Derivatives of Trig with Proof

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Proof of Trig Derivatives

The derivative of trig functions proof including proof of the trig derivatives that includes sin, cos and tan. These three are actually the most useful derivatives in trigonometric functions. That being said, the three derivatives are as below:

d/dx sin(x) = cos(x)

d/dx cos(x) = −sin(x)

d/dx tan(x) = sec2(x)

Also, remember that the trigonometric functions sin(x) and cos(x) plays an important role in calculus. The AP Calculus however doesn't need knowing the proofs of these derivatives, but as long as a proof is accessible, there's always something to learn from it.


Derivative of Sin x Proof Using Algebraic Method

Proof of (d/dx) Sin(x): With the help of algebraic Method.

Given: \[\lim_{(d\rightarrow0)}\] sin(d)/d = 1.

Solve:

(d/dx) sin(x) = \[\lim_{(d\rightarrow0)}\] {sin(x+d) - sin(x)} / d

= lim {sin(x)cos(d} + cos(x)sin(d) - sin(x)) / d

= lim {sin(x)cos(d} - sin(x) )/d + lim cos(x)sin(d)/d

= sin(x) lim (cos(d) - 1)/d + cos(x) lim sin(d)/d

= sin(x) lim ( (cos(d)-1)(cos(d)+1))/(d(cos(d)+1)) + cos(x) lim sin(d)/d

= sin(x) lim (cos² (d)-1)/(d(cos(d)+1) + cos(x) lim sin(d)/d

= sin(x) lim -sin² (d)/(d (cos(d) + 1) + cos(x) lim sin(d)/d

= sin(x) lim (-sin(d)) × lim sin (d)/d × lim 1/ (cos(d)+1) + cos(x) lim sin(d)/d

= sin(x) × 0 × 1 × ½ + cos(x) × 1 = cos(x)


Derivative of Cos x Proof Using Algebraic Method

Proof of (d/dx) Cos(x): Taking the derivative of sine.

Derivative of cos(x) can be derived in the same way as (d/dx) sin(x) was derived or more simply from the outcome of (d - dx) sin(x).

Given: (d/dx) sin(x) = cos(x); [applying the Chain Rule.]

Solve:

cos(x) = sin(x + PI/2)

(d/dx) cos(x) = (d/dx) sin(x + PI/2)

= (d/du) sin (u) × (d-dx) (x + PI/2) (Set u = x + PI/2)

= cos(u) × 1 = cos(x + PI/2) = -sin(x)

Thus, derivatives of sin and cos are performed in a similar manner.


Derivative of Tan x Proof Using Algebraic Method

Derivative of Tan Proof of (d/dx) Tan(x): Taking the derivatives of sine and cosine.

Given: (d/dx) sin(x) = cos(x); (d/dx) cos(x) = -sin(x); [applying the Quotient Rule]

Solve:

Tan(x) = sin(x) / cos(x)

(d/dx) tan(x) = (d/dx) sin(x)/cos(x)

= (cos(x) (d/dx) sin(x) - sin(x) (d/dx) cos(x))/cos²(x)

= (cos(x)cos(x) + sin(x)sin(x))/cos²(x)

= 1 + tan²(x) = sec²(x)


Solved Examples

Now that we are known of the derivative of sin, cos, tan, let's learn to solve the problems associated with derivative of trig functions proof.

Example:

Determine the derivative of:

f(x) = x² sin(3x)

Solution

We will apply the chain and the product rules.

F '(x) = (2x)(sin(3x)) + (x²)(3cos(3x))

= 2x sin (3x) + 3x² cos (3x)


Example:

The amount A of food available at day (d) of the year for an Indian bear can be modelled by the equation

A(d) = 320 cos 2 π(d - 120)/365

Find out on what day does the bear have the greatest food available?

Solution:

You are asked to find a maximum, which includes taking the 1st derivative and establishing it equivalent to zero. We have

A'(d) = -320(2 π) 2p (d - 120)/365

Establishing it equivalent to zero and solving provides us.

2p (d - 120)/365 = kπ

When k = 0, we obtain d = 120 and when k = 1, we obtain d = 302.5. The 2nd derivative test will tell us that d = 120, provides us with a maximum and d = 320 provides us with a minimum.  Other values of k will provide us the same day of the year. We thus come to the conclusion that the bear has the most food supply on April 30.


Fun Facts

  • Angles in trigonometric functions will always be given in radians.

  • Calculus and Degrees never go together.

  • If you ever hear the word "Degree" in trig functions, the appropriate question to ask would be "Do you mean Celsius or Fahrenheit"?

FAQ (Frequently Asked Questions)

Q1. How do we Prove Reciprocals in Trig Function Derivatives?

Answer: Proof of (d/dx) csc(x), (d/dx) sec(x), (d/dx) cot(x): taking derivatives of their reciprocal functions.

Given: (d/dx) sin(x) = cos(x); (d/dx) cos(x) = -sin(x); (d/dx) tan(x) = cot(x); [applying the Quotient Rule].

Solve:

(d/dx) csc(x) = (d/dx) 1/sin(x) = {sin(x) (d/dx) (1) - 1 (d/dx) sin(x)} / sin²(x) = -cos(x) / sin²(x) = -csc(x)cot(x)

(d-dx) sec(x) = (d-dx) 1/cos(x) = ( cos(x) (d-dx) (1) - 1 (d-dx) cos(x) ) / cos²(x) = sin(x) / cos²(x) = sec(x)tan(x)

(d/dx) cot(x) = (d/dx) 1/tan(x) = {tan(x) (d/dx) (1) - 1 (d/dx) tan(x)} / tan²(x) = -sec²(x) / tan²(x) = -csc²(x)

Q2. What is a Wedge in Finding Derivatives of Trig Proof?

Answer: To find tricky limits that are used in the proof, we make use of wedge. In mathematics, a wedge is something that has one thick or large side reducing to a thin edge, like a doorstop. In such a situation, the fraction of the unit circle goes from a large arc on the edge of the unit circle to a single point in the centre.

Q3. Why do we Always Use Radians in Calculus?

Answer: The proof of the formula including sine needs the angles to be in radians. If the angles are given in degrees the limit involving sine will not be 1 and thus the formulas we will derive for trig derivatives would also change.

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