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Why Does d/dx[Sin(x)] = Cos(x)? Complete Proof of the Derivative of Sin(x)

Updated on February 04, 2017
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TR Smith is a product designer and former teacher who uses math in her work every day.

Proving that the derivative of sin(x) equals cos(x) using nothing but the limit definition of the derivative and basic trigonometric and geometric principles is something many calculus students never see in class because the full proof is long. If you've always felt it was missing from your calculus education, here you can see the complete classic proof of the derivative of sin(x) worked out in detail.

To begin the proof we will need two formulas: the limit definition of derivatives and the sine angle addition identity.

Step 1

The first step is the apply the limit definition of the derivative using f(x) = sin(x). This gives us the deceptively simply expression

d/dx[sin(x)] = lim(h→0) [sin(x+h) - sin(x)]/h

Now we simplify the right-hand side using the sine angle addition formula.

lim(h→0) [sin(x+h) - sin(x)]/h
= lim(h→0) [cos(x)sin(h) + cos(h)sin(x) - sin(x)]/h
= lim(h→0) [cos(x)sin(h)]/h + lim(h→0) [sin(x)(cos(h) - 1)]/h

We can factor out any terms containing only x. This gives us the expression

cos(x)*[lim(h→0) sin(h)/h] + sin(x)*[lim(h→0) (cos(h) - 1)/h]

This appears a lot more complicated than what we started with, but it is now in a more useful form: cos(x) times a limit plus sin(x) times a limit. In order to make this expression reduce to cos(x), we just need to prove that

lim(h→0) sin(h)/h = 1, and
lim(h→0) (cos(h)-1)/h = 0

which we will tackle in the next step. We cannot plug in h = 0 into these two expressions to find the limit directly since both yield the indeterminate form 0/0. Here is a summary of what we achieved so far.

Step 2

In this step we use the squeeze theorem to prove that the limit of sin(h)/h as h goes to 0 is equal to 1. The statement of the squeeze theorem is: Suppose A(x) ≤ B(x) ≤ C(x). If lim(x→r) A(x) and lim(x→r) C(x) are equal to the same number L, then lim(x→r) B(x) also equals L.

What we want to show is that cos(θ) ≤ sin(θ)/θ ≤ 1/cos(θ). Since cos(0) = 1 and 1/cos(0) = 1, it implies that lim(h→0) sin(θ)/θ also equals 1. To demonstrate that sin(θ)/θ is bounded by cos(θ) and its reciprocal, we only need a simple diagram of a sector cut from a unit circle.

The area of the wedge is less than the area of the larger outer triangle but greater than the area of the smaller inner blue triangle. If θ is measured in radians, the area of the sector is θ/2. Using basic trigonometry and properties of similar triangles, you can see that the area of the smaller triangle is (1/2)cos(θ)sin(θ), and the area of the larger triangle is (1/2)tan(θ). Thus we have

(1/2)cos(θ)sin(θ) ≤ θ/2 ≤ (1/2)tan(θ)

Doing some algebra on this relation gives us

cos(θ)sin(θ) ≤ θ ≤ tan(θ)
cos(θ) ≤ θ/sin(θ) ≤ 1/cos(θ)
1/cos(θ) ≥ sin(θ)/θ ≥ cos(θ)

And now we can see that since cos(0) = 1 and 1/cos(0) = 1, it must be the case that the limiting value of sin(0)/0 also equals 1.

Step 3

Finally we need to prove that lim(h→0) (cos(h) - 1)/h = 0. To do this we will use algebra rather than geometry. First we transform the fraction (cos(h) - 1)/h into something else.

(cos(h) - 1)/h
= [(cos(h) - 1)/h] * [(cos(h) + 1)/(cos(h) + 1)]
= [(cos(h)^2 - 1)]/[h*(cos(h) + 1)]
= -sin(h)^2 / [h*(cos(h) + 1)]
= [sin(h)/h] * [-sin(h)/(cos(h) + 1)]

Now we have

lim(h→0) (cos(h) - 1)/h
= lim(h→0) { [sin(h)/h] * [-sin(h)/(cos(h) + 1)] }
= [ lim(h→0) sin(h)/h ] * [ lim(h→0) -sin(h)/(cos(h) + 1) ]

In the product of limits above, the factor on the left is equal to 1 as we just proved in the previous section. The factor on the right is easy because we can plug 0 directly in for h. This gives us -0/2 = 0. Thus, the product of limits is 1*0 = 0.

Step 4: Putting Everything Together

We ended Step 1 with the relation

d/dx[sin(x)] = cos(x)*[lim(h→0) sin(h)/h] + sin(x)[lim(h→0) (cos(h) - 1)/h]

and we proved in Steps 2 and 3 that the first limit is 1 and the second limit is 0. Therefore we have

d/dx[sin(x)] = cos(x)*1 + sin(x)*0 = cos(x)

And thus we prove that the derivative of sin(x) is cos(x).

Alternative Proof of the Derivative of Sin(x)

Assuming you know the derivative of a polynomial, you can use the power series for sine and cosine to quickly prove that the derivative of sin(x) equals cos(x). However, since the power series expansions of functions were developed using their derivatives, this may feel like a "cheating" in a sense with a circular proof.

Trig functions can be defined as convergent power series called Taylor series. You can think of them as single-variable polynomials with an infinite number of terms whose coefficients tend toward zero as the power of x increases. The power series formula for sin(x) is

sin(x) = (x^1)/1! - (x^3)/3! + (x^5)/5 - (x^7)/7! + ...

where x is in radians. In summation notation, it's

Taking the derivative of this series term by term gives you

= d/dx[ x - (x^3)/3! + (x^5)/5! - (x^7)/7!) + ...]
= 1 - 3(x^2)/3! + 5(x^4)/5! - 7(x^7)/7! + ...
= 1 - (x^2)/2! + (x^4)/4! - (x^6)/6!

Remarkably, the series in the line immediately above is the series for cos(x).

Thus, we can see that the derivative of sin(x) is cos(x) and the proof is complete. Of course, this short proof has several assumptions working behind the scenes. One is that the derivative of a power function f(x) = a*x^n is f'(x) = an*x^(n-1). Another is that the derivative of a sum is the sum of the derivatives of the individual pieces. And of course this proof assumes the series formulations of sine and cosine.


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