Mixing
Proof of $|sin(x)| leq 1$ and $|cos(x)| leq 1$
$begingroup$
Using the series representations of $sin(x)$ and $cos(x)$, how does one show that both $|sin(x)| leq 1$ and $|cos(x)| leq 1$? I can do this easily algebraically/trigonometrically, but I am stuck trying to determine this inequality with series.
real-analysis sequences-and-series taylor-expansion
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
asked Feb 1 at 0:40
beflyguybeflyguy
354
$endgroup$
add a comment |
$begingroup$
Using the series representations of $sin(x)$ and $cos(x)$, how does one show that both $|sin(x)| leq 1$ and $|cos(x)| leq 1$? I can do this easily algebraically/trigonometrically, but I am stuck trying to determine this inequality with series.
real-analysis sequences-and-series taylor-expansion
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
asked Feb 1 at 0:40
beflyguybeflyguy
354
$endgroup$
$begingroup$
Using the series representation, the proof I'm thinking of (from Rudin) involves defining $sin$ and $cos$ with the complex exponential. That's probably the easiest way.
$endgroup$
– Moya
Feb 1 at 0:43
5
$begingroup$
Looking at the "related" tab - see here. Not exactly the same question, but showing that the sum of squares is equal to $1$ (and that they're real for real $x$) will do it.
$endgroup$
– jmerry
Feb 1 at 0:44
add a comment |
$begingroup$
Using the series representations of $sin(x)$ and $cos(x)$, how does one show that both $|sin(x)| leq 1$ and $|cos(x)| leq 1$? I can do this easily algebraically/trigonometrically, but I am stuck trying to determine this inequality with series.
real-analysis sequences-and-series taylor-expansion
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
asked Feb 1 at 0:40
beflyguybeflyguy
354
$endgroup$
Using the series representations of $sin(x)$ and $cos(x)$, how does one show that both $|sin(x)| leq 1$ and $|cos(x)| leq 1$? I can do this easily algebraically/trigonometrically, but I am stuck trying to determine this inequality with series.
real-analysis sequences-and-series taylor-expansion
real-analysis sequences-and-series taylor-expansion
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
asked Feb 1 at 0:40
beflyguybeflyguy
354
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
asked Feb 1 at 0:40
beflyguybeflyguy
354
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
edited Feb 1 at 1:26
Theo Bendit
20.8k12354
20.8k12354
asked Feb 1 at 0:40
beflyguybeflyguy
354
asked Feb 1 at 0:40
beflyguybeflyguy
354
asked Feb 1 at 0:40
beflyguybeflyguy
354
354
$begingroup$
Using the series representation, the proof I'm thinking of (from Rudin) involves defining $sin$ and $cos$ with the complex exponential. That's probably the easiest way.
$endgroup$
– Moya
Feb 1 at 0:43
5
$begingroup$
Looking at the "related" tab - see here. Not exactly the same question, but showing that the sum of squares is equal to $1$ (and that they're real for real $x$) will do it.
$endgroup$
– jmerry
Feb 1 at 0:44
add a comment |
$begingroup$
Using the series representation, the proof I'm thinking of (from Rudin) involves defining $sin$ and $cos$ with the complex exponential. That's probably the easiest way.
$endgroup$
– Moya
Feb 1 at 0:43
5
$begingroup$
Looking at the "related" tab - see here. Not exactly the same question, but showing that the sum of squares is equal to $1$ (and that they're real for real $x$) will do it.
$endgroup$
– jmerry
Feb 1 at 0:44
$begingroup$
Using the series representation, the proof I'm thinking of (from Rudin) involves defining $sin$ and $cos$ with the complex exponential. That's probably the easiest way.
$endgroup$
– Moya
Feb 1 at 0:43
$begingroup$
Using the series representation, the proof I'm thinking of (from Rudin) involves defining $sin$ and $cos$ with the complex exponential. That's probably the easiest way.
$endgroup$
– Moya
Feb 1 at 0:43
5
5
$begingroup$
Looking at the "related" tab - see here. Not exactly the same question, but showing that the sum of squares is equal to $1$ (and that they're real for real $x$) will do it.
$endgroup$
– jmerry
Feb 1 at 0:44
$begingroup$
Looking at the "related" tab - see here. Not exactly the same question, but showing that the sum of squares is equal to $1$ (and that they're real for real $x$) will do it.
$endgroup$
– jmerry
Feb 1 at 0:44
add a comment |
1 Answer
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$begingroup$
To prove that $|sin(x)|le 1$ and $|cos(x)|le 1$ from the series definition, I would take a little "side step". From $$sin(x)= sum_{n= 0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(sin(x))'= (-1)^nsum_{n= 0}^infty frac{2n+1}{2n+ 1}!x^{2n}= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}= cos(x)$$. And from $$cos(x)= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}$$ we get $$(cos(x))'= sum_{n= 0}^infty (-1)^nfrac{2n}{(2n)!}x^{2n-1}= sum_{n=0}^infty (-1)^nfrac{1}{2n-1}x^{2n-1}$$. Let $m= n- 1$ so $n= m+ 1$ and that becomes $$sum_{m=0}^infty (-1)^{m+1}frac{1}{(2m+1)!}x^{2m+1}= -sin(x)$$.
Now, let $$f(x)= sin^2(x)+ cos^2(x)$$ then $$f'(x)= 2sin(x)(cos(x))+ 2cos(x)(-sin(x))= 0$$ for all $x$. Since the derivative is $0$ for all $x$, $f(x)$ is a constant. Taking $x= 0$ we have $$f(0)= 0+ 1= 1$$. Therefore, $$sin^2(x)+ cos^2(x)= 1$$ for all $x$.
edited Feb 1 at 2:53
D Stanley
29217
answered Feb 1 at 1:25
user247327user247327
11.5k1516
$endgroup$
add a comment |
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$begingroup$
To prove that $|sin(x)|le 1$ and $|cos(x)|le 1$ from the series definition, I would take a little "side step". From $$sin(x)= sum_{n= 0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(sin(x))'= (-1)^nsum_{n= 0}^infty frac{2n+1}{2n+ 1}!x^{2n}= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}= cos(x)$$. And from $$cos(x)= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}$$ we get $$(cos(x))'= sum_{n= 0}^infty (-1)^nfrac{2n}{(2n)!}x^{2n-1}= sum_{n=0}^infty (-1)^nfrac{1}{2n-1}x^{2n-1}$$. Let $m= n- 1$ so $n= m+ 1$ and that becomes $$sum_{m=0}^infty (-1)^{m+1}frac{1}{(2m+1)!}x^{2m+1}= -sin(x)$$.
Now, let $$f(x)= sin^2(x)+ cos^2(x)$$ then $$f'(x)= 2sin(x)(cos(x))+ 2cos(x)(-sin(x))= 0$$ for all $x$. Since the derivative is $0$ for all $x$, $f(x)$ is a constant. Taking $x= 0$ we have $$f(0)= 0+ 1= 1$$. Therefore, $$sin^2(x)+ cos^2(x)= 1$$ for all $x$.
edited Feb 1 at 2:53
D Stanley
29217
answered Feb 1 at 1:25
user247327user247327
11.5k1516
$endgroup$
add a comment |
$begingroup$
To prove that $|sin(x)|le 1$ and $|cos(x)|le 1$ from the series definition, I would take a little "side step". From $$sin(x)= sum_{n= 0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(sin(x))'= (-1)^nsum_{n= 0}^infty frac{2n+1}{2n+ 1}!x^{2n}= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}= cos(x)$$. And from $$cos(x)= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}$$ we get $$(cos(x))'= sum_{n= 0}^infty (-1)^nfrac{2n}{(2n)!}x^{2n-1}= sum_{n=0}^infty (-1)^nfrac{1}{2n-1}x^{2n-1}$$. Let $m= n- 1$ so $n= m+ 1$ and that becomes $$sum_{m=0}^infty (-1)^{m+1}frac{1}{(2m+1)!}x^{2m+1}= -sin(x)$$.
Now, let $$f(x)= sin^2(x)+ cos^2(x)$$ then $$f'(x)= 2sin(x)(cos(x))+ 2cos(x)(-sin(x))= 0$$ for all $x$. Since the derivative is $0$ for all $x$, $f(x)$ is a constant. Taking $x= 0$ we have $$f(0)= 0+ 1= 1$$. Therefore, $$sin^2(x)+ cos^2(x)= 1$$ for all $x$.
edited Feb 1 at 2:53
D Stanley
29217
answered Feb 1 at 1:25
user247327user247327
11.5k1516
$endgroup$
add a comment |
$begingroup$
To prove that $|sin(x)|le 1$ and $|cos(x)|le 1$ from the series definition, I would take a little "side step". From $$sin(x)= sum_{n= 0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(sin(x))'= (-1)^nsum_{n= 0}^infty frac{2n+1}{2n+ 1}!x^{2n}= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}= cos(x)$$. And from $$cos(x)= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}$$ we get $$(cos(x))'= sum_{n= 0}^infty (-1)^nfrac{2n}{(2n)!}x^{2n-1}= sum_{n=0}^infty (-1)^nfrac{1}{2n-1}x^{2n-1}$$. Let $m= n- 1$ so $n= m+ 1$ and that becomes $$sum_{m=0}^infty (-1)^{m+1}frac{1}{(2m+1)!}x^{2m+1}= -sin(x)$$.
Now, let $$f(x)= sin^2(x)+ cos^2(x)$$ then $$f'(x)= 2sin(x)(cos(x))+ 2cos(x)(-sin(x))= 0$$ for all $x$. Since the derivative is $0$ for all $x$, $f(x)$ is a constant. Taking $x= 0$ we have $$f(0)= 0+ 1= 1$$. Therefore, $$sin^2(x)+ cos^2(x)= 1$$ for all $x$.
edited Feb 1 at 2:53
D Stanley
29217
answered Feb 1 at 1:25
user247327user247327
11.5k1516
$endgroup$
To prove that $|sin(x)|le 1$ and $|cos(x)|le 1$ from the series definition, I would take a little "side step". From $$sin(x)= sum_{n= 0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(sin(x))'= (-1)^nsum_{n= 0}^infty frac{2n+1}{2n+ 1}!x^{2n}= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}= cos(x)$$. And from $$cos(x)= sum_{n=0}^infty (-1)^nfrac{x^{2n}}{(2n)!}$$ we get $$(cos(x))'= sum_{n= 0}^infty (-1)^nfrac{2n}{(2n)!}x^{2n-1}= sum_{n=0}^infty (-1)^nfrac{1}{2n-1}x^{2n-1}$$. Let $m= n- 1$ so $n= m+ 1$ and that becomes $$sum_{m=0}^infty (-1)^{m+1}frac{1}{(2m+1)!}x^{2m+1}= -sin(x)$$.
Now, let $$f(x)= sin^2(x)+ cos^2(x)$$ then $$f'(x)= 2sin(x)(cos(x))+ 2cos(x)(-sin(x))= 0$$ for all $x$. Since the derivative is $0$ for all $x$, $f(x)$ is a constant. Taking $x= 0$ we have $$f(0)= 0+ 1= 1$$. Therefore, $$sin^2(x)+ cos^2(x)= 1$$ for all $x$.
edited Feb 1 at 2:53
D Stanley
29217
answered Feb 1 at 1:25
user247327user247327
11.5k1516
edited Feb 1 at 2:53
D Stanley
29217
edited Feb 1 at 2:53
D Stanley
29217
edited Feb 1 at 2:53
D Stanley
29217
29217
answered Feb 1 at 1:25
user247327user247327
11.5k1516
answered Feb 1 at 1:25
user247327user247327
11.5k1516
answered Feb 1 at 1:25
user247327user247327
11.5k1516
11.5k1516
add a comment |
add a comment |
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$begingroup$
Using the series representation, the proof I'm thinking of (from Rudin) involves defining $sin$ and $cos$ with the complex exponential. That's probably the easiest way.
$endgroup$
– Moya
Feb 1 at 0:43
5
$begingroup$
Looking at the "related" tab - see here. Not exactly the same question, but showing that the sum of squares is equal to $1$ (and that they're real for real $x$) will do it.
$endgroup$
– jmerry
Feb 1 at 0:44