$p_0^2 + q_0^2 = 1$ and $-sin(s) = -p_0sin(s) + q_0cos(s)$ $rightarrow $ $p_0 = -cos2s, q_0 = -sin(2s)$
$begingroup$
How does solving $p_0^2 + q_0^2 = 1$ and $-sin(s) = -p_0sin(s) + q_0cos(s)$ give $p_0 = -cos2s $ and $q_0 = -sin(2s)$?
I can clearly see one solution is $p_0=1,q_0=0$ but can't seem to understand the other.
trigonometry systems-of-equations
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
How does solving $p_0^2 + q_0^2 = 1$ and $-sin(s) = -p_0sin(s) + q_0cos(s)$ give $p_0 = -cos2s $ and $q_0 = -sin(2s)$?
I can clearly see one solution is $p_0=1,q_0=0$ but can't seem to understand the other.
trigonometry systems-of-equations
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
How does solving $p_0^2 + q_0^2 = 1$ and $-sin(s) = -p_0sin(s) + q_0cos(s)$ give $p_0 = -cos2s $ and $q_0 = -sin(2s)$?
I can clearly see one solution is $p_0=1,q_0=0$ but can't seem to understand the other.
trigonometry systems-of-equations
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
$endgroup$
How does solving $p_0^2 + q_0^2 = 1$ and $-sin(s) = -p_0sin(s) + q_0cos(s)$ give $p_0 = -cos2s $ and $q_0 = -sin(2s)$?
I can clearly see one solution is $p_0=1,q_0=0$ but can't seem to understand the other.
trigonometry systems-of-equations
trigonometry systems-of-equations
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
edited Dec 17 '18 at 12:24
Harry Peter
5,46911439
5,46911439
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 14:41
pablo_mathscobarpablo_mathscobar
996
996
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT : Let $p_0=cos x $ and $q_0=sin x$. Then the second equation becomes $$-sin s = -cos xsin s + sin xcos s.$$
That is
$$-sin s = sin(x-s).$$
Now solve for $x$.
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
add a comment |
$begingroup$
Since $p_0^2 + q_0^2 = 1$, there is $theta$ such that $p_0 = cos(theta)$ and $q_0 = sin(theta)$. Thus your equation becomes
$$ sin(-s) = -sin(s) = - cos(theta) sin(s) + sin(theta) cos(s) = sin(theta - s)$$
Thus either $theta = 2 n pi$ or $theta = (2n+1)pi + 2 s$.
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
HINT : Let $p_0=cos x $ and $q_0=sin x$. Then the second equation becomes $$-sin s = -cos xsin s + sin xcos s.$$
That is
$$-sin s = sin(x-s).$$
Now solve for $x$.
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
add a comment |
$begingroup$
HINT : Let $p_0=cos x $ and $q_0=sin x$. Then the second equation becomes $$-sin s = -cos xsin s + sin xcos s.$$
That is
$$-sin s = sin(x-s).$$
Now solve for $x$.
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
add a comment |
$begingroup$
HINT : Let $p_0=cos x $ and $q_0=sin x$. Then the second equation becomes $$-sin s = -cos xsin s + sin xcos s.$$
That is
$$-sin s = sin(x-s).$$
Now solve for $x$.
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
HINT : Let $p_0=cos x $ and $q_0=sin x$. Then the second equation becomes $$-sin s = -cos xsin s + sin xcos s.$$
That is
$$-sin s = sin(x-s).$$
Now solve for $x$.
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
answered Dec 10 '18 at 14:50
Thomas ShelbyThomas Shelby
2,877421
2,877421
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
add a comment |
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
I still cant seem to do it
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:56
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
$begingroup$
Another hint:$sin A+sin B=0 iff 2sinfrac{A+B}2cosfrac{A-B}2=0$. This means either $sinfrac{A+B}2=0$ or $cosfrac{A-B}2=0$.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 15:12
add a comment |
$begingroup$
Since $p_0^2 + q_0^2 = 1$, there is $theta$ such that $p_0 = cos(theta)$ and $q_0 = sin(theta)$. Thus your equation becomes
$$ sin(-s) = -sin(s) = - cos(theta) sin(s) + sin(theta) cos(s) = sin(theta - s)$$
Thus either $theta = 2 n pi$ or $theta = (2n+1)pi + 2 s$.
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
$begingroup$
Since $p_0^2 + q_0^2 = 1$, there is $theta$ such that $p_0 = cos(theta)$ and $q_0 = sin(theta)$. Thus your equation becomes
$$ sin(-s) = -sin(s) = - cos(theta) sin(s) + sin(theta) cos(s) = sin(theta - s)$$
Thus either $theta = 2 n pi$ or $theta = (2n+1)pi + 2 s$.
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
$begingroup$
Since $p_0^2 + q_0^2 = 1$, there is $theta$ such that $p_0 = cos(theta)$ and $q_0 = sin(theta)$. Thus your equation becomes
$$ sin(-s) = -sin(s) = - cos(theta) sin(s) + sin(theta) cos(s) = sin(theta - s)$$
Thus either $theta = 2 n pi$ or $theta = (2n+1)pi + 2 s$.
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
$endgroup$
Since $p_0^2 + q_0^2 = 1$, there is $theta$ such that $p_0 = cos(theta)$ and $q_0 = sin(theta)$. Thus your equation becomes
$$ sin(-s) = -sin(s) = - cos(theta) sin(s) + sin(theta) cos(s) = sin(theta - s)$$
Thus either $theta = 2 n pi$ or $theta = (2n+1)pi + 2 s$.
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
answered Dec 10 '18 at 14:51
Robert IsraelRobert Israel
322k23212465
322k23212465
add a comment |
add a comment |
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