other, so why shouldn't they be the same? Actually proving that they are the same is more work than that, of course.
$endgroup$
– David K
Jan 2 at 13:29
$begingroup$
What exactly did you place on the unit circle? The two angles generally don't point to the same place on the circle, but then you take the sine of one and the cosine of the other, so why shouldn't they be the same? Actually proving that they are the same is more work than that, of course.
$endgroup$
– David K
Jan 2 at 13:29
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We know two major things about the sine and cosine function. First of all that they have a period of $2pi$ which can be written as
$$sin(x+2pi)=sin(x)text{ and }cos(x+2pi)=cos(x)$$
Secondly we know that the cosine function is the sine function which has been moved $pi/2$ along the $x$-axis and vice versa which also can be written as
$$sin(x+pi/2)=cos(x)text{ and }cos(x-pi/2)=sin(x)$$
Using these two we can rewrite your formula as the following
$$cos(a+2pi)=cos(a)text{ and }sin(-a+pi/2)=cos(-a)$$
so we are left with $cos(a)=cos(-a)$ and this turns out to be true since the cosine function is an even function and therefore has the property that for all $x$ we got $f(x)=f(-x)$.
Putting all this together yields to
$$cos(a+2pi)=cos(a)=cos(-a)=sin(-a+pi/2)=sin(pi/2-a)$$
$$therefore~cos(a+2pi)=sin(pi/2-a)$$
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
$endgroup$
add a comment |
$begingroup$
Using that $cos$ is a $2pi$-periodic function, we have $cos(a)=cos(a+2pi)$.
Now, for $ain[0,pi/2]$, imagine a right triangle with angles $angle A=a$ and $angle B=pi/2 -a$. Then,
$$begin{cases} cos(A)=frac{adj}{hyp}=frac{AC}{AB} \ sin(B)=frac{op}{hyp}=frac{AC}{AB}end{cases}$$
Finally, $cos(a+2pi)=cos(a)=cos(A)=sin(B)=sin(pi/2-a)$ for all $ain[0,pi/2]$.
Now, using the simmetries of $cos$ and $sin$ arround the axis, the equallity holds for all $ain[-pi,pi]$, and, by periodicity, holds for all $ainmathbb{R}$
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
add a comment |
$begingroup$
Cosine is called "cosine" because it's the SINE of the COmplementary angle. So, by definition, sort of, $cos(a) = sin(pi/2 -a).$
Then the fact that $cos$ has period $2pi$ gives
$$cos(a+2pi) = cos(a) = sin(pi/2-a).$$
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
$endgroup$
add a comment |
$begingroup$
Use that $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ and $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
4
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know two major things about the sine and cosine function. First of all that they have a period of $2pi$ which can be written as
$$sin(x+2pi)=sin(x)text{ and }cos(x+2pi)=cos(x)$$
Secondly we know that the cosine function is the sine function which has been moved $pi/2$ along the $x$-axis and vice versa which also can be written as
$$sin(x+pi/2)=cos(x)text{ and }cos(x-pi/2)=sin(x)$$
Using these two we can rewrite your formula as the following
$$cos(a+2pi)=cos(a)text{ and }sin(-a+pi/2)=cos(-a)$$
so we are left with $cos(a)=cos(-a)$ and this turns out to be true since the cosine function is an even function and therefore has the property that for all $x$ we got $f(x)=f(-x)$.
Putting all this together yields to
$$cos(a+2pi)=cos(a)=cos(-a)=sin(-a+pi/2)=sin(pi/2-a)$$
$$therefore~cos(a+2pi)=sin(pi/2-a)$$
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
$endgroup$
add a comment |
$begingroup$
We know two major things about the sine and cosine function. First of all that they have a period of $2pi$ which can be written as
$$sin(x+2pi)=sin(x)text{ and }cos(x+2pi)=cos(x)$$
Secondly we know that the cosine function is the sine function which has been moved $pi/2$ along the $x$-axis and vice versa which also can be written as
$$sin(x+pi/2)=cos(x)text{ and }cos(x-pi/2)=sin(x)$$
Using these two we can rewrite your formula as the following
$$cos(a+2pi)=cos(a)text{ and }sin(-a+pi/2)=cos(-a)$$
so we are left with $cos(a)=cos(-a)$ and this turns out to be true since the cosine function is an even function and therefore has the property that for all $x$ we got $f(x)=f(-x)$.
Putting all this together yields to
$$cos(a+2pi)=cos(a)=cos(-a)=sin(-a+pi/2)=sin(pi/2-a)$$
$$therefore~cos(a+2pi)=sin(pi/2-a)$$
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
$endgroup$
add a comment |
$begingroup$
We know two major things about the sine and cosine function. First of all that they have a period of $2pi$ which can be written as
$$sin(x+2pi)=sin(x)text{ and }cos(x+2pi)=cos(x)$$
Secondly we know that the cosine function is the sine function which has been moved $pi/2$ along the $x$-axis and vice versa which also can be written as
$$sin(x+pi/2)=cos(x)text{ and }cos(x-pi/2)=sin(x)$$
Using these two we can rewrite your formula as the following
$$cos(a+2pi)=cos(a)text{ and }sin(-a+pi/2)=cos(-a)$$
so we are left with $cos(a)=cos(-a)$ and this turns out to be true since the cosine function is an even function and therefore has the property that for all $x$ we got $f(x)=f(-x)$.
Putting all this together yields to
$$cos(a+2pi)=cos(a)=cos(-a)=sin(-a+pi/2)=sin(pi/2-a)$$
$$therefore~cos(a+2pi)=sin(pi/2-a)$$
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
$endgroup$
We know two major things about the sine and cosine function. First of all that they have a period of $2pi$ which can be written as
$$sin(x+2pi)=sin(x)text{ and }cos(x+2pi)=cos(x)$$
Secondly we know that the cosine function is the sine function which has been moved $pi/2$ along the $x$-axis and vice versa which also can be written as
$$sin(x+pi/2)=cos(x)text{ and }cos(x-pi/2)=sin(x)$$
Using these two we can rewrite your formula as the following
$$cos(a+2pi)=cos(a)text{ and }sin(-a+pi/2)=cos(-a)$$
so we are left with $cos(a)=cos(-a)$ and this turns out to be true since the cosine function is an even function and therefore has the property that for all $x$ we got $f(x)=f(-x)$.
Putting all this together yields to
$$cos(a+2pi)=cos(a)=cos(-a)=sin(-a+pi/2)=sin(pi/2-a)$$
$$therefore~cos(a+2pi)=sin(pi/2-a)$$
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
answered Jan 2 at 13:15
mrtaurhomrtaurho
4,06121234
4,06121234
add a comment |
add a comment |
$begingroup$
Using that $cos$ is a $2pi$-periodic function, we have $cos(a)=cos(a+2pi)$.
Now, for $ain[0,pi/2]$, imagine a right triangle with angles $angle A=a$ and $angle B=pi/2 -a$. Then,
$$begin{cases} cos(A)=frac{adj}{hyp}=frac{AC}{AB} \ sin(B)=frac{op}{hyp}=frac{AC}{AB}end{cases}$$
Finally, $cos(a+2pi)=cos(a)=cos(A)=sin(B)=sin(pi/2-a)$ for all $ain[0,pi/2]$.
Now, using the simmetries of $cos$ and $sin$ arround the axis, the equallity holds for all $ain[-pi,pi]$, and, by periodicity, holds for all $ainmathbb{R}$
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
add a comment |
$begingroup$
Using that $cos$ is a $2pi$-periodic function, we have $cos(a)=cos(a+2pi)$.
Now, for $ain[0,pi/2]$, imagine a right triangle with angles $angle A=a$ and $angle B=pi/2 -a$. Then,
$$begin{cases} cos(A)=frac{adj}{hyp}=frac{AC}{AB} \ sin(B)=frac{op}{hyp}=frac{AC}{AB}end{cases}$$
Finally, $cos(a+2pi)=cos(a)=cos(A)=sin(B)=sin(pi/2-a)$ for all $ain[0,pi/2]$.
Now, using the simmetries of $cos$ and $sin$ arround the axis, the equallity holds for all $ain[-pi,pi]$, and, by periodicity, holds for all $ainmathbb{R}$
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
add a comment |
$begingroup$
Using that $cos$ is a $2pi$-periodic function, we have $cos(a)=cos(a+2pi)$.
Now, for $ain[0,pi/2]$, imagine a right triangle with angles $angle A=a$ and $angle B=pi/2 -a$. Then,
$$begin{cases} cos(A)=frac{adj}{hyp}=frac{AC}{AB} \ sin(B)=frac{op}{hyp}=frac{AC}{AB}end{cases}$$
Finally, $cos(a+2pi)=cos(a)=cos(A)=sin(B)=sin(pi/2-a)$ for all $ain[0,pi/2]$.
Now, using the simmetries of $cos$ and $sin$ arround the axis, the equallity holds for all $ain[-pi,pi]$, and, by periodicity, holds for all $ainmathbb{R}$
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
Using that $cos$ is a $2pi$-periodic function, we have $cos(a)=cos(a+2pi)$.
Now, for $ain[0,pi/2]$, imagine a right triangle with angles $angle A=a$ and $angle B=pi/2 -a$. Then,
$$begin{cases} cos(A)=frac{adj}{hyp}=frac{AC}{AB} \ sin(B)=frac{op}{hyp}=frac{AC}{AB}end{cases}$$
Finally, $cos(a+2pi)=cos(a)=cos(A)=sin(B)=sin(pi/2-a)$ for all $ain[0,pi/2]$.
Now, using the simmetries of $cos$ and $sin$ arround the axis, the equallity holds for all $ain[-pi,pi]$, and, by periodicity, holds for all $ainmathbb{R}$
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
answered Jan 2 at 13:20
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
3,810623
add a comment |
add a comment |
$begingroup$
Cosine is called "cosine" because it's the SINE of the COmplementary angle. So, by definition, sort of, $cos(a) = sin(pi/2 -a).$
Then the fact that $cos$ has period $2pi$ gives
$$cos(a+2pi) = cos(a) = sin(pi/2-a).$$
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
$endgroup$
add a comment |
$begingroup$
Cosine is called "cosine" because it's the SINE of the COmplementary angle. So, by definition, sort of, $cos(a) = sin(pi/2 -a).$
Then the fact that $cos$ has period $2pi$ gives
$$cos(a+2pi) = cos(a) = sin(pi/2-a).$$
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
$endgroup$
add a comment |
$begingroup$
Cosine is called "cosine" because it's the SINE of the COmplementary angle. So, by definition, sort of, $cos(a) = sin(pi/2 -a).$
Then the fact that $cos$ has period $2pi$ gives
$$cos(a+2pi) = cos(a) = sin(pi/2-a).$$
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
$endgroup$
Cosine is called "cosine" because it's the SINE of the COmplementary angle. So, by definition, sort of, $cos(a) = sin(pi/2 -a).$
Then the fact that $cos$ has period $2pi$ gives
$$cos(a+2pi) = cos(a) = sin(pi/2-a).$$
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
answered Jan 2 at 13:13
B. GoddardB. Goddard
18.5k21340
18.5k21340
add a comment |
add a comment |
$begingroup$
Use that $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ and $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
4
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
add a comment |
$begingroup$
Use that $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ and $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
4
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
add a comment |
$begingroup$
Use that $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ and $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
Use that $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ and $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 13:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
4
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
add a comment |
4
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
4
4
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
$begingroup$
This is overkill
$endgroup$
– Rhys Hughes
Jan 2 at 13:37
add a comment |
2
$begingroup$
Do you see why $cos(a+2pi)=cos a$?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 13:09
$begingroup$
Use the formula for cosine of sum and sine of subtraction
$endgroup$
– Makina
Jan 2 at 13:11
1
$begingroup$
What exactly did you place on the unit circle? The two angles generally don't point to the same place on the circle, but then you take the sine of one and the cosine of the other, so why shouldn't they be the same? Actually proving that they are the same is more work than that, of course.
$endgroup$
– David K
Jan 2 at 13:29