Proof of the Derivatives of
sin, cos and tan

The three most useful derivatives in trigonometry are:

ddx sin(x) = cos(x)

ddx cos(x) = −sin(x)

ddx tan(x) = sec²(x)

Proving the Derivative of Sine

We need to go back, right back to first principles, the basic formula for derivatives:

dydx = limΔx→0 f(x+Δx)−f(x)Δx

Pop in sin(x):

ddxsin(x) = limΔx→0 sin(x+Δx)−sin(x)Δx

We can then use this trigonometric identity: sin(A+B) = sin(A)cos(B) + cos(A)sin(B) to get:

limΔx→0 sin(x)cos(Δx) + cos(x)sin(Δx) − sin(x)Δx

Regroup:

limΔx→0 sin(x)(cos(Δx)−1) + cos(x)sin(Δx)Δx

Split into two limits:

limΔx→0 sin(x)(cos(Δx)−1)Δx + limΔx→0cos(x)sin(Δx)Δx

And we can bring sin(x) and cos(x) outside the limits because they are functions of x not Δx

sin(x) limΔx→0 cos(Δx)−1Δx + cos(x) limΔx→0 sin(Δx)Δx

Now all we have to do is evaluate those two little limits. Easy, right? Ha!

Limit of sin(θ)θ

Starting with

limθ→0 sin(θ)θ

with the help of some geometry:

We can look at areas:

Area of triangle AOB < Area of sector AOB < Area of triangle AOC

12r² sin(θ) < 12r² θ < 12r² tan(θ)

Divide all terms by 12r² sin(θ)

1 < θsin(θ) < 1cos(θ)

Take the reciprocals:

1 > sin(θ)θ > cos(θ)

Now as θ→0 then cos(θ)→1

So sin(θ)θ lies between 1 and something that is tending towards 1

So as θ→0 then sin(θ)θ →1 and so:

limθ→0 sin(θ)θ = 1

(Note: we should also prove this is true from the negative side, how about you try with negative values of θ ?)

Limit of cos(θ)−1θ

So next we want to find out this one:

limθ→0 cos(θ)−1θ

When we multiply top and bottom by cos(θ)+1 we get:

(cos(θ)−1)(cos(θ)+1)θ(cos(θ)+1) = cos²(θ)−1θ(cos(θ)+1)

Now we use this trigonometric identity based on Pythagoras' Theorem:

cos²(x) + sin²(x) = 1

Rearranged to this form:

cos²(x) − 1 = −sin²(x)

And the limit we started with can become:

limθ→0 −sin²(θ)θ(cos(θ)+1)

That looks worse! But is really better because we can turn it into two limits multiplied together:

limθ→0sin(θ)θ × limθ→0−sin(θ)cos(θ)+1

We know the first limit (we worked it out above), and the second limit doesn't need much work because at θ=0 we know directly that −sin(0)cos(0)+1 = 0, so:

limθ→0sin(θ)θ × limθ→0−sin(θ)cos(θ)+1 = 1 × 0 = 0

Putting it Together

So what were we trying to do again? Oh that's right, we really wanted to work out this:

ddxsin(x) = sin(x) limΔx→0 cos(Δx)−1Δx + cos(x) limΔx→0 sin(Δx)Δx

We can now put in the values we just worked out and get:

ddxsin(x) = sin(x) × 0 + cos(x) × 1

And so (ta da!):

ddxsin(x) = cos(x)

The Derivative of Cosine

Now on to cosine!

ddxcos(x) = limΔx→0 cos(x+Δx)−cos(x)Δx

This time we will use the angle formula cos(A+B) = cos(A)cos(B) − sin(A)sin(B):

limΔx→0 cos(x)cos(Δx) − sin(x)sin(Δx) − cos(x)Δx

Rearrange to:

limΔx→0 cos(x)(cos(Δx)−1) − sin(x)sin(Δx)Δx

Split into two limits:

limΔx→0 cos(x)(cos(Δx)−1)Δx − limΔx→0sin(x)sin(Δx)Δx

We can bring cos(x) and sin(x) outside the limits because they are functions of x not Δx

cos(x) limΔx→0 cos(Δx)−1Δx − sin(x) limΔx→0 sin(Δx)Δx

And using our knowledge from above:

ddx cos(x) = cos(x) × 0 − sin(x) × 1

And so:

ddx cos(x) = −sin(x)

The Derivative of Tangent

To find the derivative of tan(x) we can use this identity:

tan(x) = sin(x)cos(x)

So we start with:

ddxtan(x) = ddx(sin(x)cos(x))

And we get:

ddxtan(x) = cos(x) × cos(x) − sin(x) × −sin(x)cos²(x)

ddxtan(x) = cos²(x) + sin²(x)cos²(x)

Then use this identity:

cos²(x) + sin²(x) = 1

To get

ddxtan(x) =1cos²(x)

Done!

But most people like to use the fact that cos = 1sec to get:

ddxtan(x) = sec²(x)

Note: we can also do this:

ddxtan(x) = cos²(x) + sin²(x)cos²(x)

ddxtan(x) = 1 + sin²(x)cos²(x) = 1 + tan²(x)

(And, yes, 1 + tan²(x) = sec²(x) anyway, see Magic Hexagon )

Taylor Series

Just on a fun side note, we can use the Taylor Series expansions and differentiate term by term.

But this is "circular reasoning" because the original expansion of the Taylor Series already use the rules "the derivative of sin(x) is cos(x)" and "the derivative of cos(x) is −sin(x)".