Proof by showing equivalence of derivatives
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It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.
But if we let $f(x) = sin(2x)$, we can differentiate two ways:
1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$
2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$
Is this a viable proof?
Thanks in Advance!
derivatives trigonometry proof-verification proof-writing
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
add a comment |
up vote
4
down vote
favorite
It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.
But if we let $f(x) = sin(2x)$, we can differentiate two ways:
1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$
2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$
Is this a viable proof?
Thanks in Advance!
derivatives trigonometry proof-verification proof-writing
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
– Janitha357
Dec 6 at 4:56
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.
But if we let $f(x) = sin(2x)$, we can differentiate two ways:
1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$
2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$
Is this a viable proof?
Thanks in Advance!
derivatives trigonometry proof-verification proof-writing
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.
But if we let $f(x) = sin(2x)$, we can differentiate two ways:
1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$
2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$
Is this a viable proof?
Thanks in Advance!
derivatives trigonometry proof-verification proof-writing
derivatives trigonometry proof-verification proof-writing
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
asked Dec 6 at 4:48
Aniruddh Venkatesan
1189
1189
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
– Janitha357
Dec 6 at 4:56
add a comment |
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
– Janitha357
Dec 6 at 4:56
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
– Janitha357
Dec 6 at 4:56
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
– Janitha357
Dec 6 at 4:56
add a comment |
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.
A bit overkill, obviously, but 100% valid, as with your method.
answered Dec 6 at 4:50
Eevee Trainer
3,320225
add a comment |
up vote
3
down vote
The ideas and work shown are valid.
In order to format the work here into the format of a proof, we need to identify our hypotheses.
The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.
Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.
answered Dec 6 at 5:18
John B
1766
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.
A bit overkill, obviously, but 100% valid, as with your method.
answered Dec 6 at 4:50
Eevee Trainer
3,320225
add a comment |
up vote
3
down vote
accepted
Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.
A bit overkill, obviously, but 100% valid, as with your method.
answered Dec 6 at 4:50
Eevee Trainer
3,320225
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.
A bit overkill, obviously, but 100% valid, as with your method.
answered Dec 6 at 4:50
Eevee Trainer
3,320225
Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.
A bit overkill, obviously, but 100% valid, as with your method.
answered Dec 6 at 4:50
Eevee Trainer
3,320225
answered Dec 6 at 4:50
Eevee Trainer
3,320225
answered Dec 6 at 4:50
Eevee Trainer
3,320225
answered Dec 6 at 4:50
Eevee Trainer
3,320225
3,320225
add a comment |
add a comment |
up vote
3
down vote
The ideas and work shown are valid.
In order to format the work here into the format of a proof, we need to identify our hypotheses.
The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.
Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.
answered Dec 6 at 5:18
John B
1766
add a comment |
up vote
3
down vote
The ideas and work shown are valid.
In order to format the work here into the format of a proof, we need to identify our hypotheses.
The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.
Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.
answered Dec 6 at 5:18
John B
1766
add a comment |
up vote
3
down vote
up vote
3
down vote
The ideas and work shown are valid.
In order to format the work here into the format of a proof, we need to identify our hypotheses.
The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.
Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.
answered Dec 6 at 5:18
John B
1766
The ideas and work shown are valid.
In order to format the work here into the format of a proof, we need to identify our hypotheses.
The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.
Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.
answered Dec 6 at 5:18
John B
1766
answered Dec 6 at 5:18
John B
1766
answered Dec 6 at 5:18
John B
1766
answered Dec 6 at 5:18
John B
1766
1766
add a comment |
add a comment |
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I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
– Janitha357
Dec 6 at 4:56