Differentiation sin and cos from First principles.

I recently watched a TLmaths video on this topic and he says that sinh/h=1 and cosh-1/h =0 is there a reason for this?

When h tends to 0:

sinh/h = 1 (e.g. if we were to say that h=0.5, then sin(0.5) / 0.5 = 0.9588.... and as h gets closer to 0, the value of sinh/h gets closer to 1 i.e. if you substitute smaller values of h into sinh / h you'll find that you get closer and closer to a value of 1)

cosh - 1 / h = 0 (e.g. if we were to again say that h=0.5, then cos(0.5) - 1 / 0.5 = -0.2448... and as h gets closer to 0, the value of cosh-1/h gets closer to 0 i.e. if you substitute smaller values of h into cosh - 1 / h, you'll find you get closer and closer to 0)

Edit: All of this is calculated in radians btw!

(edited 1 year ago)

Reply 2

TypicalNerd

Original post by Bennayy

I recently watched a TLmaths video on this topic and he says that sinh/h=1 and cosh-1/h =0 is there a reason for this?

As h —> 0, if h is in radians, then you can use the small angle approximations (sin(h) ≈ h, tan(h) ≈ h and cos(h) ≈ 1 - h^2/2).

So sin(h)/h ≈ h/h = 1

And (cos(h) - 1)/h ≈ -h/2 ≈ 0 (since h —> 0)

This is also why differentiation of trig functions in degrees does not work, as the reliance on small angle approximations means inputs must be in radians.

(edited 1 year ago)

Reply 3

Sinnoh

Original post by TypicalNerd

The problem is that you can only arrive at those expressions by already taking the derivatives as given

Reply 4

mqb2766

Original post by Bennayy

I recently watched a TLmaths video on this topic and he says that sinh/h=1 and cosh-1/h =0 is there a reason for this?

You can think of the unit circle and the usual right triangle with legs x=cos(h) and y=sin(h) and unit hypotenuse, then cosnder as the angle h goes to zero
https://en.wikipedia.org/wiki/Small-angle_approximation#Geometric
So h is the arc length (measured in radians), sin is the vertical height and cos is the horizontal length. As h goes to zero, the vertical height and the arc length tend to the same value and the horizontal length tends to 1 (base tends to hypotenuse).

Making it a bit more precise
https://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions
You can also make the cos() argument using the sin() result and pythagoras/binomial

(edited 1 year ago)

Reply 5

TypicalNerd

Original post by Sinnoh

The problem is that you can only arrive at those expressions by already taking the derivatives as given

My guess is that you have only seen the small angle approximations derived from the Maclaurin series expansions of cosx and sinx.

You can arrive at the small angle approximations geometrically using SOH CAH TOA and with the cosine law.

Spoiler

(edited 1 year ago)

Reply 6

username6101774