Need Help with Absolute Value Equation involving Trig Function
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I am asked to show that |sin(x)|=|AC|, where x=theta and -pi/2< x< pi/2 and AC corresponds to the length of the vertical line that falls from the point (cosx,sinx) on the unit circle to the horizontal x-axis. I am also asked to show both cases, positive and negative, for x=theta.
The positive case, sin(x)=AC, is trivial to show as sin(x)=y/r. That simplifies to just y=AC, so sin(x)=AC and, making the substitution, AC=AC.
The negative case however, trips me up a bit. I chose to tack the negative sign on the left side of the equation to get -sin(x)=AC, but -sin(x)=-y/r. Thus -sin(x)=-y=-AC. -AC does not equal AC. Could someone show me step by step how to do the negative case please?
trigonometry absolute-value
asked Dec 30 '18 at 20:25
user596476user596476
133
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add a comment |
$begingroup$
I am asked to show that |sin(x)|=|AC|, where x=theta and -pi/2< x< pi/2 and AC corresponds to the length of the vertical line that falls from the point (cosx,sinx) on the unit circle to the horizontal x-axis. I am also asked to show both cases, positive and negative, for x=theta.
The positive case, sin(x)=AC, is trivial to show as sin(x)=y/r. That simplifies to just y=AC, so sin(x)=AC and, making the substitution, AC=AC.
The negative case however, trips me up a bit. I chose to tack the negative sign on the left side of the equation to get -sin(x)=AC, but -sin(x)=-y/r. Thus -sin(x)=-y=-AC. -AC does not equal AC. Could someone show me step by step how to do the negative case please?
trigonometry absolute-value
asked Dec 30 '18 at 20:25
user596476user596476
133
$endgroup$
$begingroup$
Should the point be $(cos(x),sin(x))$?
$endgroup$
– Dave
Dec 30 '18 at 20:28
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are you sure you understood how absolute values work? I mean besides from having a negative length does not make to much sense anyways.
$endgroup$
– Riquelme
Dec 30 '18 at 20:28
$begingroup$
Yes sorry, I've corrected the point to (cos(x), sin(x)). Could someone explain to me how absolute values work in this particular case please? How should I be thinking about this problem?
$endgroup$
– user596476
Dec 30 '18 at 20:40
$begingroup$
I believe I've figured it out on my own. The case where both sin(x) and AC are positive is perfectly fine. For the negative case, you can tack on the negative sign on either the left expression or the right expression. Doing it on the left side produces -(sin(x))=AC. Since you assume that sin(x) is negative, we can rewrite the left side as -(sin(-x))=AC because in this case sin(x) is only negative for negative values of x=theta. Since sin(-x)=-sin(x), we have -(-sin(x))=AC, or sin(x)=AC. The rest of the proof or explanation is trivial.
$endgroup$
– user596476
Dec 30 '18 at 23:01
add a comment |
$begingroup$
I am asked to show that |sin(x)|=|AC|, where x=theta and -pi/2< x< pi/2 and AC corresponds to the length of the vertical line that falls from the point (cosx,sinx) on the unit circle to the horizontal x-axis. I am also asked to show both cases, positive and negative, for x=theta.
The positive case, sin(x)=AC, is trivial to show as sin(x)=y/r. That simplifies to just y=AC, so sin(x)=AC and, making the substitution, AC=AC.
The negative case however, trips me up a bit. I chose to tack the negative sign on the left side of the equation to get -sin(x)=AC, but -sin(x)=-y/r. Thus -sin(x)=-y=-AC. -AC does not equal AC. Could someone show me step by step how to do the negative case please?
trigonometry absolute-value
asked Dec 30 '18 at 20:25
user596476user596476
133
$endgroup$
I am asked to show that |sin(x)|=|AC|, where x=theta and -pi/2< x< pi/2 and AC corresponds to the length of the vertical line that falls from the point (cosx,sinx) on the unit circle to the horizontal x-axis. I am also asked to show both cases, positive and negative, for x=theta.
The positive case, sin(x)=AC, is trivial to show as sin(x)=y/r. That simplifies to just y=AC, so sin(x)=AC and, making the substitution, AC=AC.
The negative case however, trips me up a bit. I chose to tack the negative sign on the left side of the equation to get -sin(x)=AC, but -sin(x)=-y/r. Thus -sin(x)=-y=-AC. -AC does not equal AC. Could someone show me step by step how to do the negative case please?
trigonometry absolute-value
trigonometry absolute-value
asked Dec 30 '18 at 20:25
user596476user596476
133
asked Dec 30 '18 at 20:25
user596476user596476
133
edited Dec 30 '18 at 20:32
user596476
asked Dec 30 '18 at 20:25
user596476user596476
133
asked Dec 30 '18 at 20:25
user596476user596476
133
asked Dec 30 '18 at 20:25
user596476user596476
133
133
$begingroup$
Should the point be $(cos(x),sin(x))$?
$endgroup$
– Dave
Dec 30 '18 at 20:28
$begingroup$
are you sure you understood how absolute values work? I mean besides from having a negative length does not make to much sense anyways.
$endgroup$
– Riquelme
Dec 30 '18 at 20:28
$begingroup$
Yes sorry, I've corrected the point to (cos(x), sin(x)). Could someone explain to me how absolute values work in this particular case please? How should I be thinking about this problem?
$endgroup$
– user596476
Dec 30 '18 at 20:40
$begingroup$
I believe I've figured it out on my own. The case where both sin(x) and AC are positive is perfectly fine. For the negative case, you can tack on the negative sign on either the left expression or the right expression. Doing it on the left side produces -(sin(x))=AC. Since you assume that sin(x) is negative, we can rewrite the left side as -(sin(-x))=AC because in this case sin(x) is only negative for negative values of x=theta. Since sin(-x)=-sin(x), we have -(-sin(x))=AC, or sin(x)=AC. The rest of the proof or explanation is trivial.
$endgroup$
– user596476
Dec 30 '18 at 23:01
add a comment |
$begingroup$
Should the point be $(cos(x),sin(x))$?
$endgroup$
– Dave
Dec 30 '18 at 20:28
$begingroup$
are you sure you understood how absolute values work? I mean besides from having a negative length does not make to much sense anyways.
$endgroup$
– Riquelme
Dec 30 '18 at 20:28
$begingroup$
Yes sorry, I've corrected the point to (cos(x), sin(x)). Could someone explain to me how absolute values work in this particular case please? How should I be thinking about this problem?
$endgroup$
– user596476
Dec 30 '18 at 20:40
$begingroup$
I believe I've figured it out on my own. The case where both sin(x) and AC are positive is perfectly fine. For the negative case, you can tack on the negative sign on either the left expression or the right expression. Doing it on the left side produces -(sin(x))=AC. Since you assume that sin(x) is negative, we can rewrite the left side as -(sin(-x))=AC because in this case sin(x) is only negative for negative values of x=theta. Since sin(-x)=-sin(x), we have -(-sin(x))=AC, or sin(x)=AC. The rest of the proof or explanation is trivial.
$endgroup$
– user596476
Dec 30 '18 at 23:01
$begingroup$
Should the point be $(cos(x),sin(x))$?
$endgroup$
– Dave
Dec 30 '18 at 20:28
$begingroup$
Should the point be $(cos(x),sin(x))$?
$endgroup$
– Dave
Dec 30 '18 at 20:28
$begingroup$
are you sure you understood how absolute values work? I mean besides from having a negative length does not make to much sense anyways.
$endgroup$
– Riquelme
Dec 30 '18 at 20:28
$begingroup$
are you sure you understood how absolute values work? I mean besides from having a negative length does not make to much sense anyways.
$endgroup$
– Riquelme
Dec 30 '18 at 20:28
$begingroup$
Yes sorry, I've corrected the point to (cos(x), sin(x)). Could someone explain to me how absolute values work in this particular case please? How should I be thinking about this problem?
$endgroup$
– user596476
Dec 30 '18 at 20:40
$begingroup$
Yes sorry, I've corrected the point to (cos(x), sin(x)). Could someone explain to me how absolute values work in this particular case please? How should I be thinking about this problem?
$endgroup$
– user596476
Dec 30 '18 at 20:40
$begingroup$
I believe I've figured it out on my own. The case where both sin(x) and AC are positive is perfectly fine. For the negative case, you can tack on the negative sign on either the left expression or the right expression. Doing it on the left side produces -(sin(x))=AC. Since you assume that sin(x) is negative, we can rewrite the left side as -(sin(-x))=AC because in this case sin(x) is only negative for negative values of x=theta. Since sin(-x)=-sin(x), we have -(-sin(x))=AC, or sin(x)=AC. The rest of the proof or explanation is trivial.
$endgroup$
– user596476
Dec 30 '18 at 23:01
$begingroup$
I believe I've figured it out on my own. The case where both sin(x) and AC are positive is perfectly fine. For the negative case, you can tack on the negative sign on either the left expression or the right expression. Doing it on the left side produces -(sin(x))=AC. Since you assume that sin(x) is negative, we can rewrite the left side as -(sin(-x))=AC because in this case sin(x) is only negative for negative values of x=theta. Since sin(-x)=-sin(x), we have -(-sin(x))=AC, or sin(x)=AC. The rest of the proof or explanation is trivial.
$endgroup$
– user596476
Dec 30 '18 at 23:01
add a comment |
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$begingroup$
Should the point be $(cos(x),sin(x))$?
$endgroup$
– Dave
Dec 30 '18 at 20:28
$begingroup$
are you sure you understood how absolute values work? I mean besides from having a negative length does not make to much sense anyways.
$endgroup$
– Riquelme
Dec 30 '18 at 20:28
$begingroup$
Yes sorry, I've corrected the point to (cos(x), sin(x)). Could someone explain to me how absolute values work in this particular case please? How should I be thinking about this problem?
$endgroup$
– user596476
Dec 30 '18 at 20:40
$begingroup$
I believe I've figured it out on my own. The case where both sin(x) and AC are positive is perfectly fine. For the negative case, you can tack on the negative sign on either the left expression or the right expression. Doing it on the left side produces -(sin(x))=AC. Since you assume that sin(x) is negative, we can rewrite the left side as -(sin(-x))=AC because in this case sin(x) is only negative for negative values of x=theta. Since sin(-x)=-sin(x), we have -(-sin(x))=AC, or sin(x)=AC. The rest of the proof or explanation is trivial.
$endgroup$
– user596476
Dec 30 '18 at 23:01