Mixing
Proving $sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$
$begingroup$
Knowing
$$sin^2theta +cos^2theta equiv 1$$
how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$
Can I substitute the first equation to prove the second one? If so, how can I?
Please help.
trigonometry
edited Jan 15 at 13:46
Blue
48.5k870154
asked Jan 15 at 13:42
user8469209user8469209
122
$endgroup$
add a comment |
$begingroup$
Knowing
$$sin^2theta +cos^2theta equiv 1$$
how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$
Can I substitute the first equation to prove the second one? If so, how can I?
Please help.
trigonometry
edited Jan 15 at 13:46
Blue
48.5k870154
asked Jan 15 at 13:42
user8469209user8469209
122
$endgroup$
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47
6
$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47
$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54
add a comment |
$begingroup$
Knowing
$$sin^2theta +cos^2theta equiv 1$$
how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$
Can I substitute the first equation to prove the second one? If so, how can I?
Please help.
trigonometry
edited Jan 15 at 13:46
Blue
48.5k870154
asked Jan 15 at 13:42
user8469209user8469209
122
$endgroup$
Knowing
$$sin^2theta +cos^2theta equiv 1$$
how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$
Can I substitute the first equation to prove the second one? If so, how can I?
Please help.
trigonometry
trigonometry
edited Jan 15 at 13:46
Blue
48.5k870154
asked Jan 15 at 13:42
user8469209user8469209
122
edited Jan 15 at 13:46
Blue
48.5k870154
asked Jan 15 at 13:42
user8469209user8469209
122
edited Jan 15 at 13:46
Blue
48.5k870154
edited Jan 15 at 13:46
Blue
48.5k870154
edited Jan 15 at 13:46
Blue
48.5k870154
48.5k870154
asked Jan 15 at 13:42
user8469209user8469209
122
asked Jan 15 at 13:42
user8469209user8469209
122
asked Jan 15 at 13:42
user8469209user8469209
122
122
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47
6
$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47
$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54
add a comment |
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47
6
$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47
$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47
6
6
$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47
$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47
$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54
$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
$endgroup$
add a comment |
$begingroup$
Yes you can substitute the first equation into the second one.
$sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$
Substitute to get,
$begin{eqnarray}
sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
&=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
&=& cos^2y - cos^2x
end{eqnarray}$
answered Jan 15 at 14:54
E.NoleE.Nole
178114
$endgroup$
add a comment |
$begingroup$
$sin^2x cos^2y - cos^2x sin^2y \
= sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
= (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
= cos^2y - cos^2x$
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
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active
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active
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votes
$begingroup$
Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
$endgroup$
add a comment |
$begingroup$
Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
$endgroup$
add a comment |
$begingroup$
Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
$endgroup$
Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
edited 10 hours ago
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
answered Jan 15 at 14:40
CuhrazateeCuhrazatee
370110
370110
add a comment |
add a comment |
$begingroup$
Yes you can substitute the first equation into the second one.
$sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$
Substitute to get,
$begin{eqnarray}
sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
&=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
&=& cos^2y - cos^2x
end{eqnarray}$
answered Jan 15 at 14:54
E.NoleE.Nole
178114
$endgroup$
add a comment |
$begingroup$
Yes you can substitute the first equation into the second one.
$sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$
Substitute to get,
$begin{eqnarray}
sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
&=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
&=& cos^2y - cos^2x
end{eqnarray}$
answered Jan 15 at 14:54
E.NoleE.Nole
178114
$endgroup$
add a comment |
$begingroup$
Yes you can substitute the first equation into the second one.
$sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$
Substitute to get,
$begin{eqnarray}
sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
&=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
&=& cos^2y - cos^2x
end{eqnarray}$
answered Jan 15 at 14:54
E.NoleE.Nole
178114
$endgroup$
Yes you can substitute the first equation into the second one.
$sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$
Substitute to get,
$begin{eqnarray}
sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
&=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
&=& cos^2y - cos^2x
end{eqnarray}$
answered Jan 15 at 14:54
E.NoleE.Nole
178114
answered Jan 15 at 14:54
E.NoleE.Nole
178114
answered Jan 15 at 14:54
E.NoleE.Nole
178114
answered Jan 15 at 14:54
E.NoleE.Nole
178114
178114
add a comment |
add a comment |
$begingroup$
$sin^2x cos^2y - cos^2x sin^2y \
= sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
= (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
= cos^2y - cos^2x$
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
$endgroup$
add a comment |
$begingroup$
$sin^2x cos^2y - cos^2x sin^2y \
= sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
= (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
= cos^2y - cos^2x$
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
$endgroup$
add a comment |
$begingroup$
$sin^2x cos^2y - cos^2x sin^2y \
= sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
= (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
= cos^2y - cos^2x$
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
$endgroup$
$sin^2x cos^2y - cos^2x sin^2y \
= sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
= (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
= cos^2y - cos^2x$
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
answered Jan 15 at 15:31
gandalf61gandalf61
8,741725
8,741725
add a comment |
add a comment |
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$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47
6
$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47
$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54