Trig Derivatives: Questions Answered

In summary, the derivatives of sin(x) and cos(x) can be found using the product rule, quotient rule, and other derivative rules. They are not equal to each other, as seen in their graphs and behavior at different points. The derivatives were not "made up" but were derived using trigonometric identities and the definition of a derivative.
  • #1
Robokapp
218
0
Okay. i know that
d/dx of SIn(x)=Cos(X)

d/dx of COs(x) = -Sin(x)

and the rest of them you get by product rule, quotent rule etc using the rules of the derivatives and setting the fractions up correctly in regard to sin and cos.

but i have three questions:

1) If a sin is a trig function of the smae power as the cos becasue they have the same number of direction changes...how can a cos be the derivative of sin.

2) if d/dx of Sin(x)=cos(x) why is D/dx of Cos(x) not = to Sin(x) since sin and cos are same behavior equations.

3) How do you get the derivative of a sin or cos...i mean how did they come up with them? did they just made them up and make the match around it match?

Like i saw how they came up with 0!=1 and i think they kinda "made it fit"...with the whole 5!=5*4! explanation...
 
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  • #2
Robokapp said:
Okay. i know that
d/dx of SIn(x)=Cos(X)

d/dx of COs(x) = -Sin(x)

and the rest of them you get by product rule, quotent rule etc using the rules of the derivatives and setting the fractions up correctly in regard to sin and cos.

but i have three questions:

1) If a sin is a trig function of the smae power as the cos becasue they have the same number of direction changes...how can a cos be the derivative of sin.

2) if d/dx of Sin(x)=cos(x) why is D/dx of Cos(x) not = to Sin(x) since sin and cos are same behavior equations.

3) How do you get the derivative of a sin or cos...i mean how did they come up with them? did they just made them up and make the match around it match?

Like i saw how they came up with 0!=1 and i think they kinda "made it fit"...with the whole 5!=5*4! explanation...
When you say you know what the derivatives of sin(x) and cos(x) are, what do you mean? Do you mean you memorized them? Do you know how to show what their derivatives are? If you do, the answers to your questions should be apparent. For the sine, use this (similar can be done for cosine):

[tex]\lim_{h\to 0}\frac{\sin{\left(x+h\right)}-\sin{\left(x\right)}}{h}[/tex]

Alex
 
Last edited:
  • #3
I'll take 3), which includes answers to 1) and 2) as well:

Let us find the derivative of sin(x).
x is measured in radians. That is IMPORTANT!
By the definition of the derivative, it must be given by (if existing)
(\sin(x))'=\lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}[/tex]
We now use the trigonometric identity:
[tex]\sin(x+h)=\sin(x)\cos(h)+\sin(h)\cos(x)[/tex]
that is, we have:
[tex]\frac{\sin(x+h)-\sin(x)}{h}=\sin(x)\frac{\cos(h)-1}{h}+\cos(x)\frac{\sin(h)}{h}=\frac{\sin(h)}{h}(\cos(x)-\sin(x)\frac{\sin(h)}{1+\cos(h}})[/tex]
Now, assuming sin(x), cos(x) is continuous at h=0, we see that the ugly term within the big parenthesis must go to zero as h goes to zero.

Hence, the derivative of sin(x) should be given by:
[tex](sin(x))'=\lim_{h\to{0}}\frac{\sin(h)}{h}\cos(x)[/tex]

It is relatively simple to show that we have the limit [tex]\lim_{h\to{0}}\frac{\sin(h)}{h}=1[/tex], for example by geometry.

Do you follow that?
 
  • #4
arildno said:
I'll take 3), which includes answers to 1) and 2) as well:

Let us find the derivative of sin(x).
x is measured in radians. That is IMPORTANT!
By the definition of the derivative, it must be given by (if existing)
(\sin(x))'=\lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}[/tex]
We now use the trigonometric identity:
[tex]\sin(x+h)=\sin(x)\cos(h)+\sin(h)\cos(x)[/tex]
that is, we have:
[tex]\frac{\sin(x+h)-\sin(x)}{h}=\sin(x)\frac{\cos(h)-1}{h}+\cos(x)\frac{\sin(h)}{h}=\frac{\sin(h)}{h}(\cos(x)-\sin(x)\frac{\sin(h)}{1+\cos(h}})[/tex]
Now, assuming sin(x), cos(x) is continuous at h=0, we see that the ugly term within the big parenthesis must go to zero as h goes to zero.

Hence, the derivative of sin(x) should be given by:
[tex](sin(x))'=\lim_{h\to{0}}\frac{\sin(h)}{h}\cos(x)[/tex]

It is relatively simple to show that we have the limit [tex]\lim_{h\to{0}}\frac{\sin(h)}{h}=1[/tex], for example by geometry.
Good point. Usually (at least how I've seen it), the limit is shown geometrically using the fact:

[tex]\sin{x}\cos{x}\leq x\leq \tan{x}[/tex]

Alex
 
  • #5
i'm looking at it and understand the work. you used the limit and used the product sisusoid rule to break sin(ab) into Sin(a)Cos(B)+Sin(b)Cos(a) and you simplified it.

it makes sense in a way...but still...although logically it works, out, it's hard to swallow that sin(x) derives into Cos(x)...i mean...you know what I'm saying, right?
 
  • #6
Robokapp said:
i'm looking at it and understand the work. you used the limit and used the product sisusoid rule to break sin(ab) into Sin(a)Cos(B)+Sin(b)Cos(a) and you simplified it.

it makes sense in a way...but still...although logically it works, out, it's hard to swallow that sin(x) derives into Cos(x)...i mean...you know what I'm saying, right?
Not really.

Look at a few values: At x=0, cos(x)=1, that is, it predicts that the slope of sin(x) at the origin is 1
At [itex]x=\frac{\pi}{2}[/itex] the sine function has its maximum value, and thus its derivative should be 0, which agrees with [itex]\cos\frac{\pi}{2}=0[/itex]
and so on..
 
  • #7
Robokapp said:
2) if d/dx of Sin(x)=cos(x) why is D/dx of Cos(x) not = to Sin(x) since sin and cos are same behavior equations.

That could be immediately seen as wrong just by looking at the graphs of the sine and cosine functions. Start with cosine at x=0+ (that means slightly to the right of zero). Is it increasing or decreasing? What does this imply for the sign of the derivative? And what is the sign of sin(x) at x=0+?

If you answer those questions correctly you'll see why sin(x) cannot be the derivative of cos(x).
 
  • #8
Tom Mattson said:
That could be immediately seen as wrong just by looking at the graphs of the sine and cosine functions. Start with cosine at x=0+ (that means slightly to the right of zero). Is it increasing or decreasing? What does this imply for the sign of the derivative? And what is the sign of sin(x) at x=0+?

If you answer those questions correctly you'll see why sin(x) cannot be the derivative of cos(x).

yea...i know some basic limits properties. i follow you. i get it now. Thank you guys.

also, i succesfully worked out the other 4 of them and memorised them becasue it saves me time...It looks like an interesting chapter in math, especially when you know the Chain Rule.
 
  • #9
Robokapp said:
yea...i know some basic limits properties. i follow you. i get it now. Thank you guys.
also, i succesfully worked out the other 4 of them and memorised them becasue it saves me time...It looks like an interesting chapter in math, especially when you know the Chain Rule.
Memorizing saves time, understanding saves brain capacity.
 
  • #10
the power of the mind is infinite. not properly focused...but infinite.
 
  • #11
Robokapp,
This may help:
If
[tex]\frac {d \sin x}{dx} = \cos x[/tex]
and
[tex]\frac {d \cos x}{dx} = -\sin x[/tex]
then
[tex]\sin x \frac {d \sin x}{dx} + \cos x \frac {d \cos x}{dx} = 0[/tex]
and it follows that
[tex]\frac {d}{dx} \left(\sin^2 x + \cos^2 x\right) = 0[/tex]
with the obvious conclusion that the sine, cosine and their derivatives are rather intimately related.

P.S. The power of the mind is not infinite since only a finite number of neural connections are possible with a finite number of neurons! :)
 

FAQ: Trig Derivatives: Questions Answered

What are trigonometric derivatives?

Trigonometric derivatives are a type of derivative in calculus that involves finding the rate of change of a trigonometric function at a specific point. This is done by using the rules of differentiation to find the derivative of the trigonometric function.

Why are trigonometric derivatives important?

Trigonometric derivatives are important because they are used to solve real-world problems involving rates of change, such as in physics and engineering. They also play a crucial role in higher level math courses, such as calculus and differential equations.

What are the basic trigonometric derivatives?

The basic trigonometric derivatives are the derivatives of the six trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. These derivatives follow specific rules, such as the derivative of sine being cosine and the derivative of cosine being negative sine.

How do you find the derivative of a composite trigonometric function?

To find the derivative of a composite trigonometric function, you can use the chain rule. This involves finding the derivative of the outer function and multiplying it by the derivative of the inner function. For example, to find the derivative of sin(3x), you would take the derivative of sin(x) (cos(x)) and multiply it by the derivative of 3x (3), giving you a final answer of 3cos(3x).

Can you find the derivative of inverse trigonometric functions?

Yes, the derivative of inverse trigonometric functions can be found using the inverse function theorem. This involves finding the derivative of the inverse function and taking the reciprocal of that value to find the derivative of the original inverse trigonometric function. For example, the derivative of arctan(x) is 1/(1+x^2).

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