Prove that the sum of squares of sin and cos equals 1
I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:
$r_{i}(theta)= cos(theta_{1}) $if $i=1$
$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$
$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$
So my question is, how do I prove that:
$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$
It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.
trigonometry
asked Jan 5 at 22:22
HannahHannah
82
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add a comment |
I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:
$r_{i}(theta)= cos(theta_{1}) $if $i=1$
$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$
$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$
So my question is, how do I prove that:
$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$
It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.
trigonometry
asked Jan 5 at 22:22
HannahHannah
82
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
– Dominik Kutek
Jan 5 at 22:26
add a comment |
I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:
$r_{i}(theta)= cos(theta_{1}) $if $i=1$
$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$
$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$
So my question is, how do I prove that:
$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$
It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.
trigonometry
asked Jan 5 at 22:22
HannahHannah
82
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:
$r_{i}(theta)= cos(theta_{1}) $if $i=1$
$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$
$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$
So my question is, how do I prove that:
$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$
It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.
trigonometry
trigonometry
asked Jan 5 at 22:22
HannahHannah
82
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 22:22
HannahHannah
82
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 22:22
HannahHannah
82
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 22:22
HannahHannah
82
asked Jan 5 at 22:22
HannahHannah
82
82
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hannah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
– Dominik Kutek
Jan 5 at 22:26
add a comment |
1
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
– Dominik Kutek
Jan 5 at 22:26
1
1
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
– Dominik Kutek
Jan 5 at 22:26
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
– Dominik Kutek
Jan 5 at 22:26
add a comment |
2 Answers
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active
oldest
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Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):
$(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =
$(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =
$(prod_{k=1}^{n-2}sin(theta_k))^2$
Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).
Continue analogously, combine the new term with $n-2$'th one:
$(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =
=$(prod_{k=1}^{n-3}sin(theta_k))^2$
By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:
$cos^2(theta_1) + sin^2(theta_1) = 1 $
$1 = 1 $
Q.E.D
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
add a comment |
I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
$$begin{align}
(r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
&= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
end{align}$$
Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
$$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
by induction hypothesis.
answered Jan 5 at 22:49
0x5390x539
1,057316
add a comment |
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2 Answers
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2 Answers
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Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):
$(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =
$(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =
$(prod_{k=1}^{n-2}sin(theta_k))^2$
Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).
Continue analogously, combine the new term with $n-2$'th one:
$(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =
=$(prod_{k=1}^{n-3}sin(theta_k))^2$
By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:
$cos^2(theta_1) + sin^2(theta_1) = 1 $
$1 = 1 $
Q.E.D
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
add a comment |
Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):
$(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =
$(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =
$(prod_{k=1}^{n-2}sin(theta_k))^2$
Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).
Continue analogously, combine the new term with $n-2$'th one:
$(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =
=$(prod_{k=1}^{n-3}sin(theta_k))^2$
By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:
$cos^2(theta_1) + sin^2(theta_1) = 1 $
$1 = 1 $
Q.E.D
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
add a comment |
Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):
$(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =
$(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =
$(prod_{k=1}^{n-2}sin(theta_k))^2$
Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).
Continue analogously, combine the new term with $n-2$'th one:
$(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =
=$(prod_{k=1}^{n-3}sin(theta_k))^2$
By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:
$cos^2(theta_1) + sin^2(theta_1) = 1 $
$1 = 1 $
Q.E.D
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):
$(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =
$(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =
$(prod_{k=1}^{n-2}sin(theta_k))^2$
Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).
Continue analogously, combine the new term with $n-2$'th one:
$(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =
=$(prod_{k=1}^{n-3}sin(theta_k))^2$
By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:
$cos^2(theta_1) + sin^2(theta_1) = 1 $
$1 = 1 $
Q.E.D
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
answered Jan 5 at 22:44
Dominik KutekDominik Kutek
3045
3045
add a comment |
add a comment |
I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
$$begin{align}
(r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
&= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
end{align}$$
Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
$$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
by induction hypothesis.
answered Jan 5 at 22:49
0x5390x539
1,057316
add a comment |
I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
$$begin{align}
(r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
&= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
end{align}$$
Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
$$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
by induction hypothesis.
answered Jan 5 at 22:49
0x5390x539
1,057316
add a comment |
I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
$$begin{align}
(r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
&= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
end{align}$$
Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
$$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
by induction hypothesis.
answered Jan 5 at 22:49
0x5390x539
1,057316
I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
$$begin{align}
(r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
&= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
end{align}$$
Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
$$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
by induction hypothesis.
answered Jan 5 at 22:49
0x5390x539
1,057316
answered Jan 5 at 22:49
0x5390x539
1,057316
answered Jan 5 at 22:49
0x5390x539
1,057316
answered Jan 5 at 22:49
0x5390x539
1,057316
1,057316
add a comment |
add a comment |
Hannah is a new contributor. Be nice, and check out our Code of Conduct.
Hannah is a new contributor. Be nice, and check out our Code of Conduct.
Hannah is a new contributor. Be nice, and check out our Code of Conduct.
Hannah is a new contributor. Be nice, and check out our Code of Conduct.
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1
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
– Dominik Kutek
Jan 5 at 22:26