Finding the graph defined by $x = sin theta$ and $y = 3 - 2cos(2theta)$
$begingroup$
The question is as follows:
Find the graph of the parametric equations defined by
$$
x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)
$$
We are supposed to use the identity that
$sin^2theta + cos^2theta = 1$
However, that identity requires that sin and cos both have the same theta, and in this instance they are different.
trigonometry parametric
edited Jan 11 at 5:27
Blue
47.9k870153
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Find the graph of the parametric equations defined by
$$
x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)
$$
We are supposed to use the identity that
$sin^2theta + cos^2theta = 1$
However, that identity requires that sin and cos both have the same theta, and in this instance they are different.
trigonometry parametric
edited Jan 11 at 5:27
Blue
47.9k870153
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
$endgroup$
2
$begingroup$
Hint: $cos(2theta) = cos^2 (theta) - sin^2(theta)$. Now apply your identity to get rid of the cosine term.
$endgroup$
– John Hughes
Jan 11 at 2:58
add a comment |
$begingroup$
The question is as follows:
Find the graph of the parametric equations defined by
$$
x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)
$$
We are supposed to use the identity that
$sin^2theta + cos^2theta = 1$
However, that identity requires that sin and cos both have the same theta, and in this instance they are different.
trigonometry parametric
edited Jan 11 at 5:27
Blue
47.9k870153
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
$endgroup$
The question is as follows:
Find the graph of the parametric equations defined by
$$
x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)
$$
We are supposed to use the identity that
$sin^2theta + cos^2theta = 1$
However, that identity requires that sin and cos both have the same theta, and in this instance they are different.
trigonometry parametric
trigonometry parametric
edited Jan 11 at 5:27
Blue
47.9k870153
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
edited Jan 11 at 5:27
Blue
47.9k870153
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
edited Jan 11 at 5:27
Blue
47.9k870153
edited Jan 11 at 5:27
Blue
47.9k870153
edited Jan 11 at 5:27
Blue
47.9k870153
47.9k870153
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
asked Jan 11 at 2:56
Nik GautamNik Gautam
32
32
2
$begingroup$
Hint: $cos(2theta) = cos^2 (theta) - sin^2(theta)$. Now apply your identity to get rid of the cosine term.
$endgroup$
– John Hughes
Jan 11 at 2:58
add a comment |
2
$begingroup$
Hint: $cos(2theta) = cos^2 (theta) - sin^2(theta)$. Now apply your identity to get rid of the cosine term.
$endgroup$
– John Hughes
Jan 11 at 2:58
2
2
$begingroup$
Hint: $cos(2theta) = cos^2 (theta) - sin^2(theta)$. Now apply your identity to get rid of the cosine term.
$endgroup$
– John Hughes
Jan 11 at 2:58
$begingroup$
Hint: $cos(2theta) = cos^2 (theta) - sin^2(theta)$. Now apply your identity to get rid of the cosine term.
$endgroup$
– John Hughes
Jan 11 at 2:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)$$
Note that $$ cos (2theta) = 1-2sin ^2 (theta)$$
The expression for $y(theta )$ simplifies to $$y(theta) = 3 - 2cos(2theta)=1+4sin ^2 (theta) = 1+4 x^2$$
Thus your parabola is simply $y=1+4x^2$ where, $-1le xle 1$
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$$x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)$$
Note that $$ cos (2theta) = 1-2sin ^2 (theta)$$
The expression for $y(theta )$ simplifies to $$y(theta) = 3 - 2cos(2theta)=1+4sin ^2 (theta) = 1+4 x^2$$
Thus your parabola is simply $y=1+4x^2$ where, $-1le xle 1$
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
$$x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)$$
Note that $$ cos (2theta) = 1-2sin ^2 (theta)$$
The expression for $y(theta )$ simplifies to $$y(theta) = 3 - 2cos(2theta)=1+4sin ^2 (theta) = 1+4 x^2$$
Thus your parabola is simply $y=1+4x^2$ where, $-1le xle 1$
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
$$x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)$$
Note that $$ cos (2theta) = 1-2sin ^2 (theta)$$
The expression for $y(theta )$ simplifies to $$y(theta) = 3 - 2cos(2theta)=1+4sin ^2 (theta) = 1+4 x^2$$
Thus your parabola is simply $y=1+4x^2$ where, $-1le xle 1$
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$$x(theta) = sin theta \
y(theta) = 3 - 2cos(2theta)$$
Note that $$ cos (2theta) = 1-2sin ^2 (theta)$$
The expression for $y(theta )$ simplifies to $$y(theta) = 3 - 2cos(2theta)=1+4sin ^2 (theta) = 1+4 x^2$$
Thus your parabola is simply $y=1+4x^2$ where, $-1le xle 1$
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 11 at 3:20
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
add a comment |
add a comment |
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$begingroup$
Hint: $cos(2theta) = cos^2 (theta) - sin^2(theta)$. Now apply your identity to get rid of the cosine term.
$endgroup$
– John Hughes
Jan 11 at 2:58