Mixing
Remove theta from these of trigonometric equations (basic)
$begingroup$
I have a very simple problem but am self-teaching and have become a bit stuck in finding an answer anywhere. The question is below and I have proceeded as follows:
'Please remove theta from the following pairs of equations:'
$ x = 4 sectheta\\ y = 4 tantheta$
$x = 4(1+tan^2theta)\\
y=4tantheta$
$x/4 = 1 + tan^2theta\\
y^2/4^4 = tan^2theta$
$x/4 - y^2/4^4 = 1$
Now in this situation I would always expand the square and multiply x/4 by 4/4 to get 4x/16, and my final answer was $4x-y^2 = 16$ however the book answer was $x^2 - y^2 = 16$. Are these equivalent (it doesn't seem like they are...) or have I gone wrong somewhere?
Many thanks for your help.
trigonometry
edited Jan 29 at 22:38
Bernard
124k741118
asked Jan 29 at 22:34
JosephJoseph
304
$endgroup$
add a comment |
$begingroup$
I have a very simple problem but am self-teaching and have become a bit stuck in finding an answer anywhere. The question is below and I have proceeded as follows:
'Please remove theta from the following pairs of equations:'
$ x = 4 sectheta\\ y = 4 tantheta$
$x = 4(1+tan^2theta)\\
y=4tantheta$
$x/4 = 1 + tan^2theta\\
y^2/4^4 = tan^2theta$
$x/4 - y^2/4^4 = 1$
Now in this situation I would always expand the square and multiply x/4 by 4/4 to get 4x/16, and my final answer was $4x-y^2 = 16$ however the book answer was $x^2 - y^2 = 16$. Are these equivalent (it doesn't seem like they are...) or have I gone wrong somewhere?
Many thanks for your help.
trigonometry
edited Jan 29 at 22:38
Bernard
124k741118
asked Jan 29 at 22:34
JosephJoseph
304
$endgroup$
add a comment |
$begingroup$
I have a very simple problem but am self-teaching and have become a bit stuck in finding an answer anywhere. The question is below and I have proceeded as follows:
'Please remove theta from the following pairs of equations:'
$ x = 4 sectheta\\ y = 4 tantheta$
$x = 4(1+tan^2theta)\\
y=4tantheta$
$x/4 = 1 + tan^2theta\\
y^2/4^4 = tan^2theta$
$x/4 - y^2/4^4 = 1$
Now in this situation I would always expand the square and multiply x/4 by 4/4 to get 4x/16, and my final answer was $4x-y^2 = 16$ however the book answer was $x^2 - y^2 = 16$. Are these equivalent (it doesn't seem like they are...) or have I gone wrong somewhere?
Many thanks for your help.
trigonometry
edited Jan 29 at 22:38
Bernard
124k741118
asked Jan 29 at 22:34
JosephJoseph
304
$endgroup$
I have a very simple problem but am self-teaching and have become a bit stuck in finding an answer anywhere. The question is below and I have proceeded as follows:
'Please remove theta from the following pairs of equations:'
$ x = 4 sectheta\\ y = 4 tantheta$
$x = 4(1+tan^2theta)\\
y=4tantheta$
$x/4 = 1 + tan^2theta\\
y^2/4^4 = tan^2theta$
$x/4 - y^2/4^4 = 1$
Now in this situation I would always expand the square and multiply x/4 by 4/4 to get 4x/16, and my final answer was $4x-y^2 = 16$ however the book answer was $x^2 - y^2 = 16$. Are these equivalent (it doesn't seem like they are...) or have I gone wrong somewhere?
Many thanks for your help.
trigonometry
trigonometry
edited Jan 29 at 22:38
Bernard
124k741118
asked Jan 29 at 22:34
JosephJoseph
304
edited Jan 29 at 22:38
Bernard
124k741118
asked Jan 29 at 22:34
JosephJoseph
304
edited Jan 29 at 22:38
Bernard
124k741118
edited Jan 29 at 22:38
Bernard
124k741118
edited Jan 29 at 22:38
Bernard
124k741118
124k741118
asked Jan 29 at 22:34
JosephJoseph
304
asked Jan 29 at 22:34
JosephJoseph
304
asked Jan 29 at 22:34
JosephJoseph
304
304
add a comment |
add a comment |
3 Answers
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$begingroup$
The third line is wrong. It should be
$$x^2 = 16(1 + tan^2theta)$$
Then
$$frac{x^2}{4^2} - frac{y^2}{4^2} = 1$$
Which is the answer in the book.
answered Jan 29 at 22:59
KY TangKY Tang
50436
$endgroup$
add a comment |
$begingroup$
You made an error in the third line:
$dfrac x4=sec theta$, so $sec^2;theta=1+tan^2theta=dfrac{x^2}{16}$.
Can you proceed?
answered Jan 29 at 22:42
BernardBernard
124k741118
$endgroup$
add a comment |
$begingroup$
$$begin{array} \ x = 4 tan theta \ y = 4 sec theta end{array}$$
Dividing both equations by $4$ to isolate the trigonometric functions...
$$begin{array} \ dfrac {x} {4} = tan theta \ dfrac {y} {4} = sec theta end{array}$$
Squaring both sides...
$$begin{array} \ dfrac {x^2} {16} = tan^2 theta \ dfrac {y^2} {16} = sec^2 theta end{array}$$
Since $tan^2 theta - sec^2 theta = 1$ $$dfrac {x^2} {16} - dfrac {y^2} {16} = 1$$
and multiplying by $16$ gives us $$x^2 - y^2 = 16$$
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
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3 Answers
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3 Answers
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$begingroup$
The third line is wrong. It should be
$$x^2 = 16(1 + tan^2theta)$$
Then
$$frac{x^2}{4^2} - frac{y^2}{4^2} = 1$$
Which is the answer in the book.
answered Jan 29 at 22:59
KY TangKY Tang
50436
$endgroup$
add a comment |
$begingroup$
The third line is wrong. It should be
$$x^2 = 16(1 + tan^2theta)$$
Then
$$frac{x^2}{4^2} - frac{y^2}{4^2} = 1$$
Which is the answer in the book.
answered Jan 29 at 22:59
KY TangKY Tang
50436
$endgroup$
add a comment |
$begingroup$
The third line is wrong. It should be
$$x^2 = 16(1 + tan^2theta)$$
Then
$$frac{x^2}{4^2} - frac{y^2}{4^2} = 1$$
Which is the answer in the book.
answered Jan 29 at 22:59
KY TangKY Tang
50436
$endgroup$
The third line is wrong. It should be
$$x^2 = 16(1 + tan^2theta)$$
Then
$$frac{x^2}{4^2} - frac{y^2}{4^2} = 1$$
Which is the answer in the book.
answered Jan 29 at 22:59
KY TangKY Tang
50436
answered Jan 29 at 22:59
KY TangKY Tang
50436
answered Jan 29 at 22:59
KY TangKY Tang
50436
answered Jan 29 at 22:59
KY TangKY Tang
50436
50436
add a comment |
add a comment |
$begingroup$
You made an error in the third line:
$dfrac x4=sec theta$, so $sec^2;theta=1+tan^2theta=dfrac{x^2}{16}$.
Can you proceed?
answered Jan 29 at 22:42
BernardBernard
124k741118
$endgroup$
add a comment |
$begingroup$
You made an error in the third line:
$dfrac x4=sec theta$, so $sec^2;theta=1+tan^2theta=dfrac{x^2}{16}$.
Can you proceed?
answered Jan 29 at 22:42
BernardBernard
124k741118
$endgroup$
add a comment |
$begingroup$
You made an error in the third line:
$dfrac x4=sec theta$, so $sec^2;theta=1+tan^2theta=dfrac{x^2}{16}$.
Can you proceed?
answered Jan 29 at 22:42
BernardBernard
124k741118
$endgroup$
You made an error in the third line:
$dfrac x4=sec theta$, so $sec^2;theta=1+tan^2theta=dfrac{x^2}{16}$.
Can you proceed?
answered Jan 29 at 22:42
BernardBernard
124k741118
answered Jan 29 at 22:42
BernardBernard
124k741118
answered Jan 29 at 22:42
BernardBernard
124k741118
answered Jan 29 at 22:42
BernardBernard
124k741118
124k741118
add a comment |
add a comment |
$begingroup$
$$begin{array} \ x = 4 tan theta \ y = 4 sec theta end{array}$$
Dividing both equations by $4$ to isolate the trigonometric functions...
$$begin{array} \ dfrac {x} {4} = tan theta \ dfrac {y} {4} = sec theta end{array}$$
Squaring both sides...
$$begin{array} \ dfrac {x^2} {16} = tan^2 theta \ dfrac {y^2} {16} = sec^2 theta end{array}$$
Since $tan^2 theta - sec^2 theta = 1$ $$dfrac {x^2} {16} - dfrac {y^2} {16} = 1$$
and multiplying by $16$ gives us $$x^2 - y^2 = 16$$
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
$endgroup$
add a comment |
$begingroup$
$$begin{array} \ x = 4 tan theta \ y = 4 sec theta end{array}$$
Dividing both equations by $4$ to isolate the trigonometric functions...
$$begin{array} \ dfrac {x} {4} = tan theta \ dfrac {y} {4} = sec theta end{array}$$
Squaring both sides...
$$begin{array} \ dfrac {x^2} {16} = tan^2 theta \ dfrac {y^2} {16} = sec^2 theta end{array}$$
Since $tan^2 theta - sec^2 theta = 1$ $$dfrac {x^2} {16} - dfrac {y^2} {16} = 1$$
and multiplying by $16$ gives us $$x^2 - y^2 = 16$$
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
$endgroup$
add a comment |
$begingroup$
$$begin{array} \ x = 4 tan theta \ y = 4 sec theta end{array}$$
Dividing both equations by $4$ to isolate the trigonometric functions...
$$begin{array} \ dfrac {x} {4} = tan theta \ dfrac {y} {4} = sec theta end{array}$$
Squaring both sides...
$$begin{array} \ dfrac {x^2} {16} = tan^2 theta \ dfrac {y^2} {16} = sec^2 theta end{array}$$
Since $tan^2 theta - sec^2 theta = 1$ $$dfrac {x^2} {16} - dfrac {y^2} {16} = 1$$
and multiplying by $16$ gives us $$x^2 - y^2 = 16$$
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
$endgroup$
$$begin{array} \ x = 4 tan theta \ y = 4 sec theta end{array}$$
Dividing both equations by $4$ to isolate the trigonometric functions...
$$begin{array} \ dfrac {x} {4} = tan theta \ dfrac {y} {4} = sec theta end{array}$$
Squaring both sides...
$$begin{array} \ dfrac {x^2} {16} = tan^2 theta \ dfrac {y^2} {16} = sec^2 theta end{array}$$
Since $tan^2 theta - sec^2 theta = 1$ $$dfrac {x^2} {16} - dfrac {y^2} {16} = 1$$
and multiplying by $16$ gives us $$x^2 - y^2 = 16$$
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
answered Jan 30 at 1:45
bjcolby15bjcolby15
1,51711016
1,51711016
add a comment |
add a comment |
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