Error in proving Lagrange's Identity for $1+costheta+cdots+cos ntheta$
$begingroup$
Note: I was originally trying essentially to prove the same thing as this Finalising proof of Lagrange's Trig Identity. However, I do not consider my question to be a duplicate. I am looking for someone to point out my mistake in my "proof," not to prove it for
me.
My attempted "proof"
My goal was to show
$$1+costheta+cdots+cos ntheta= frac12+frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}$$ when $0<theta<2 pi$.
I see that
$$begin{align}LHS
&= Releft(frac {1-e^{i(n+1)theta}} {1-e^{itheta}}right) tag{1}\[4pt]
&= frac {(costheta-1)(cos((n+1)theta)-1)+sin((n+1)theta)sintheta} {(costheta-1)^2+sin^2theta} tag{2}\[4pt]
&=frac{(costheta-1)(cos ntheta cos theta-sin ntheta sintheta-1)+(sin ntheta costheta + cos ntheta sintheta)sintheta} {2-2costheta} tag{3} \[4pt]
&= frac {cos ntheta-costheta-cos ntheta costheta +sin ntheta sintheta +1} {2-2costheta} tag{4}\[4pt]
&=frac12+frac12cos ntheta+ frac{sin ntheta sintheta} {2-2costheta} tag{5}
end{align}$$
To finish proof, it suffices to show that
$$frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}=frac12cos ntheta+frac{sin ntheta sintheta} {2-2costheta} tag{6}$$
By our restriction on $theta$, we must have $sinfractheta2=sqrt{frac12(1-cos theta)}$.
We see that $cos^2fractheta{2}=frac12(1+cos theta)$.
Case 1: Suppose $cosfrac{theta}{2}=sqrt{frac12(1+cos theta)}$. Then
$$begin{align}
frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}
&= frac{sin ntheta cosfrac{theta}{2}+cos ntheta sinfrac{theta}{2}} {2sinfrac{theta}{2}} tag{7} \[4pt]
&=frac{cos ntheta}{2}+frac {sin n theta sqrt {1+cos theta}} {2 sqrt {1-cos theta}} tag{8} \[4pt]
&=frac12cos ntheta + frac{sin n theta sin theta} {2-2cos theta} tag{9}
end{align}$$
as needed.
Case 2: Suppose $cosfrac{theta}{2}=-sqrt{frac12(1+costheta)}$. Then, by similar logic, I obtained
$$frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}= frac12 cos n theta- frac{sin n theta sin theta} {2-2cos theta} tag{10}$$
but this seems wrong to me.
Where's my mistake?
complex-analysis trigonometry proof-verification
edited Jan 8 at 19:37
Blue
49.7k870158
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
$endgroup$
add a comment |
$begingroup$
Note: I was originally trying essentially to prove the same thing as this Finalising proof of Lagrange's Trig Identity. However, I do not consider my question to be a duplicate. I am looking for someone to point out my mistake in my "proof," not to prove it for
me.
My attempted "proof"
My goal was to show
$$1+costheta+cdots+cos ntheta= frac12+frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}$$ when $0<theta<2 pi$.
I see that
$$begin{align}LHS
&= Releft(frac {1-e^{i(n+1)theta}} {1-e^{itheta}}right) tag{1}\[4pt]
&= frac {(costheta-1)(cos((n+1)theta)-1)+sin((n+1)theta)sintheta} {(costheta-1)^2+sin^2theta} tag{2}\[4pt]
&=frac{(costheta-1)(cos ntheta cos theta-sin ntheta sintheta-1)+(sin ntheta costheta + cos ntheta sintheta)sintheta} {2-2costheta} tag{3} \[4pt]
&= frac {cos ntheta-costheta-cos ntheta costheta +sin ntheta sintheta +1} {2-2costheta} tag{4}\[4pt]
&=frac12+frac12cos ntheta+ frac{sin ntheta sintheta} {2-2costheta} tag{5}
end{align}$$
To finish proof, it suffices to show that
$$frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}=frac12cos ntheta+frac{sin ntheta sintheta} {2-2costheta} tag{6}$$
By our restriction on $theta$, we must have $sinfractheta2=sqrt{frac12(1-cos theta)}$.
We see that $cos^2fractheta{2}=frac12(1+cos theta)$.
Case 1: Suppose $cosfrac{theta}{2}=sqrt{frac12(1+cos theta)}$. Then
$$begin{align}
frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}
&= frac{sin ntheta cosfrac{theta}{2}+cos ntheta sinfrac{theta}{2}} {2sinfrac{theta}{2}} tag{7} \[4pt]
&=frac{cos ntheta}{2}+frac {sin n theta sqrt {1+cos theta}} {2 sqrt {1-cos theta}} tag{8} \[4pt]
&=frac12cos ntheta + frac{sin n theta sin theta} {2-2cos theta} tag{9}
end{align}$$
as needed.
Case 2: Suppose $cosfrac{theta}{2}=-sqrt{frac12(1+costheta)}$. Then, by similar logic, I obtained
$$frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}= frac12 cos n theta- frac{sin n theta sin theta} {2-2cos theta} tag{10}$$
but this seems wrong to me.
Where's my mistake?
complex-analysis trigonometry proof-verification
edited Jan 8 at 19:37
Blue
49.7k870158
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
$endgroup$
2
$begingroup$
How about this one? It's shorter and starts the same way.
$endgroup$
– rtybase
Jan 8 at 21:08
1
$begingroup$
@rtybase +1 That was really helpful and solved my problem, even if it doesn't technically answer my "question."
$endgroup$
– Pascal's Wager
Jan 8 at 21:43
add a comment |
$begingroup$
Note: I was originally trying essentially to prove the same thing as this Finalising proof of Lagrange's Trig Identity. However, I do not consider my question to be a duplicate. I am looking for someone to point out my mistake in my "proof," not to prove it for
me.
My attempted "proof"
My goal was to show
$$1+costheta+cdots+cos ntheta= frac12+frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}$$ when $0<theta<2 pi$.
I see that
$$begin{align}LHS
&= Releft(frac {1-e^{i(n+1)theta}} {1-e^{itheta}}right) tag{1}\[4pt]
&= frac {(costheta-1)(cos((n+1)theta)-1)+sin((n+1)theta)sintheta} {(costheta-1)^2+sin^2theta} tag{2}\[4pt]
&=frac{(costheta-1)(cos ntheta cos theta-sin ntheta sintheta-1)+(sin ntheta costheta + cos ntheta sintheta)sintheta} {2-2costheta} tag{3} \[4pt]
&= frac {cos ntheta-costheta-cos ntheta costheta +sin ntheta sintheta +1} {2-2costheta} tag{4}\[4pt]
&=frac12+frac12cos ntheta+ frac{sin ntheta sintheta} {2-2costheta} tag{5}
end{align}$$
To finish proof, it suffices to show that
$$frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}=frac12cos ntheta+frac{sin ntheta sintheta} {2-2costheta} tag{6}$$
By our restriction on $theta$, we must have $sinfractheta2=sqrt{frac12(1-cos theta)}$.
We see that $cos^2fractheta{2}=frac12(1+cos theta)$.
Case 1: Suppose $cosfrac{theta}{2}=sqrt{frac12(1+cos theta)}$. Then
$$begin{align}
frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}
&= frac{sin ntheta cosfrac{theta}{2}+cos ntheta sinfrac{theta}{2}} {2sinfrac{theta}{2}} tag{7} \[4pt]
&=frac{cos ntheta}{2}+frac {sin n theta sqrt {1+cos theta}} {2 sqrt {1-cos theta}} tag{8} \[4pt]
&=frac12cos ntheta + frac{sin n theta sin theta} {2-2cos theta} tag{9}
end{align}$$
as needed.
Case 2: Suppose $cosfrac{theta}{2}=-sqrt{frac12(1+costheta)}$. Then, by similar logic, I obtained
$$frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}= frac12 cos n theta- frac{sin n theta sin theta} {2-2cos theta} tag{10}$$
but this seems wrong to me.
Where's my mistake?
complex-analysis trigonometry proof-verification
edited Jan 8 at 19:37
Blue
49.7k870158
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
$endgroup$
Note: I was originally trying essentially to prove the same thing as this Finalising proof of Lagrange's Trig Identity. However, I do not consider my question to be a duplicate. I am looking for someone to point out my mistake in my "proof," not to prove it for
me.
My attempted "proof"
My goal was to show
$$1+costheta+cdots+cos ntheta= frac12+frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}$$ when $0<theta<2 pi$.
I see that
$$begin{align}LHS
&= Releft(frac {1-e^{i(n+1)theta}} {1-e^{itheta}}right) tag{1}\[4pt]
&= frac {(costheta-1)(cos((n+1)theta)-1)+sin((n+1)theta)sintheta} {(costheta-1)^2+sin^2theta} tag{2}\[4pt]
&=frac{(costheta-1)(cos ntheta cos theta-sin ntheta sintheta-1)+(sin ntheta costheta + cos ntheta sintheta)sintheta} {2-2costheta} tag{3} \[4pt]
&= frac {cos ntheta-costheta-cos ntheta costheta +sin ntheta sintheta +1} {2-2costheta} tag{4}\[4pt]
&=frac12+frac12cos ntheta+ frac{sin ntheta sintheta} {2-2costheta} tag{5}
end{align}$$
To finish proof, it suffices to show that
$$frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}=frac12cos ntheta+frac{sin ntheta sintheta} {2-2costheta} tag{6}$$
By our restriction on $theta$, we must have $sinfractheta2=sqrt{frac12(1-cos theta)}$.
We see that $cos^2fractheta{2}=frac12(1+cos theta)$.
Case 1: Suppose $cosfrac{theta}{2}=sqrt{frac12(1+cos theta)}$. Then
$$begin{align}
frac {sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}
&= frac{sin ntheta cosfrac{theta}{2}+cos ntheta sinfrac{theta}{2}} {2sinfrac{theta}{2}} tag{7} \[4pt]
&=frac{cos ntheta}{2}+frac {sin n theta sqrt {1+cos theta}} {2 sqrt {1-cos theta}} tag{8} \[4pt]
&=frac12cos ntheta + frac{sin n theta sin theta} {2-2cos theta} tag{9}
end{align}$$
as needed.
Case 2: Suppose $cosfrac{theta}{2}=-sqrt{frac12(1+costheta)}$. Then, by similar logic, I obtained
$$frac{sinfrac{(2n+1)theta}{2}} {2sinfrac{theta}{2}}= frac12 cos n theta- frac{sin n theta sin theta} {2-2cos theta} tag{10}$$
but this seems wrong to me.
Where's my mistake?
complex-analysis trigonometry proof-verification
complex-analysis trigonometry proof-verification
edited Jan 8 at 19:37
Blue
49.7k870158
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
edited Jan 8 at 19:37
Blue
49.7k870158
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
edited Jan 8 at 19:37
Blue
49.7k870158
edited Jan 8 at 19:37
Blue
49.7k870158
edited Jan 8 at 19:37
Blue
49.7k870158
49.7k870158
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
asked Jan 8 at 18:50
Pascal's WagerPascal's Wager
381315
381315
2
$begingroup$
How about this one? It's shorter and starts the same way.
$endgroup$
– rtybase
Jan 8 at 21:08
1
$begingroup$
@rtybase +1 That was really helpful and solved my problem, even if it doesn't technically answer my "question."
$endgroup$
– Pascal's Wager
Jan 8 at 21:43
add a comment |
2
$begingroup$
How about this one? It's shorter and starts the same way.
$endgroup$
– rtybase
Jan 8 at 21:08
1
$begingroup$
@rtybase +1 That was really helpful and solved my problem, even if it doesn't technically answer my "question."
$endgroup$
– Pascal's Wager
Jan 8 at 21:43
2
2
$begingroup$
How about this one? It's shorter and starts the same way.
$endgroup$
– rtybase
Jan 8 at 21:08
$begingroup$
How about this one? It's shorter and starts the same way.
$endgroup$
– rtybase
Jan 8 at 21:08
1
1
$begingroup$
@rtybase +1 That was really helpful and solved my problem, even if it doesn't technically answer my "question."
$endgroup$
– Pascal's Wager
Jan 8 at 21:43
$begingroup$
@rtybase +1 That was really helpful and solved my problem, even if it doesn't technically answer my "question."
$endgroup$
– Pascal's Wager
Jan 8 at 21:43
add a comment |
0
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2
$begingroup$
How about this one? It's shorter and starts the same way.
$endgroup$
– rtybase
Jan 8 at 21:08
1
$begingroup$
@rtybase +1 That was really helpful and solved my problem, even if it doesn't technically answer my "question."
$endgroup$
– Pascal's Wager
Jan 8 at 21:43