Why derive the cosine derivative from first principles?
A differentiation rule is useful, but understanding where it comes from makes the rule more reliable. For cosine, the key is to examine what happens to \(\cos x\) when its input changes by a small amount \(h\). The difference quotient records that change per unit of input change.
We will use the cosine subtraction identity to rewrite the numerator, and then apply the standard trigonometric limits. The argument works for every real \(x\), provided angles are measured in radians.
The derivative as a limit
For \(f(x)=\cos x\), the definition gives
if this limit exists. We will evaluate it rather than assume a derivative formula for cosine.
The standard trigonometric limits used below are
These limits require angles measured in radians. The second limit can also be seen from the first. Using the identity \(\cos h-1=-2\sin^2(h/2)\), for \(h\ne 0\) we have
As \(h\to 0\), the squared factor tends to \(1\), while \(h/2\) tends to zero. Therefore the whole expression tends to zero. This justifies the second standard limit from the first.
Applying the cosine subtraction identity
The cosine subtraction identity states that
Substitute this identity into the difference quotient. Keep the terms involving \(\cos h\) and \(\sin h\) separate so each can be handled by a standard limit:
For a fixed real number \(x\), both \(\cos x\) and \(\sin x\) are constants with respect to \(h\). Thus the standard limits can be applied to the two factors that depend on \(h\).
The derivative formula
Here is the complete limit argument. It uses the difference quotient definition, the cosine subtraction identity, and the two standard trigonometric limits stated above.
For \(f(x)=\cos x\), consider \(\lim_{h\to 0}[\cos(x+h)-\cos x]/h\), with \(h\ne 0\).
The subtraction identity gives \(\cos(x+h)-\cos x=\cos x(\cos h-1)-\sin x\sin h\). Dividing by \(h\) separates the quotient into two terms.
The first term tends to \(\cos x\cdot 0=0\). The second tends to \(-\sin x\cdot 1=-\sin x\).
The difference quotient has a limit, and that limit is \(-\sin x\). Therefore \(\cos x\) is differentiable at every real \(x\), with derivative \(-\sin x\).
In equation form, the full calculation is
The key point is that the two parts of the quotient behave differently: the cosine term tends to zero, while the sine term tends to one before its coefficient \(-\sin x\) is applied. This is why the result is negative sine, not positive sine.
Worked Example: Derivative at zero
Worked Example: Derivative at zero
Find the derivative of cosine at \(x=0\) directly from the difference quotient.
Begin with the definition and substitute \(x=0\):
The last equality is the standard limit for \((\cos h-1)/h\). Check the result against the formula just proved: \(-\sin 0=-(0)=0\). The direct calculation and the general formula agree.
Worked Example: Derivative at \(\pi/3\)
Worked Example: Derivative at \(\pi/3\)
Evaluate the difference quotient limit at \(x=\pi/3\). The exact values are \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt{3}/2\).
Apply the cosine subtraction identity before taking the limit:
Check: the general formula gives \(-\sin(\pi/3)=-\sqrt3/2\), exactly the same value.
Worked Example: Derivative at \(\pi/2\)
Worked Example: Derivative at \(\pi/2\)
Evaluate the derivative at \(x=\pi/2\) from its defining limit. Since \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\), the subtraction identity gives
Substitute this into the quotient, showing each simplification:
Check: the formula gives \(-\sin(\pi/2)=-1\). This agrees with the limit calculation.
A common sign mistake
A frequent error is to use the wrong sign in the cosine subtraction identity, writing \(\cos(x+h)=\cos x\cos h+\sin x\sin h\). That is the cosine subtraction identity for \(\cos(x-h)\), not the addition identity needed here; the addition identity is \(\cos(x+h)=\cos x\cos h-\sin x\sin h\). It produces a wrong sign in the derivative.
For example, with that incorrect identity the quotient would be wrongly simplified as
The correction is to start with the correct identity, \(\cos(x+h)=\cos x\cos h-\sin x\sin h\). Subtracting \(\cos x\) and dividing by \(h\) then gives a negative sine term:
The sign should be checked in the identity before simplifying the limit. A small sign error at that stage changes the final derivative.
Why radians matter
The limit \(\lim_{h\to 0}\sin h/h=1\) is true when \(h\) is measured in radians. The derivation above directly uses that value. If angles are measured in degrees, the corresponding small-angle limit has a different constant, so the same calculation would not produce this derivative formula without a conversion factor.
Radians are therefore part of the conditions of the theorem, not just a convenient way to write angles. In calculus, trigonometric derivative formulas use radian measure.
What the derivation establishes
This proof establishes the derivative at an arbitrary real input, rather than checking only a collection of particular angles. It also identifies exactly which ingredients are needed: the derivative definition, the cosine subtraction identity, and the standard trigonometric limits. Since the resulting limit exists for every real \(x\), cosine is differentiable throughout its domain \(\mathbb R\).
Once established, the formula \(\frac{d}{dx}\cos x=-\sin x\) can be used as a differentiation rule. The first-principles derivation remains valuable because it explains the origin of the minus sign and shows why radian measure is essential.
Check Your Understanding
Use the difference quotient, the cosine subtraction identity, and the stated limits to answer these questions.
- Write the difference quotient whose limit defines the derivative of \(\cos x\) at an arbitrary real number \(x\).
- After applying the cosine subtraction identity, which quotient tends to zero, and which quotient tends to one?
- Why does the sine term contribute \(-\sin x\) rather than \(+\sin x\)?
- Evaluate the derivative at \(x=\pi\) using the formula, and state the exact value.
- Why must the standard trigonometric limit be used with angles measured in radians?
