Is this a valid way of finding the an answer using the unit circle?
$begingroup$
If I want to find what the value for $sin(-3 pi) over 4$ is, can I just find $sin(3 pi) over 4$ because it's easier (having a positive x) and then just multiply it by $-1$?
calculus
asked Dec 7 '18 at 7:56
mingming
3415
$endgroup$
add a comment |
$begingroup$
If I want to find what the value for $sin(-3 pi) over 4$ is, can I just find $sin(3 pi) over 4$ because it's easier (having a positive x) and then just multiply it by $-1$?
calculus
asked Dec 7 '18 at 7:56
mingming
3415
$endgroup$
2
$begingroup$
Do you mean $sin frac{-3pi}{4}$? If so, then yes, because $sin x$ is an odd function, meaning $sin (-x) = -sin x$. (Even if you didn’t mean that, it still applies, but it’s probably not what you meant.)
$endgroup$
– KM101
Dec 7 '18 at 8:00
$begingroup$
Yeah it wasn't what I meant I think I just got lucky by using that method. So I just learned about how $sin(x) = -sin(x)$ because it's an odd function but how come $cos(x) = cos(-x)$? That means that $cos(x)$ is an even function but isn't $sin$ and $cos$ sorta the same graph?
$endgroup$
– ming
Dec 7 '18 at 15:36
$begingroup$
I’ll clarify on that right now.
$endgroup$
– KM101
Dec 7 '18 at 15:39
$begingroup$
They aren’t the same graph. They have the exact same shape but they’re $frac{pi}{2}$ radians apart (in the case of $sin x$ and $cos x$). That’s like saying $y = 2x$ and $y = 2x+3$ are the same.
$endgroup$
– KM101
Dec 7 '18 at 16:01
$begingroup$
You are right that $sin x$ is an odd function, and $cos x$ is an even function. And yes, the graph of one is a phase shift of the graph of the other. The behavior of the graph around $0$ will show that one is even and one is odd.
$endgroup$
– Doug M
Dec 7 '18 at 16:02
add a comment |
$begingroup$
If I want to find what the value for $sin(-3 pi) over 4$ is, can I just find $sin(3 pi) over 4$ because it's easier (having a positive x) and then just multiply it by $-1$?
calculus
asked Dec 7 '18 at 7:56
mingming
3415
$endgroup$
If I want to find what the value for $sin(-3 pi) over 4$ is, can I just find $sin(3 pi) over 4$ because it's easier (having a positive x) and then just multiply it by $-1$?
calculus
calculus
asked Dec 7 '18 at 7:56
mingming
3415
asked Dec 7 '18 at 7:56
mingming
3415
asked Dec 7 '18 at 7:56
mingming
3415
asked Dec 7 '18 at 7:56
mingming
3415
asked Dec 7 '18 at 7:56
mingming
3415
3415
2
$begingroup$
Do you mean $sin frac{-3pi}{4}$? If so, then yes, because $sin x$ is an odd function, meaning $sin (-x) = -sin x$. (Even if you didn’t mean that, it still applies, but it’s probably not what you meant.)
$endgroup$
– KM101
Dec 7 '18 at 8:00
$begingroup$
Yeah it wasn't what I meant I think I just got lucky by using that method. So I just learned about how $sin(x) = -sin(x)$ because it's an odd function but how come $cos(x) = cos(-x)$? That means that $cos(x)$ is an even function but isn't $sin$ and $cos$ sorta the same graph?
$endgroup$
– ming
Dec 7 '18 at 15:36
$begingroup$
I’ll clarify on that right now.
$endgroup$
– KM101
Dec 7 '18 at 15:39
$begingroup$
They aren’t the same graph. They have the exact same shape but they’re $frac{pi}{2}$ radians apart (in the case of $sin x$ and $cos x$). That’s like saying $y = 2x$ and $y = 2x+3$ are the same.
$endgroup$
– KM101
Dec 7 '18 at 16:01
$begingroup$
You are right that $sin x$ is an odd function, and $cos x$ is an even function. And yes, the graph of one is a phase shift of the graph of the other. The behavior of the graph around $0$ will show that one is even and one is odd.
$endgroup$
– Doug M
Dec 7 '18 at 16:02
add a comment |
2
$begingroup$
Do you mean $sin frac{-3pi}{4}$? If so, then yes, because $sin x$ is an odd function, meaning $sin (-x) = -sin x$. (Even if you didn’t mean that, it still applies, but it’s probably not what you meant.)
$endgroup$
– KM101
Dec 7 '18 at 8:00
$begingroup$
Yeah it wasn't what I meant I think I just got lucky by using that method. So I just learned about how $sin(x) = -sin(x)$ because it's an odd function but how come $cos(x) = cos(-x)$? That means that $cos(x)$ is an even function but isn't $sin$ and $cos$ sorta the same graph?
$endgroup$
– ming
Dec 7 '18 at 15:36
$begingroup$
I’ll clarify on that right now.
$endgroup$
– KM101
Dec 7 '18 at 15:39
$begingroup$
They aren’t the same graph. They have the exact same shape but they’re $frac{pi}{2}$ radians apart (in the case of $sin x$ and $cos x$). That’s like saying $y = 2x$ and $y = 2x+3$ are the same.
$endgroup$
– KM101
Dec 7 '18 at 16:01
$begingroup$
You are right that $sin x$ is an odd function, and $cos x$ is an even function. And yes, the graph of one is a phase shift of the graph of the other. The behavior of the graph around $0$ will show that one is even and one is odd.
$endgroup$
– Doug M
Dec 7 '18 at 16:02
2
2
$begingroup$
Do you mean $sin frac{-3pi}{4}$? If so, then yes, because $sin x$ is an odd function, meaning $sin (-x) = -sin x$. (Even if you didn’t mean that, it still applies, but it’s probably not what you meant.)
$endgroup$
– KM101
Dec 7 '18 at 8:00
$begingroup$
Do you mean $sin frac{-3pi}{4}$? If so, then yes, because $sin x$ is an odd function, meaning $sin (-x) = -sin x$. (Even if you didn’t mean that, it still applies, but it’s probably not what you meant.)
$endgroup$
– KM101
Dec 7 '18 at 8:00
$begingroup$
Yeah it wasn't what I meant I think I just got lucky by using that method. So I just learned about how $sin(x) = -sin(x)$ because it's an odd function but how come $cos(x) = cos(-x)$? That means that $cos(x)$ is an even function but isn't $sin$ and $cos$ sorta the same graph?
$endgroup$
– ming
Dec 7 '18 at 15:36
$begingroup$
Yeah it wasn't what I meant I think I just got lucky by using that method. So I just learned about how $sin(x) = -sin(x)$ because it's an odd function but how come $cos(x) = cos(-x)$? That means that $cos(x)$ is an even function but isn't $sin$ and $cos$ sorta the same graph?
$endgroup$
– ming
Dec 7 '18 at 15:36
$begingroup$
I’ll clarify on that right now.
$endgroup$
– KM101
Dec 7 '18 at 15:39
$begingroup$
I’ll clarify on that right now.
$endgroup$
– KM101
Dec 7 '18 at 15:39
$begingroup$
They aren’t the same graph. They have the exact same shape but they’re $frac{pi}{2}$ radians apart (in the case of $sin x$ and $cos x$). That’s like saying $y = 2x$ and $y = 2x+3$ are the same.
$endgroup$
– KM101
Dec 7 '18 at 16:01
$begingroup$
They aren’t the same graph. They have the exact same shape but they’re $frac{pi}{2}$ radians apart (in the case of $sin x$ and $cos x$). That’s like saying $y = 2x$ and $y = 2x+3$ are the same.
$endgroup$
– KM101
Dec 7 '18 at 16:01
$begingroup$
You are right that $sin x$ is an odd function, and $cos x$ is an even function. And yes, the graph of one is a phase shift of the graph of the other. The behavior of the graph around $0$ will show that one is even and one is odd.
$endgroup$
– Doug M
Dec 7 '18 at 16:02
$begingroup$
You are right that $sin x$ is an odd function, and $cos x$ is an even function. And yes, the graph of one is a phase shift of the graph of the other. The behavior of the graph around $0$ will show that one is even and one is odd.
$endgroup$
– Doug M
Dec 7 '18 at 16:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
That's fine since sine is an odd function, it satisfies $f(-x)=-f(x)$.
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
In response to your comment, $cos x$ and $sin x$ are even and odd functions, respectively. If you think about it in terms of the unit circle, things become clear.
For a straight line of length one drawn at angle $theta$ from the origin of a unit circle to a point $B$ on the circumference, point $B$ will have the coordinates $(cos theta, sin theta)$. This is obvious because
$$cos theta = frac{adj}{hyp} = frac{adj}{1} implies adj = costheta$$
$$sin theta = frac{opp}{hyp} = frac{opp}{1} implies opp = sintheta$$
So, $cos x$ corresponds with the $x$-axis and $sin x$ corresponds with the $y$-axis. In quadrants $I$ and $II$, sine functions are positive as they deal with the upper portion of the $y$-axis. In quadrants $III$ and $IV$, sine functions are negative as they deal with the lower portion of the $y$-axis.
In quadrants $IV$ and $I$, cosine functions are positive since they deal with the right portion of the $x$-axis. In quadrants $II$ and $III$, cosine functions are positive since they deal with the left portion of the $x$-axis.
Combining this with the fact that $-theta = 2pi-theta$, it becomes clear that $cos(-theta) = costheta$ and $sin(-theta) = -sintheta$.
You can think of this visually. Imagine you draw an angle $theta$ in the first quadrant and the exact same angle again but in the fourth quadrant ($-theta$). You form two congruent right triangles. They share the same adjacent leg. Therefore, their cosine values are the same. However, their opposite legs extend the same amount but in opposite directions: upward for quadrant $I$ and downward for quadrant $IV$. Hence, their sine values are opposites.
This is also consistent with the fact that $sin x$ and $cos x$ have the same graph but with phase shift $frac{pi}{2}$.
Or, yet again, you could use well-known angle identities.
$$sin (apm b) = sin acos bpmcos asin b$$
$$sin (0-theta) = sin 0cos theta-cos 0sintheta = color{blue}{-sintheta}$$
$$cos(apm b) = cos acos bmpsin asin b$$
$$cos (0-theta) = cos 0costheta+sin 0sintheta = color{blue}{costheta}$$
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
$endgroup$
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
That's fine since sine is an odd function, it satisfies $f(-x)=-f(x)$.
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
That's fine since sine is an odd function, it satisfies $f(-x)=-f(x)$.
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
That's fine since sine is an odd function, it satisfies $f(-x)=-f(x)$.
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
That's fine since sine is an odd function, it satisfies $f(-x)=-f(x)$.
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 7 '18 at 8:01
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
add a comment |
add a comment |
$begingroup$
In response to your comment, $cos x$ and $sin x$ are even and odd functions, respectively. If you think about it in terms of the unit circle, things become clear.
For a straight line of length one drawn at angle $theta$ from the origin of a unit circle to a point $B$ on the circumference, point $B$ will have the coordinates $(cos theta, sin theta)$. This is obvious because
$$cos theta = frac{adj}{hyp} = frac{adj}{1} implies adj = costheta$$
$$sin theta = frac{opp}{hyp} = frac{opp}{1} implies opp = sintheta$$
So, $cos x$ corresponds with the $x$-axis and $sin x$ corresponds with the $y$-axis. In quadrants $I$ and $II$, sine functions are positive as they deal with the upper portion of the $y$-axis. In quadrants $III$ and $IV$, sine functions are negative as they deal with the lower portion of the $y$-axis.
In quadrants $IV$ and $I$, cosine functions are positive since they deal with the right portion of the $x$-axis. In quadrants $II$ and $III$, cosine functions are positive since they deal with the left portion of the $x$-axis.
Combining this with the fact that $-theta = 2pi-theta$, it becomes clear that $cos(-theta) = costheta$ and $sin(-theta) = -sintheta$.
You can think of this visually. Imagine you draw an angle $theta$ in the first quadrant and the exact same angle again but in the fourth quadrant ($-theta$). You form two congruent right triangles. They share the same adjacent leg. Therefore, their cosine values are the same. However, their opposite legs extend the same amount but in opposite directions: upward for quadrant $I$ and downward for quadrant $IV$. Hence, their sine values are opposites.
This is also consistent with the fact that $sin x$ and $cos x$ have the same graph but with phase shift $frac{pi}{2}$.
Or, yet again, you could use well-known angle identities.
$$sin (apm b) = sin acos bpmcos asin b$$
$$sin (0-theta) = sin 0cos theta-cos 0sintheta = color{blue}{-sintheta}$$
$$cos(apm b) = cos acos bmpsin asin b$$
$$cos (0-theta) = cos 0costheta+sin 0sintheta = color{blue}{costheta}$$
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
$endgroup$
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
add a comment |
$begingroup$
In response to your comment, $cos x$ and $sin x$ are even and odd functions, respectively. If you think about it in terms of the unit circle, things become clear.
For a straight line of length one drawn at angle $theta$ from the origin of a unit circle to a point $B$ on the circumference, point $B$ will have the coordinates $(cos theta, sin theta)$. This is obvious because
$$cos theta = frac{adj}{hyp} = frac{adj}{1} implies adj = costheta$$
$$sin theta = frac{opp}{hyp} = frac{opp}{1} implies opp = sintheta$$
So, $cos x$ corresponds with the $x$-axis and $sin x$ corresponds with the $y$-axis. In quadrants $I$ and $II$, sine functions are positive as they deal with the upper portion of the $y$-axis. In quadrants $III$ and $IV$, sine functions are negative as they deal with the lower portion of the $y$-axis.
In quadrants $IV$ and $I$, cosine functions are positive since they deal with the right portion of the $x$-axis. In quadrants $II$ and $III$, cosine functions are positive since they deal with the left portion of the $x$-axis.
Combining this with the fact that $-theta = 2pi-theta$, it becomes clear that $cos(-theta) = costheta$ and $sin(-theta) = -sintheta$.
You can think of this visually. Imagine you draw an angle $theta$ in the first quadrant and the exact same angle again but in the fourth quadrant ($-theta$). You form two congruent right triangles. They share the same adjacent leg. Therefore, their cosine values are the same. However, their opposite legs extend the same amount but in opposite directions: upward for quadrant $I$ and downward for quadrant $IV$. Hence, their sine values are opposites.
This is also consistent with the fact that $sin x$ and $cos x$ have the same graph but with phase shift $frac{pi}{2}$.
Or, yet again, you could use well-known angle identities.
$$sin (apm b) = sin acos bpmcos asin b$$
$$sin (0-theta) = sin 0cos theta-cos 0sintheta = color{blue}{-sintheta}$$
$$cos(apm b) = cos acos bmpsin asin b$$
$$cos (0-theta) = cos 0costheta+sin 0sintheta = color{blue}{costheta}$$
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
$endgroup$
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
add a comment |
$begingroup$
In response to your comment, $cos x$ and $sin x$ are even and odd functions, respectively. If you think about it in terms of the unit circle, things become clear.
For a straight line of length one drawn at angle $theta$ from the origin of a unit circle to a point $B$ on the circumference, point $B$ will have the coordinates $(cos theta, sin theta)$. This is obvious because
$$cos theta = frac{adj}{hyp} = frac{adj}{1} implies adj = costheta$$
$$sin theta = frac{opp}{hyp} = frac{opp}{1} implies opp = sintheta$$
So, $cos x$ corresponds with the $x$-axis and $sin x$ corresponds with the $y$-axis. In quadrants $I$ and $II$, sine functions are positive as they deal with the upper portion of the $y$-axis. In quadrants $III$ and $IV$, sine functions are negative as they deal with the lower portion of the $y$-axis.
In quadrants $IV$ and $I$, cosine functions are positive since they deal with the right portion of the $x$-axis. In quadrants $II$ and $III$, cosine functions are positive since they deal with the left portion of the $x$-axis.
Combining this with the fact that $-theta = 2pi-theta$, it becomes clear that $cos(-theta) = costheta$ and $sin(-theta) = -sintheta$.
You can think of this visually. Imagine you draw an angle $theta$ in the first quadrant and the exact same angle again but in the fourth quadrant ($-theta$). You form two congruent right triangles. They share the same adjacent leg. Therefore, their cosine values are the same. However, their opposite legs extend the same amount but in opposite directions: upward for quadrant $I$ and downward for quadrant $IV$. Hence, their sine values are opposites.
This is also consistent with the fact that $sin x$ and $cos x$ have the same graph but with phase shift $frac{pi}{2}$.
Or, yet again, you could use well-known angle identities.
$$sin (apm b) = sin acos bpmcos asin b$$
$$sin (0-theta) = sin 0cos theta-cos 0sintheta = color{blue}{-sintheta}$$
$$cos(apm b) = cos acos bmpsin asin b$$
$$cos (0-theta) = cos 0costheta+sin 0sintheta = color{blue}{costheta}$$
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
$endgroup$
In response to your comment, $cos x$ and $sin x$ are even and odd functions, respectively. If you think about it in terms of the unit circle, things become clear.
For a straight line of length one drawn at angle $theta$ from the origin of a unit circle to a point $B$ on the circumference, point $B$ will have the coordinates $(cos theta, sin theta)$. This is obvious because
$$cos theta = frac{adj}{hyp} = frac{adj}{1} implies adj = costheta$$
$$sin theta = frac{opp}{hyp} = frac{opp}{1} implies opp = sintheta$$
So, $cos x$ corresponds with the $x$-axis and $sin x$ corresponds with the $y$-axis. In quadrants $I$ and $II$, sine functions are positive as they deal with the upper portion of the $y$-axis. In quadrants $III$ and $IV$, sine functions are negative as they deal with the lower portion of the $y$-axis.
In quadrants $IV$ and $I$, cosine functions are positive since they deal with the right portion of the $x$-axis. In quadrants $II$ and $III$, cosine functions are positive since they deal with the left portion of the $x$-axis.
Combining this with the fact that $-theta = 2pi-theta$, it becomes clear that $cos(-theta) = costheta$ and $sin(-theta) = -sintheta$.
You can think of this visually. Imagine you draw an angle $theta$ in the first quadrant and the exact same angle again but in the fourth quadrant ($-theta$). You form two congruent right triangles. They share the same adjacent leg. Therefore, their cosine values are the same. However, their opposite legs extend the same amount but in opposite directions: upward for quadrant $I$ and downward for quadrant $IV$. Hence, their sine values are opposites.
This is also consistent with the fact that $sin x$ and $cos x$ have the same graph but with phase shift $frac{pi}{2}$.
Or, yet again, you could use well-known angle identities.
$$sin (apm b) = sin acos bpmcos asin b$$
$$sin (0-theta) = sin 0cos theta-cos 0sintheta = color{blue}{-sintheta}$$
$$cos(apm b) = cos acos bmpsin asin b$$
$$cos (0-theta) = cos 0costheta+sin 0sintheta = color{blue}{costheta}$$
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
answered Dec 7 '18 at 15:55
KM101KM101
5,9251524
5,9251524
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
add a comment |
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Thanks! This helped
$endgroup$
– ming
Dec 7 '18 at 16:29
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
$begingroup$
Glad to have helped. Doug M’s comment is also very useful. If you have any doubts, checking the graphs for $sin x$ and $cos x$ should resolve them.
$endgroup$
– KM101
Dec 7 '18 at 16:48
add a comment |
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2
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Do you mean $sin frac{-3pi}{4}$? If so, then yes, because $sin x$ is an odd function, meaning $sin (-x) = -sin x$. (Even if you didn’t mean that, it still applies, but it’s probably not what you meant.)
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– KM101
Dec 7 '18 at 8:00
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Yeah it wasn't what I meant I think I just got lucky by using that method. So I just learned about how $sin(x) = -sin(x)$ because it's an odd function but how come $cos(x) = cos(-x)$? That means that $cos(x)$ is an even function but isn't $sin$ and $cos$ sorta the same graph?
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– ming
Dec 7 '18 at 15:36
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I’ll clarify on that right now.
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– KM101
Dec 7 '18 at 15:39
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They aren’t the same graph. They have the exact same shape but they’re $frac{pi}{2}$ radians apart (in the case of $sin x$ and $cos x$). That’s like saying $y = 2x$ and $y = 2x+3$ are the same.
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– KM101
Dec 7 '18 at 16:01
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You are right that $sin x$ is an odd function, and $cos x$ is an even function. And yes, the graph of one is a phase shift of the graph of the other. The behavior of the graph around $0$ will show that one is even and one is odd.
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– Doug M
Dec 7 '18 at 16:02