Derivative of sine without using limits?

[apple2357]

Is there a way of finding the derivative of sin x without using first principles and therefore the limits that $$sinx \approx x$$ and $$cosx \approx 1$$ ?
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 $$sinx \approx x$$

 

[topsquark]

Is there a way of finding the derivative of sin x without using first principles and therefore the limits that $$sinx \approx x$$ and $$cosx \approx 1$$ ?
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 $$sinx \approx x$$

The gradient of tan(x) near x = 0 is x, not 1. Was this a typo?

-Dan

 

[apple2357]

The gradient of tan(x) near x = 0 is x, not 1. Was this a typo?

-Dan

I didnt mean tanx i meant the tangent to the curve at that point.

 

[BigBeachBanana]

Is there a way of finding the derivative of sin x without using first principles and therefore the limits that $$sinx \approx x$$ and $$cosx \approx 1$$ ?
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 $$sinx \approx x$$

I'm confused by your question. Is it to show $(\sin(x))'=\cos(x)$ or $\sin(x)\approx x?$

 

[Steven G]

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

 

[apple2357]

I'm confused by your question. Is it to show $(\sin(x))'=\cos(x)$ or $\sin(x)\approx x?$

My fault! The derivative without using small angle approx !

 

[apple2357]

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

Sorry I wasn’t clear, I do get this bit.