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Linearization and Local Change · Tutorial 313 of 999

Linearizing Trigonometric Functions Near Standard Angles

Estimate nearby trigonometric values by using their derivatives at familiar angles, with careful attention to angle units and domain restrictions.

Intermediate 9 min read

What You'll Learn

  • Build linearizations for sine, cosine, and tangent at standard angles
  • Use radians correctly in the derivative and the angle change
  • Convert degree changes into radians when estimating
  • Check that tangent’s base point and target lie in a valid interval
  • Verify an estimate by checking the model’s value and slope at its base point

Why Linearize Trigonometric Functions?

A trigonometric value at a nearby angle can often be estimated without evaluating the function directly. The idea is to use the tangent line at a familiar angle, such as \(\pi/6\), \(\pi/4\), or \(\pi/3\), where sine, cosine, and tangent have known values.

This method is useful only when the angle change is expressed in the right units. The standard derivative formulas for trigonometric functions assume that the input is measured in radians. A change of \(1\) degree is not a change of \(1\) radian, so mixing those units changes the calculation.

Key idea: Near a standard angle \(a\), approximate a trigonometric function \(f\) by its linearization \(L_a(x)=f(a)+f'(a)(x-a)\). Use radians for \(x\) and \(a\) when applying the standard trigonometric derivative formulas.

The Linearization Formulas

The linearization of a differentiable function \(f\) at a point \(a\) is the linear function that has the same value and the same derivative as \(f\) at \(a\). For trigonometric functions with radian inputs, the standard derivative formulas are \(\frac{d}{dx}\sin x=\cos x\), \(\frac{d}{dx}\cos x=-\sin x\), and \(\frac{d}{dx}\tan x=\sec^2 x\), with tangent defined only where \(\cos x\ne0\).

$$ \begin{aligned} L_{\sin,a}(x)&=\sin a+\cos a\,(x-a),\\ L_{\cos,a}(x)&=\cos a-\sin a\,(x-a),\\ L_{\tan,a}(x)&=\tan a+\sec^2 a\,(x-a). \end{aligned} $$

The tangent formula requires \(\cos a\ne0\), so its base point cannot be an angle where tangent is undefined. The target angle must also be in the same interval of definition as the base point: do not use a tangent linearization across an asymptote. Sine and cosine are defined for all real inputs, so they have no corresponding restriction.

For any of these formulas, \(x-a\) is the signed change in angle. A positive change moves to a larger angle; a negative change moves to a smaller angle. Because the derivative gives change in function value per radian, \(x-a\) must be in radians when the displayed formulas are used.

Why the Formulas Work

At the base point \(a\), the linearization must match the original function’s value. Its slope must match the original function’s derivative there. These two requirements determine the line: a line with value \(f(a)\) at \(a\) and slope \(f'(a)\) has equation \(f(a)+f'(a)(x-a)\).

$$ L_a(a)=f(a)+f'(a)(a-a)=f(a) $$

Also, differentiating the linear function gives \(L_a'(x)=f'(a)\), so \(L_a'(a)=f'(a)\). Thus it agrees with \(f\) in both value and slope at \(a\). When \(x\) is near \(a\), this matching tangent line provides a local estimate.

Applying this reasoning to sine gives \(f(a)=\sin a\) and \(f'(a)=\cos a\), producing \(L_{\sin,a}(x)\). For cosine, \(f(a)=\cos a\) and \(f'(a)=-\sin a\), producing \(L_{\cos,a}(x)\). For tangent, \(f(a)=\tan a\) and \(f'(a)=\sec^2 a\), producing \(L_{\tan,a}(x)\), provided tangent is defined at \(a\).

Worked Examples

Worked Example: Estimate Sine Near \(\pi/6\)

Estimate \(\sin(\pi/6+0.04)\), where the angle change \(0.04\) is in radians.

Use \(a=\pi/6\). The standard values are \(\sin(\pi/6)=1/2\) and \(\cos(\pi/6)=\sqrt{3}/2\). The sine linearization is therefore

$$ L_{\sin,\pi/6}(x) =\frac12+\frac{\sqrt3}{2}\left(x-\frac{\pi}{6}\right). $$

The target angle is \(x=\pi/6+0.04\), so the change from the base point is \(x-\pi/6=0.04\). Substituting gives

$$ \begin{aligned} \sin\left(\frac{\pi}{6}+0.04\right) &\approx L_{\sin,\pi/6}\left(\frac{\pi}{6}+0.04\right)\\ &=\frac12+\frac{\sqrt3}{2}(0.04)\\ &=0.5+0.02\sqrt3\\ &\approx0.5+0.0346410162\\ &\approx0.5346410162. \end{aligned} $$

Check the model at its base point: \(L_{\sin,\pi/6}(\pi/6)=1/2=\sin(\pi/6)\). Its slope is \(\sqrt3/2=\cos(\pi/6)\), as required. The requested estimate is approximately \(0.534641\).

Worked Example: Estimate Cosine Near \(\pi/3\)

Estimate \(\cos(\pi/3+0.03)\), with the angle change measured in radians.

At \(a=\pi/3\), \(\cos a=1/2\) and \(\sin a=\sqrt3/2\). Since the derivative of cosine is negative sine,

$$ L_{\cos,\pi/3}(x) =\frac12-\frac{\sqrt3}{2}\left(x-\frac{\pi}{3}\right). $$

Here \(x-\pi/3=0.03\). Thus

$$ \begin{aligned} \cos\left(\frac{\pi}{3}+0.03\right) &\approx L_{\cos,\pi/3}\left(\frac{\pi}{3}+0.03\right)\\ &=\frac12-\frac{\sqrt3}{2}(0.03)\\ &=0.5-0.015\sqrt3\\ &\approx0.5-0.0259807621\\ &\approx0.4740192379. \end{aligned} $$

For the check, \(L_{\cos,\pi/3}(\pi/3)=1/2=\cos(\pi/3)\), and the model’s slope is \(-\sqrt3/2=-\sin(\pi/3)\). Both match the cosine function at the base point. The estimate is approximately \(0.474019\).

Worked Example: Estimate Tangent Near \(\pi/4\)

Estimate \(\tan(\pi/4+0.02)\), with angles in radians.

At \(a=\pi/4\), \(\tan a=1\) and \(\sec^2 a=2\). The tangent linearization is

$$ L_{\tan,\pi/4}(x) =1+2\left(x-\frac{\pi}{4}\right). $$

The target is only \(0.02\) radians from \(\pi/4\), so the estimate is

$$ \begin{aligned} \tan\left(\frac{\pi}{4}+0.02\right) &\approx L_{\tan,\pi/4}\left(\frac{\pi}{4}+0.02\right)\\ &=1+2(0.02)\\ &=1.04. \end{aligned} $$

We must check the domain. The next tangent asymptote to the right of \(\pi/4\) is at \(\pi/2\), and \(0.02<\pi/4\), so the target \(\pi/4+0.02\) remains below that asymptote. At the base point, \(L_{\tan,\pi/4}(\pi/4)=1=\tan(\pi/4)\), and its slope is \(2=\sec^2(\pi/4)\). The estimate is \(1.04\).

When Angles Are Given in Degrees

The formulas above use radians. If the angle is given in degrees, first convert its change to radians, or explicitly include the conversion factor in the derivative with respect to degrees. Since \(x^\circ=\pi x/180\) radians, a change of \(h\) degrees corresponds to a radian change of \(h\pi/180\).

Unit conversion: For a degree change of \(h^\circ\), use the radian change \(h\pi/180\) in the radian-based linearization. Equivalently, when the input variable is measured in degrees, the derivatives of sine and cosine include a factor of \(\pi/180\).

Worked Example: Estimate Sine at \(31^\circ\)

Estimate \(\sin(31^\circ)\) by linearizing at \(30^\circ\). Write the base point in radians: \(30^\circ=\pi/6\). The change is \(1^\circ\), which equals \(\pi/180\) radians. Using the sine linearization at \(\pi/6\),

$$ \begin{aligned} \sin(31^\circ) &\approx \sin\left(\frac{\pi}{6}+\frac{\pi}{180}\right)\\ &\approx \frac12+\frac{\sqrt3}{2}\left(\frac{\pi}{180}\right)\\ &=\frac12+\frac{\pi\sqrt3}{360}\\ &\approx0.5+0.0151149947\\ &\approx0.5151149947. \end{aligned} $$

For the check, setting the change equal to zero gives the base value \(1/2=\sin(\pi/6)\), and the coefficient of the change is \(\sqrt3/2=\cos(\pi/6)\), the correct slope per radian. The one-degree change was converted before it was used. The estimate is approximately \(0.515115\).

Common Mistake: Using Degrees as if They Were Radians

A frequent error is to insert a degree change directly into a formula whose derivative assumes radians. For instance, using the sine linearization at \(30^\circ\) but treating the change to \(31^\circ\) as the number \(1\) would give

$$ \frac12+\frac{\sqrt3}{2}(1)\approx1.366. $$

This is not the correct linearization calculation: the input change is one degree, not one radian. Convert the change first:

$$ 1^\circ=\frac{\pi}{180}\text{ radians}, \qquad \frac12+\frac{\sqrt3}{2}\left(\frac{\pi}{180}\right) \approx0.515115. $$

The wrong calculation does not merely round poorly; it uses a slope with the wrong angle unit. Always identify the units of both the base point and the change before substituting.

Choosing a Base Point and Respecting the Domain

Use a standard angle close to the target and with known trigonometric values. A closer base point generally makes the tangent-line estimate more local, but a familiar value is also useful because it keeps the arithmetic manageable. The estimate is a local model, not an identity, so it should not be treated as an exact value or extended arbitrarily far from its base point.

For tangent, domain checks are essential. Tangent is undefined when \(\cos x=0\), at angles \(\pi/2+k\pi\) for integers \(k\). Its graph has a vertical asymptote at each such angle. A tangent line at a valid base point cannot provide a model across one of these breaks. For sine and cosine, there are no excluded real angles, though their linearizations are still intended for nearby inputs.

1
Select a standard base angle.
Choose a nearby angle \(a\) with a familiar function value and derivative.
2
Check units.
Express the angle change in radians before using the standard trigonometric derivative formulas.
3
Check the function’s domain.
For tangent, confirm that the base point and target remain in an interval that contains no asymptote.
4
Evaluate the linearization.
Compute the function value at the base point and add the derivative there times the signed angle change.
5
Verify the model at the base point.
Check that the line gives the known function value and has the correct slope.

Check Your Understanding

For each question, keep track of whether the angle change is in radians or degrees.

  1. Write the linearization of \(\sin x\) at \(a=\pi/3\).
  2. Use a linearization at \(\pi/4\) to estimate \(\cos(\pi/4+0.01)\), with the change in radians.
  3. Use the tangent linearization at \(\pi/6\) to estimate \(\tan(\pi/6+0.02)\). State why the estimate is within the domain interval of the base point.
  4. What radian change should be used to estimate \(\sin(29^\circ)\) by linearizing at \(30^\circ\)?
  5. Explain why substituting a degree change directly into the standard sine linearization can give an incorrect estimate.

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