Why Use Tangent Lines for Trigonometric Estimates?
Trigonometric functions can be difficult to evaluate exactly at many angles. When an angle is close to one whose sine, cosine, or tangent is known, the tangent line at that known angle gives a simple nearby estimate. This is called linearization.
For example, \(\sin 0=0\) and \(\cos 0=1\). A tangent line at \(0\) can therefore give estimates for sine and cosine at small angles. Similarly, \(\tan(\pi/4)=1\), so a tangent line at \(\pi/4\) can estimate tangent values close to that angle.
These estimates are useful only when we also understand their limitations. The function value need not equal its tangent-line estimate. We will track the signed error, determine whether the estimate is above or below the true value, and bound the error's size.
Linearization in Radians
For the trigonometric functions, the derivative formulas used here are \(\frac{d}{dx}\sin x=\cos x\), \(\frac{d}{dx}\cos x=-\sin x\), and \(\frac{d}{dx}\tan x=\sec^2 x\). These formulas assume angles are measured in radians. The tangent function is defined only where \(\cos x\ne 0\), so its estimate must be made on an interval that does not cross a vertical asymptote.
Substituting the derivative formulas into the tangent-line definition gives the three linearizations:
The symbol \(\approx\) indicates an estimate, not an identity. The base angle \(a\) should be chosen so that the function value and derivative there are known and \(x\) is close to \(a\).
How Concavity Determines the Error Direction
The second derivative describes the concavity of a function. If \(f''\) is nonnegative throughout an interval, \(f\) is concave up there and its graph lies above its tangent line. Consequently, \(L_a(x)\le f(x)\) on that interval. If \(f''\) is nonpositive, the graph lies below its tangent line, so \(f(x)\le L_a(x)\). These statements apply when the whole interval from \(a\) to \(x\) has the stated sign of \(f''\).
For the functions in this tutorial, the second derivatives are
These signs give a quick direction check. For instance, sine is concave down on \((0,\pi)\), so its tangent line overestimates sine when both the base angle and the estimated angle lie in that interval. Tangent is concave up on \((0,\pi/2)\), so its tangent line underestimates there.
A Bound for the Size of the Error
This bound makes the role of “nearby” precise: if the distance from the base point is halved, the bound on the error becomes one quarter as large, provided the same bound \(M\) applies. The estimate requires a bound on the second derivative over the entire interval between \(a\) and \(x\), not just at one endpoint.
Here is a justification. The Fundamental Theorem of Calculus gives \(f(x)-f(a)=\int_a^x f'(t)\,dt\). Subtracting the linear term \(f'(a)(x-a)\) yields
By the Mean Value Theorem applied to \(f'\), the bound on \(f''\) implies \(|f'(t)-f'(a)|\le M|t-a|\). Taking absolute values in the integral and integrating this bound along the interval gives
This proves the stated error bound for either order of \(a\) and \(x\). The sign of the error, when known from concavity, complements this size bound.
Worked Estimate for Sine
Worked Example: Estimate \(\sin(0.1)\)
Use the tangent line at \(a=0\). The angle \(0.1\) is in radians, and \(\sin 0=0\), \(\cos 0=1\). Thus the linearization is \(L_0(x)=0+1(x-0)=x\), giving the estimate \(\sin(0.1)\approx 0.1\).
To check the direction, \((\sin x)''=-\sin x<0\) for \(0<x\le 0.1\). The function is concave down between \(0\) and \(0.1\), so the tangent line overestimates. The signed error \(E=\sin(0.1)-0.1\) should be negative.
For the size, \(|(\sin x)''|=|\sin x|\le 0.1\) on \([0,0.1]\). The error bound with \(M=0.1\) is
A numerical check gives \(\sin(0.1)\approx 0.0998334166\), so \(E\approx 0.0998334166-0.1=-0.0001665834\). This is negative, as predicted, and its magnitude \(0.0001665834\) is less than \(0.0005\). The estimate \(0.1\) is therefore a slight overestimate, with its direction and size both consistent with the checks.
Worked Estimate for Cosine
Worked Example: Estimate \(\cos(0.1)\)
Again use the base point \(a=0\). Since \(\cos 0=1\) and \(-\sin 0=0\), the tangent line is \(L_0(x)=1+0(x-0)=1\). Therefore the estimate is \(\cos(0.1)\approx 1\).
On \([0,0.1]\), \((\cos x)''=-\cos x<0\). The tangent line overestimates, so the signed error \(E=\cos(0.1)-1\) should be negative. For the size bound, \(|(\cos x)''|=|\cos x|\le 1\), so \(M=1\) gives
Numerically, \(\cos(0.1)\approx 0.9950041653\). Subtracting the estimate gives \(E\approx 0.9950041653-1=-0.0049958347\). This error has the predicted negative sign, and its magnitude is about \(0.0049958347\), just below the bound \(0.005\). Thus the tangent-line estimate is high by about \(0.005\).
Worked Estimate for Tangent
Worked Example: Estimate \(\tan(\pi/4+0.02)\)
Choose \(a=\pi/4\), where \(\tan a=1\) and \(\sec^2 a=2\). The input differs from the base angle by \(h=0.02\), so
To check the direction, the interval \([\pi/4,\pi/4+0.02]\) lies inside \((0,\pi/2)\), where \((\tan x)''=2\sec^2 x\tan x>0\). Tangent is concave up there, so the tangent-line estimate underestimates: \(E=\tan(\pi/4+0.02)-1.04\) should be positive.
For an error-size check, \(f''(x)=2\sec^2 x\tan x\) is increasing on this interval: its derivative is \(4\sec^2 x\tan^2 x+2\sec^4 x>0\). At the right endpoint, \(\tan(\pi/4+0.02)\approx1.04082<1.1\), so \(\sec^2 x=1+\tan^2 x<2.21\) there. Monotonicity then gives \(f''(x)<2(2.21)(1.1)=4.862<5\) throughout the interval. Using \(M=5\),
For a numerical check, the tangent addition formula gives \(\tan(\pi/4+0.02)=\frac{1+\tan(0.02)}{1-\tan(0.02)}\). Using \(\tan(0.02)\approx0.0200026671\) gives \(\tan(\pi/4+0.02)\approx1.0408218807\). Thus \(E\approx1.0408218807-1.04=0.0008218807\). The error is positive as predicted, and \(0.0008218807<0.001\). The estimate \(1.04\) is low by about \(0.00082\).
A Common Mistake: Ignoring the Error Sign
A numerical estimate alone does not tell you whether it is high or low. For example, after finding the tangent-line estimate \(1.04\) in the tangent calculation, it would be incorrect to call it an overestimate merely because it is close to the function value. The concavity check gives the opposite conclusion: tangent is concave up on the interval, so its graph lies above the tangent line.
The numerical comparison confirms the correction: \(\tan(\pi/4+0.02)\approx1.0408218807>1.04\). Therefore, \(1.04\) is an underestimate, and the signed error \(f(x)-L_a(x)\) is positive. When checking any trigonometric estimate, keep the sign convention explicit and verify the predicted direction against the function's concavity.
A Reliable Checking Process
Use a nearby angle where the function value and derivative are known. Record the displacement \(h=x-a\) in radians.
Substitute \(f(a)\), \(f'(a)\), and \(h\) into \(L_a(a+h)=f(a)+f'(a)h\).
Determine the sign of \(f''\) throughout the interval. Concave down means an overestimate; concave up means an underestimate.
Find a valid upper bound \(M\) for \(|f''|\) over the interval, use \(|E|\le M|h|^2/2\), and compare with a numerical function value when available.
Each check answers a different question. The tangent-line formula produces the estimate; concavity predicts the error's direction; and the second-derivative bound limits its size. A reliable result is supported by all three.
Check Your Understanding
For each question, keep the base point, interval, and signed-error convention clear.
- What is the tangent-line estimate for \(\sin(0.05)\) using the base point \(a=0\)?
- On the interval from \(0\) to \(0.2\), does the tangent line to cosine at \(0\) overestimate or underestimate \(\cos(0.2)\)? Explain using the second derivative.
- Why must a bound on \(|f''|\) apply over the whole interval between the base point and the input?
- For \(\tan x\) on an interval contained in \((0,\pi/2)\), what does its concavity imply about the direction of its tangent-line estimate?
- If a valid second-derivative bound is \(M=3\) and \(|x-a|=0.04\), what upper bound does the error formula give?
