Derivative of sine without using limits?

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apple2357

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Is there a way of finding the derivative of sin x without using first principles and therefore the limits that [math]sinx \approx x[/math] and [math]cosx \approx 1[/math] ?
Dont ask me why but i want to use the fact that sinx differentiates to cosx to show that the gradient of the tangent at x=0 is 1 so the equation of the tangent is y=x which leads to [math]sinx \approx x[/math]
 
apple2357 said:
Is there a way of finding the derivative of sin x without using first principles and therefore the limits that [math]sinx \approx x[/math] and [math]cosx \approx 1[/math] ?
Dont ask me why but i want to use the fact that sinx differentiates to cosx to show that the gradient of the tangent at x=0 is 1 so the equation of the tangent is y=x which leads to [math]sinx \approx x[/math]
Click to expand...
The gradient of tan(x) near x = 0 is x, not 1. Was this a typo?

-Dan
 
apple2357 said:
Is there a way of finding the derivative of sin x without using first principles and therefore the limits that [math]sinx \approx x[/math] and [math]cosx \approx 1[/math] ?
Dont ask me why but i want to use the fact that sinx differentiates to cosx to show that the gradient of the tangent at x=0 is 1 so the equation of the tangent is y=x which leads to [math]sinx \approx x[/math]
Click to expand...
I'm confused by your question. Is it to show [imath](\sin(x))'=\cos(x)[/imath] or [imath]\sin(x)\approx x?[/imath]
 
What are you talking about?

sinx≈x is not true at all. Draw the line y=x and y=sinx and see for yourself if they are close to one another.
For example, let x= 10,000. Then for y=x, y would be 10,000. However, for y = sinx, y would be sin 10,000 which is a number between -1 and 1---nowhere near 10,000
Same for cosx≈ 1. cosx goes between -1 and 1 and is not always close to 1.

It is true that \(\displaystyle \lim_{x\rightarrow 0} \dfrac{sinx}{x} = 1\) but that shows that sinx≈x ONLY for x-values near 0
 
Steven G said:
What are you talking about?

sinx≈x is not true at all. Draw the line y=x and y=sinx and see for yourself if they are close to one another.
For example, let x= 10,000. Then for y=x, y would be 10,000. However, for y = sinx, y would be sin 10,000 which is a number between -1 and 1---nowhere near 10,000
Same for cosx≈ 1. cosx goes between -1 and 1 and is not always close to 1.

It is true that \(\displaystyle \lim_{x\rightarrow 0} \dfrac{sinx}{x} = 1\) but that shows that sinx≈x ONLY for x-values near 0
Click to expand...
Sorry I wasn’t clear, I do get this bit.
 
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